Slashdot Mirror
The Tuesday Birthday Problem
An anonymous reader sends in a mathematical puzzle introduced at the recent Gathering 4 Gardner, a convention of mathematicians, magicians, and puzzle enthusiasts held biannually in Atlanta. The Tuesday Birthday Problem is simply stated, but tends to mislead both intuitive and mathematically informed guesses. "I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?" The submitter adds, "Believe it or not, the Tuesday thing is relevant. Well, sort of. It's ambiguous."
13/27
First off, I am a huge Martin Gardner fan and if this puzzle intrigues you and you haven't heard of him, get one of his books.
This problem hinges very greatly on how it is phrased and I think it's more a trick of English converting to statistics than it is a true puzzle. If you were to rephrase this problem as "My first child was a boy born on Tuesday now what are the odds that my next child is a boy?" But they don't. They phrase it as after the fact of both births we are now studying a set of two objects that have been determined prior to me asking the question. This ordering causes the set to be enumerable. Which brings in an interesting piece of information theory to this game. Whenever you say something about one child that is exclusive to that child and that trait is enumerable than you have just affected the outcome of that second child. In the original two childs problem this is just gender. But in the above problem it pairs gender with day of week the child was born on. Now since ordering matters you recognize that whether or not the older or younger child is the one in question creates a different scenario and you can't have twins because only one was born on Tuesday. The article does a good job of explaining this.
The interesting thing is that the answer to this comes down to 13/27. And the larger the enumerable set is of possibilities, the closer this converges to 1/2. If you did this with a specific day of the year, your answer would be 729/1459 which is even closer to 1/2. The general rule is that for a set with N possibilities you would have (N*2 - 1)/(N*4 - 1). Now, what's interesting is if N is unbounded or unenumerable? What if I said "I have two children, one of whom is a boy that likes the number 1835736583. What's the probability that my other child is a boy?" Wouldn't it converge to 1/2?
My work here is dung.
and statistics (umm... probability juggling based on hind-sights).
Questions raise, answers kill. Raise questions to stay alive.
Let's assume that, if I have two children, it is equally probable that they are born as boy+boy, boy+girl, girl+boy, girl+girl. If I have one boy, girl+girl has probability 0, and the other options are equally likely, so they have probability 1/3. If it is known that I have a boy, there is 1/3 probability that the other is also a boy.
X=one boy is born on a tuesday
P(X|boyboy) = 1/7 + (6/7*1/7) = 13/49
P(X|boygirl) = 1/7
P(X|girlboy) = 1/7
P(boyboy) = P(boygirl) = P(girlboy) = 1/3
P(X) = (1/7 + 1/7 + 13/49)/3 = 9/49
Using Bayes's theorem:
P(boyboy|X) = P(X|boyboy)*P(boyboy)/P(X) = 13/49 * 1/3 * 49/9 = 13/27
Which is different from 1/3. So yes, the weekday of birth is significant.
This reminds me of a famous joke and variations thereof, (at least around eastern europe):
A man is asked on the street: What is the probability you will come across a dinosaur on the street today?
The man replies: less than 0,000000001%
When a woman is asked the same question, she replies:
50% - I either will or I won't.
So, really, it depends on who you ask.
> The submitter adds, "Believe it or not, the Tuesday thing is relevant. Well, sort of. It's ambiguous."
Why even say that? Anyone who does these puzzles likes to figure it out by themselves.
Maths fail..
No it isn't. Go and read the article again.
wot no sig
I have absolutely no knowledge of statistics, but why would you assume that just because one of the boys where born on tuesday, that one of his siblings then couldn't be?
Emotions! In your brain!
Nice work! I read the article and it agrees with your math. 13/27 is exactly what the author concludes.
Big apple, new Yorik, undig it, something's unrotting in Edenmark.
As the article notes, it depends what you mean by "one of", (specific one vs "at least one"), and quibbling mathematicians don't always pick the most common interpretation.
In other news, an aeroplane carrying a hundred mathematicians crashed with no survivors; their university made a press release stating that one of its mathematicians died in the crash.
Well, if it isn't the same as saying that, then it's a distortion of information expression or an error in method, in the sense that the mathematics don't fit reality when it's your intention that they do so?
Emotions! In your brain!
"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?" .. this is the same as saying "I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?
No, it isn't. It's the same as saying, "I have just tossed a quarter twice, and given that it came up heads at least once, what's the chance it that it came up head both times?"
We can do an analogue, but not in the way you're saying. I.e., "I have just tossed a coin and it came up heads, what's the chance it will come up heads if I toss it again" == "I have just had a kid and it's a boy, if I have another kid, what's the chance it will be a boy" == 50%
If you don't understand the problem, that's fine. It's counterintuitive, these things usually take a second look. If you don't understand the problem and you want to claim the solution is incorrect because you don't get it, well, that's something else entirely.
I don't believe in time. It's a grand conspiracy designed to sell watches.
This is kind of like saying "I flip a coin. What is the chance it lands heads facing up?"
And you say "50%."
And I say, "Incorrect. There is a very small chance it will land balanced perfectly on it's side, so both the chance of heads and the chance of tails is under 50%."
Big apple, new Yorik, undig it, something's unrotting in Edenmark.
The problem stated in the article is: "Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?" This is a different scenario than what is stated in the summary: "I have two children, one of whom is a boy. What's the probability that my other child is a boy?" In the first scenario, the probabilities are dependent on each other because it is not stated whether the first or the second child is the boy. In the second problem, it is given that the FIRST child is a boy. But that does not affect the odds of the second child which should therefore be 0.5.
Football Odds
. this is the same as saying "I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?"
Nope, it is equivalent to "I have just tossed a 10 pence coin twice, and I tell you that it has come up heads at least once, what is the probability that it has come up heads twice".
The 2/3 vs 1/3 probability hinges on the fact that the ordering of the kids is not defined.
If the kid's father told you "my oldest child is a boy", then you would be right.
Unfortunately, any defined order can play that role ("the first of his kids that I met in person", "the first of his kids that he mentioned", ...), which makes this problem so hard to grasp. Depending on exactly in which context he mentioned that one of his kids was a boy may change the probability of the other being a boy too from 1/2 to 1/3 or any value in between.
Actually, I made a mistake up there...on the first line, I meant to take the Tuesday thing out. For reasons that are explained in the article.
I don't believe in time. It's a grand conspiracy designed to sell watches.
Nope.
"I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?"
Is *not* the same as:
"I just tossed two coins and one of them was heads, what is the probability that the other one was heads as well?"
No sig today...
I think he is leaving out an option when working out the permutations The given boy (born on Tuesday) is different entity from his sibling so that gives 4 options and a 1/2 chance?
Named Boy, girl
Named Boy, boy
boy, Named Boy
girl, Named Boy
What a silly problem! Of course the answer is 0.5 and all the other information is irrelevant to the problem! Can you imagine what kind of silly math the author would do in the following "problem"? "I have two children, one of whom is a boy born on a Tuesday, in November, on a full moon night. What's the probability that my other child is a boy?"
I'm trying to understand this in laymans terms. Is this a valid way to interpret it? If you know that a specific child is a boy (e.g. the youngest), then the probability the other one is a boy is 1/2. If you only know that one of the two children is a boy, and you don't know which, then the probability they are both boys is 1/3. So the more you pin down details towards identifying a specific child is a boy the closer the probability tends towards 1/2 for both boys. In this case you're specifying a day, and as that reduces the ambiguity a lot it gets you quite close to 1/2.
"I have two children, one of whom is a boy born in the first day of the year. What's the probability that my other child is a boy?"
"I have two children, one of whom is a boy born in January. What's the probability that my other child is a boy?"
"I have two children, one of whom is a boy born in Winter. What's the probability that my other child is a boy?"
Do they give different probabilities?
This is related to the Principle of Restricted Choice often seen in Contract Bridge.
If the parent has two boys born on a Tuesday, he could equally have declared the other boy as being born on a Tuesday. In a parallel universe, the other boy would have been declared as being born on a Tuesday, whereas if only one of the child was a boy born on Tuesday nothing would have changed in any of the other parallel universes. Therefore the effect is the probability of 2 boys borne on Tuesday has been halved, resulting in 13/27 probability of the second child being a boy.
Ah right.. I get it.. while previous statement is 100% correct and true, the Tuesday statement does put down restrictions on the second child and thus reducing the 50% chance..
I used to tend bar and this is not a math puzzle, but fun for messing with the barflies when they've had a beer or 5 and start wanting to tell you their life story. First, as you pose the question, take out 3 coins (this only translates well using USA coins, one being a nickle, the other a penny, the third a quarter, dime or fifty cent piece) and state that "Johnny's Mom has 3 kids, the first one is named 'Penny,' (point to the penny) the second one is name is 'Nicky' (point to the nickle) and then point to the third coin (doesn't matter which you use) and ask What is the third child's name?" Then see how long it takes them to figure it out. And then whether or not they leave you a tip.
I don't know about the chance of it landing heads up, but if you threw it up, it has a rather high chance of being damaged by your stomach acid...
I completely don't get the explanation at the initial stage. Namely that the possibility of the second child being a boy is 1/3.
Once you're told that there's at least one boy, doesn't that mean that one result is fixed and has no outcome on the second result?
The calculations for a 1/3 chance would make sense in the following situation: You flip two coins, if you get two tails, you flip again. What is the chance of you ending up with two heads?
To me the question doesn't match the answer, the question is saying: This result happened, what is the chances of the second try being the same result? The answer is saying: One outcome is impossible, what is the chances of a certain outcome?
Maybe I am being thick and not grasping basic probability...
OK, so I'm not a mathematician but I have done a lot of maths and logic in my time and I believe the description of the Two Birthdays problem is wrong but maybe a mathematician here can correct me. In that simpler problem, Devlin and therefore I assume Gardner, make an issue of the order in which the children were born, creating the possibilities:
If I annotate this it may become more clear why I think they are wrong.
Now, where is the situation where the known male child is the younger child, i.e. the Boy (unknown, older), Boy (known, younger) scenario. Does the order somehow become unimportant just because they are both boys? It we add this 4th possibility then the possibility of the "other" child being a boy returns to a half and sanity is preserved.
If I've missed something completely here I'd love to hear why, these problems fascinate me. To really understand variable change I had to write myself a little simulator and it wasn't until I was putting it into code that I understood it properly :-P
Take a thousand families, with two children, where one of the children was a boy born on a Tuesday.
I don't mean a thousand theoretical families. I mean, lets say you straight up took one thousand real families, that matched the above constraints, straight out of the census. No joke, you break out the SQL.
When you check the gender of the other child, you are going to see the breakdown of gender being 50% male, 50% female.
Now, I know there's a lot of fun handwaving going on. Here's the flaw, in a nutshell. There are indeed three possibilities, when one child is constrained to be a boy:
boy, girl
girl, boy
boy, boy
The mistake -- and it is a mistake, because when you actually run the experiment, the hypothesis is invalidated -- is thinking that each of the above cases is equally likely. Specifically, order of birth has been incorrectly elevated as a determining factor. So we see:
boy, girl: 33%
girl, boy: 33%
boy, boy: 33%
When we really should be seeing:
boy, boy: 50%
boy, girl: 25%
girl, boy: 25%
Or, more accurately:
same-gender, both male: 50%
different-gender: 50%
boy first: 25%
girl first: 25%
Another way to frame the query, with similar results, is to say:
Select the gender of all second children where the first child was born on a Tuesday and the first child was male.
Select the gender of all first children where the second child was born on a Tuesday and the second child was male.
You'll note the girl, girl families will show up in neither result set. So they can do nothing to skew the numbers.
The results of both queries will, predictably, be 50/50 male and female.
This is a good example of why framing a problem correctly is so difficult and critical. It's only because this problem is so amenable to experimental formulation that it's easily defensible.
(Note that the use of Tuesday was an excellent DoS against math geeks.)
(Note also, by the way, this is the exact opposite of the Monty Hall problem. In that problem, people are expecting:
Door 2: 50% ...when, really, we have:
Door 3: 50%
Host Told You Where The Car Was: 66%
Was Behind 3, Therefore Exposed 2: 33%
Was Behind 2, Therefore Exposed 3: 33%
Host Didn't Tell You Where The Car Was: 33%
Randomly Exposed 2: 16.5%
Randomly Exposed 3: 16.5%
If you modify the Monty Hall problem, such that he opens a random door *which might actually expose the car*, then when he opens the door and you see a goat, it doesn't matter whether you switch or not.)
Take an abstract mathematical problem, invent a pseudo-real-world context, rephrase the problem very sloppily and ambiguously in plain English then laugh smugly when people get the wrong answer.
The correct answer to the question, by the way, is "I don't know - you have not given me enough information, and I'd have to go check that the gender of successive offspring from the same couple is actually independent, but its probably gonna be somewhere between 1/3 and 2/3 - and since you'd have to somehow re-formulate it as a viable experiment and run it 100 times to confirm that result, only the Bayesians give a flying fuck what the precise value is".
In other news: you can't actually build a hotel with an infinite number of rooms - you'd run out of bricks - so don't try and engage my interest in all the weird thing that would happen if you did something impossible. And stop hiding goats behind my door!
In a survey of 100 programmers, 111111 thought that duck-typing was a good idea.
Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?
This is the question posed without the birth weekday specified. TFA actually tries to say that there are 4 outcomes for the pair of children, one of which is impossible, so they remove it. Since "boy, boy" is only one of the 3 outcomes, then the probability must be 1/3. Right?
Wrong.
The boy (let's call him Peter) being a boy is a given of the problem, so it has P = 1. The other child -- we don't care about it being born before or after Peter -- is independent, so the probability that it's a boy is 0.5*. The 4 outcomes are as follows:
Peter, Boy = 0.25*
Peter, Girl = 0.25*
Boy, Peter = 0.25*
Girl, Peter = 0.25*
So, whichever way we slice this problem, the solution is 0.5*.
P(Peter, Boy) + P(Boy, Peter) = 0.5*
1 * P(Other is Boy) = 0.5*
- - - - - -
* May slightly differ due to the male:female ratio at birth. It is assumed here to be 1:1.
""I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?""
Umm no, if you look at the original problem, "one of my 2 kinds is a boy, what's the probability both are boys" is like saying "Ive've tossed a coin twice, at least one of which was heads, what;s the probability that both were heads?" - so here the possibilities are H H / H T/ T H of which only one of the 3 is both heads so it's 1/3rd.
It's not saying, "the first guy was a boy - or in your analagy, the first coin toss was heads" - where you would be right, the chance of the second child being a boy, or the second coin toss being heads would be 50% - it's a different problem.
And the people shall be oppressed, every one by another, and every one by his neighbour Isaiah 3:5
Exactly. The second child is not a boy born on a Tuesday.
Help! I'm a slashdot refugee.
Only if you believe in randomness. If the other child in fact is a boy the probablility for it is 1.
I agree - by saying "one of whom" doesn't in English preclude the fact that both of them could be boys born on Tuesday!
no, go read the article again. the solution DOES include the possiibility the the other child is also a boy born on a tuesday.
And the people shall be oppressed, every one by another, and every one by his neighbour Isaiah 3:5
This is an interesting and highly technical application of math and grammar, but biology is too messy for it. Probabilities are for when you don't have all of the information. You might be able to refine the estimate slightly, but if you knew everything then you have a 100% "chance" of the actual outcome (assuming a deterministic universe... or maybe not). Since the Y chromosome is lighter, more males are born than females. OTOH, females are more likely to survive. Apparently birth order also affects the distribution of sexes. Birthdays also aren't entirely randomly distributed, so did a Tuesday fall nine months after a holiday 10-15 years ago? There's probably some more epidemiology you could throw in, so this quickly rises outside the scope of a mathematical problem. It just depends on how technical you want to get, and what your area of expertise is. But it's just refinement, and at some point you'll run into the L'effet Tetris.
"it doesn't however influence the probability of the sex of the next child."
Except the problem doesn;t state that the son born on a Tuesday is the FIRST child. It could be the second, Actually the original prblem of "at least one of my two sons is a boy" is more like asking "I threw a coin up yesterday and also today, on at least one of the throws it landed up heads, what is the chance both of the throws landed up heads" giving the set {HH, HT, TH} of the 3 members, only one is HH so it's 1/3
And the people shall be oppressed, every one by another, and every one by his neighbour Isaiah 3:5
> I have never seen "DOES not disallow" in my entire life. ... never emphasize a positive immediately before a negative
We mathematicians do this all the time (in our proofs, etc.)!
> Normal humans
Ah, sorry, forget it...
If the dad is Schrödinger the other kid is both born and unborn at the same time.
Don't fight for your country, if your country does not fight for you.
Reading the article they talk about an older simpler problem not mentioning a birthdate resulting in probability of 1/3. This was wrong just as well as the Tuesday not being relevant.
The calculations in both cases failed to take into account that we did not know which boy where older as well. The list should not have been:
boy / boy
boy / girl
girl / boy
girl / girl ( impossible)
resulting in 2/3 probability the other was a girl and 1/3 for a boy. This is however wrong... It should have been listed as:
boy (1) / boy (2)
boy (2) / boy (1)
boy / girl
girl / boy
girl (1) / girl (2) (impossible)
girl (2) / girl (1) (impossible)
Resulting in 2/4 for both boy and girl giving the real result of 50%.
There are no indication as to whether the boy listed was the younger or older.
Adding the Tuesday information will now correctly not make any difference.
The number of days in a week is arbitrary. It's a pure human historical convention. In another time or another culture, the number of days in the "week" could be 10, or 5 or anything else people decide to agree on.
True, but it's hardly likely that they would call one of the days of their "week" Tuesday, then, is it?
I find it fascinating, though, that you pick on that particular part of the ambiguity of the language of the question, and totally ignore the much more fundamental question, which is what is the exact meaning of "one" (i.e., did he mean "exactly one" or "at least one").
Sorry but I don't get this. The boy born on tuesday is just as relevant as the favorite color of the boy or whatever. The probability does not change because irrelevant details are added to the story. Otherwise, I'd suggest to also factor in the place the child was born, the year, month and the position of any random set of celestial bodies at the time of birth of both children as well as all their known ancestors. The chance her other child is a boy is equal to the chance of a child being a boy in general. So it is about 50% or probably a bit off since there's a slight difference between the number of boys and girls being born. It's not like children are like little balls blindly picked from a vessel filled with a finite amount of balls. Unless stated otherwise, which might happen because the family may not have been randomly selected, but in a way one picks balls from a vessel, but that is not stated in the problem so it is not the case and it should not be bothered about. Why is this on slashdot? And why are mathematicians discussing about this?! Am I a moron?!
0x or or snor perron?!
You are right in what you say, but your second example is exactly the same as the first. Itym:
"I have two children, the first of which is a boy. What's the probability that my second child is a boy?"
I'm at a company picnic and I have 5 employees sitting around. I need 2 more players for the women's volleyball team, so I take 2 of the women away. What is the sex distribution of the rest? The issue is I'm cherrypicking based on a condition. Here the condition is clear: I'm picking based on sex.
The confusion in the Tuesday problem comes in because the condition doesn't appear relevant. Who cares about Tuesday?? The assumption is he picked one of his kids to talk about.
You're misreading the question. If you exclude the Tuesday part (which I still don't get, and I still doubt), the question is not what's the chance that the second child is a boy, which would be 50%. The question is what is the chance that both kids have the same sex. Since the fact that one of them is a boy doesn't bring any new information to the table about they being the same sex or not, the probability remains the same -- 1/3.
From TFA:
Still, Gardner’s initial overly narrow interpretation warns of the dangers of over-hasty analysis of probability questions — and shows the wonder that can come from them.
Is it "wonder", or does it mostly reveal how non-mathematical and unscientific probability is?
The constraints are not defined.
"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?"
I have three children. But I also have 2 children. See the problem? I have a son born on a Tuesday, and I have another son born on a Tuesday. See the problem?
It doesn't say I have *only* two children. It doesn't say the other child can't be a son born on a Tuesday. It assumes the birth rate is 50/50, but most statistics agree it's not even. FTA, it assumes there's no such thing as twins. It assumes you have only one wife. But none of this shit is specified.
Pisses me off. Use coins and cards. Not assumed biblical customs.
How many more years will slashdot have an off-by-one error on your Score in your profile?
"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?" and "Now suppose that the older child isn’t a boy born on Tuesday."
It is futile to discuss that one is not the other. That was already a fact in the article.
Regardless of that, I think the article is very misleading and the probabilities are all wrong for the wording given. Even the original "The Two Children Problem" he refers to was phrased ambiguously, so taking only one of the possible probabilities into consideration seems silly.
Kate is 21 years older than her son Tom.
In 6 years from now, Kate will be 5 times as old as Tom.
Where is Tom's father?
[answer below]
[answer below]
[answer below]
[answer below]
[answer below]
Ans : Tom's father is inside Kate now concieving Tom.
But yet we interpret the "two children" as meaning exactly two.
Confucius say, "Find worm in apple - bad. Find half a worm - worse."
No, that outcome is not specifically disallowed.
It only states that one is a boy, he was born on a Tuesday.
...
This so-called problem depends on a counter-intuitive interpretation of the language. Yes, if you're a math nerd you can find an obscure way of interpreting a simple question; BFD.
I'm a demography nerd instead, so the answer, if your second child is under about two years old, is 51%. The ratio of male to female births is about 51-49.
I piss off bigots.
"Select the gender of all second children where the first child was born on a Tuesday and the first child was male." Yes, it will be 50/50 male and female. "Select the gender of all first children where the second child was born on a Tuesday and the second child was male." Again it will be 50/50.
But the gender split in the union of those two groups will *not* be 50/50. You have counted the families with two boys born on a Tuesday twice. 1 in 14 of the first group will also be in the second group. Taking the union correctly, 7 out of 14 in the first group will have two boys, and 6 out of 13 of those in the second group *not already counted* will be boys. In total, 13 out of 27 will be boys.
Biologically, in this case, the chance is always 50%-50%!
\m/
Umm, kdawson posted something without be flamed as being a moron. All of his other posts I've read the comments for contain at least one, usually more like 5+ which state that he's a waste of oxygen who never has, nor ever will contribute anything of value to this site of "News for Nerds" .So, to help correct this oversight, I will provide the following:
kdawson was all like "I r smrt!", but we know to hear him going all like "I r....durr...what?". Now get off my lawn.
This covers my membership dues to the groupthink, right?
I needed a sig so people would know who I am, but I was too drunk to make something witty, so you get this instead.
25%?
Chance of one children being a boy = Roughly 50% (F or M)
Chances of two children both being boys = Roughly 25% (FF, FM, MF or MM)
One is a boy => Requirement for the second to be a boy is that both are boys = 25% chance?
Because us mathematicians come up with these really simple problems, and when we tell you that your naive answer is wrong, you grab your ball, shove reality in our face, and run home...
WARNING! This girl exceeds the MAXIMUM SAFE standards established by the FDA for BRATTINESS
Indeed, is everyone drunk or something...
First, The question doesn't say the other (this does not mean older or younger...) child was not born on a Tuesday, maybe the questioner meant to include this info but they failed to.
Second, the probability that the other child is a boy is either 1 or 0, it's something that has already occurred... The questioner probably meant to ask "What is the probability that if you guess the other one is a boy you will be correct?"
So if we correct the question to read as it was likely intended to be read:
"I have two children, one of whom is a boy born on a Tuesday. The other child is not a boy born on a Tuesday. What is the probability that you will be correct if you guess that my other child is a boy?"
So you think to yourself, well assuming boys pop out of that mom just as easily of girls and doesnt prefer to do it any particular day of the week, that means its 1/7 chance it might have happened on any given day and the likelihood it was a boy is 50:50 for 6 days of the week and then 0 for tuesday. So you multiply .5 by 6, add zero, and take the average (divide by 7) to get 3/7.
The real confusion occurs due to the use of odd numbers... Imagine a world where everything was found in sets of twos, people had 2 heads, 4 arms, etc. They would always be dealing with eating animals that were siamese, if they wanted to hunt by throwing rocks or whatever each siamese would throw a rock so they would use two rocks. In this world I would say that what we call the number 2 would actually be like their number 1, and what we use as unity, or one, would be for the siamese called a half. Therefore their numberline would go 0, .5, 2, 2.5, 4, 4.5, 6, 6.5, 8, etc.
This is actually more reflective of reality in that, deep down, math and counting are extensions of logic, and the fundamental unit of logic is a true-false statement which is basically a set of 2. True is only 1/2 of the total possibilities for any given logical statement. For example say you have counted one rock, what that actually represents is both having one rock in your presence butt also, concurrently, not having counted other than one rock, so in essence you have counted two different things and are representing them with a number supposed to correspond with one thing. Wouldnt it make more sense to just use "two" to represent the one thing youve counted?
The probability of guessing correctly by saying the second child is a boy would therefore be 1/2(6), or 3, divided by 6 and a half, which gives you 6 out of 12 and 1/2 odds.
If you ignore the Tuesday bit, then eliminating the option that you know is not possible (FF won't work since one child is a boy) you have three options remaining, each with equal probability, giving 33% chance.
Or to put it another way: If it was 25% chance like you say then so is FM and MF. There are no other possibilities so it should add up to 100%, but you only get 75%...
He might be. The question does not say one way or the other.
The idea is that there are 198 posibillities for two children and the day they are born (as in: the first is a boy on a monday the second is a girl on sunday - there are 198 different statements like that). If you go through it you will see that excatly 27 of them contains a boy on a tuesday. Of those 27, 13 had two boys. Each of the events are equally likely - well not in the real world, but it would require a bit more info otherwise. So you end up with 13/27.
A non-numerical version goes as follows. To easen the reading I will use unordered pairs:
There are three sets. A set where the boy on a tuesday is born first, F, one there he is born second, S, and one where there are two boys on a tuesday, B. The S and F are equally large and the number of girl boy pairs are equally large in both (similary for boy boy pairs), but they contain a slightly larger probabillity for boy, girl pairs over boy, boy pairs(because neither S nor F contains two boys on a tuesday - they are in B - but they DO contain the posbillity that it is a boy on a tuesday and a girl on a tuesday). The difference is excatly such that if you add B to e.g. F the numbers of boy, girl pairs are equal to the number of boy, boy pairs. Therefore you will have a slightly larger probabillity for girl, boy pairs (because the number of boy, boy in F+B is equal to the number of boy, girl in F+B, but the number of girl, boy pairs are slightly larger than the number of boy, boy pairs in S).
Don't know if that made much sense?
But:
Chance of one child to be a boy? 50%
Chance of two children to both be boys? 25%
Chance that the second of two children is a boy? 50%
Chance that the second of two children is a boy AND that the first one is a boy? 25%
But things like this always fuck me up, because I find it hard to tell if the other statement matters or not.
Oh the shame to be wrong on Slashdot if I where (was? ;D) :D ... because one also want to ask oneself "but what are the chance that a children, regardless of what has happened is a boy?"
(FF, FM, MF or MM)
FF is not possible
FM is not possible
http://p8ste.com - Web based Clipboard
First, The question doesn't say the other (this does not mean older or younger...) child was not born on a Tuesday, maybe the questioner meant to include this info but they failed to.
It's not part of the problem that only one of the children can be born on a Tuesday, you might be dealing with two boys of differing ages, both of whom were born on a Tuesday. In Keith Devlin's analysis (which is correct, given certain assumptions about the problem), there is only one outcome where there are two boys born on a Tuesday. This is why, once you've accounted for this by enumerating the outcomes where the elder child is a boy born on Tuesday, you can't count it again when enumerating the outcomes for the younger child.
It would be a bit clearer if the question was rephrased "I have two children, ONLY one of whom is a boy born on a Tuesday..." Then the answer is a bit more obvious.
The answer from the story is that the Tuesday bit matters if you let it matter. But really, it doesn't SAY anything. it becomes a word game.
Is there any meaning, any information in the fact that one of the kids, a boy, was born on a Tuesday?
Does it preclude the other child having been born on a Tuesday? No, it does not. If both kids were born on a Tuesday you PROBABLY wouldn't say that "one of them was born on Tuesday" but that information is being deduced. It has not been stated as a fact.
For instance, "I have two children, one of whom pees standing up born on a Tuesday...."
Does this tell you that one of the kids is a boy? No. Girls can pee standing up. Yes, most kids who pee standing up are boys BUT it ain't an absolute.
It is indeed funny to see how strongly assumptions become in even such a place as math where only cold hard facts should matter.
In reality, the answer 1/2 is the correct one. It either is or it isn't. You can calculate the true randomness, number of girls vs boys, distribution of gender among siblings etc etc but ultimately for a REAL case, the chances boil down to 50/50.
And why does only the 1/2 matter? Because the people who over calculate the case happily leave out hemaprodites, who are neither boy or girl. Or that the other child is a chimera or that we are talking about siamese twins.
If you spend some time calculating all the variables, did you REALLY include ALL the variables? Every single one of them? Including such that one of the kids might have been adopted? A kid might have been placed out of care? person might have kids they don't know about? With different people?
No? Then you are just guessing and 50/50 is as good a guess as any.
MMO Quests are like orgasms:
You may solo them, I prefer them in a group.
Here is the list of outcomes if it is possible for both boys to have been born on a Tuesday:
Girl on Monday, Boy on Tuesday
Girl on Tuesday, Boy on Tuesday
Girl on Wednesday Boy on Tuesday
Girl on Thursday, Boy on Tuesday
Girl on Friday, Boy on Tuesday
Girl on Saturday, Boy on Tuesday
Girl on Sunday, Boy on Tuesday
Boy on Monday, Boy on Tuesday
Boy on Tuesday, Boy on Tuesday
Boy on Wednesday Boy on Tuesday
Boy on Thursday, Boy on Tuesday
Boy on Friday, Boy on Tuesday
Boy on Saturday, Boy on Tuesday
Boy on Sunday, Boy on Tuesday
If having a boy or girl is equally likely (we dont do any weighting), then the chance its a boy is therefore 7/14=1/2. If you couldnt have two boys on Tuesday it would be 6/14=3/7. Please point out where this analysis goes astray.
Boho, crap, so wrong as always :D
I was going to say this needs to be modified by the probability of having a second child of the same gender; i.e. if you have one boy, does that increase the probability of the second child being a boy? But it turns out it doesn't (according to the first website I found, but hey, I'm not a professional researcher!).
There is, overall, apparently a 51% chance of having a boy, which marginally skews the result, but this apparently doesn't vary for subsequent children; i.e. there is no overall predisposition to having children of a certain gender (having children of an uncertain gender is a different matter, and just plain bad luck...)
Sigs are so 1990s. No way would I be seen dead with one.
What's misssing is the statistic that the child will actually be born intersex. He's using data assuming biological birth results in only male/female. I know it's nitpicky, but that's the whole point of the puzzle, to look past intuitive results.
The actual outcomes are
Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl
Boy/Intersex
Girl/Intersex
Intersex/Boy
Intersex/Girl
We can eliminate Girl/Girl, Girl/Intersex, Intersex/Girl. Someone else can figure out the math.
meep!
Wow. The article lays out the analysis and correct answer quite clearly. I am stunned at the number of /.ers who STILL post diatribes that are completely off the mark. Well done slashdot!!
Obi-Wan: "I felt a great disturbance in the Force, as if millions of voices suddenly cried out in terror and were sudden
Sorry, but that is why the conclusion in the summary is that is ambigous.
The tuesday statement does NOT preclude the other child from being a boy born on a tuesday.
The trick here is that you are making an ASSUMPTION based on abstract language conventions, not on hard factual statements.
You PRESUME that if the other child was a boy on a tuesday you would not state it like this, but that is just because you expect people to use language in a certain way.
I have two cars, one is red. What is the color of the other car?
It could be any color. That one is red gives you no information about the other. I might really like red.
London and Amsterdam have firetrucks, Amsterdam fire-trucks are red, what color are London fire-trucks?
See?
That is the entire puzzle in fact. Does the Tuesday bit matter? A lot of these type of riddles work like that. What part is factual statement and what is dressing up?
They should have asked this question to a 10 yr old. They are good at this. Adults tend to be so experienced at getting hidden meaning out of sentences, to read between the lines that they are unable to stop doing it.
You attach meaning to the tuesday statement. But this is math, and it has none.
MMO Quests are like orgasms:
You may solo them, I prefer them in a group.
But can one assume they wasn't possible when one look at the bigger picture / the chance of both being boys?
Or is that just wrong and one only should care about the current situation?
And if you where to decide on 50% would that be because the first kid being female options isn't valid longer since the first one was a boy, or is it rather because it's 50% chance that the second one is a boy regardless of anything else?
My older brother and I were both born on Tuesdays.
But yet we interpret the "two children" as meaning exactly two.
"two children" is an unambiguous statement ... it can't mean one child, it can't meant three children, neither can it mean two dogs.
"one of whom" can be ambiguous ... it can mean only one (of the children), or just the one I am describing. Nowhere in the original statement is it said that the second child was not a boy born on a Tuesday. You can argue it's implied, but it's not stated.
Just because you say that the first child was a boy born on a Tuesday doesn't mean that the second can't be the way the statement is worded. This is a mathematician using English badly to prove his point!
This is a gambler's fallacy problem. The more tangents you throw at it, the closer you get to .5 (50%), while never reaching it. This is the limit, why? Because there's only two potential outcomes for the other child: boy or girl.
What you (or the website you copied and pasted the ratio from) fail to take into account (and why it's a Gambler's fallacy problem) is that when involving chance, anything that happened in the past is completely irrelevant to future probables. I could roll a die 99 times, and get 6, the probability of getting 100 6's when I've already got 99 6's is still 1 out of 6, not 6^100.
The reason the chi square doesn't come into play here is because it doesn't MATTER the order. Has she said "What is the probability my SECOND-BORN was a boy?" it would be perfectly logical to write the square because the boy who was born on Tuesday could be either the first born or the second born, she never stipulated.
We can say that the boy, who was born on a tuesday, was also a Gemini. Does this change the ratio? No, the probability of having two boys is still 50-50%, because the unknown only has two possible outcomes: boy or girl.
When modding "Informative", please make sure it both has a source and IS actually informative.
The slashdot summary got it all wrong, stating "What's the probability that my other child is a boy?".
This is completely different than:
"What's the probability that I have two boys?"
The Slashdot summary makes the logic puzzle much easier, by simplifying the combinations to two children, one of whom is a boy. The real question has no such limitation.
If I have exactly eight children, logic follows that "I have two children" as well. The problem doesn't not specifically state that there are exactly two children, only that there are two children, one of whom is a boy. What are the chances there are two boys? Well, we'd have to know how many children there are total.
I hate to be pedantic like this, but isn't that the entire point of a finely crafted puzzle like this?
The math logic can't be solved until the English logic is correctly understood first. I see this more as a language/logic question than a math question.
The remarkable thing that Foshee’s variation points out is that any piece of information that affects the selection will also affect the probability.
Sort of sounds like (but probably isn't) how probability works with quantum mechanics and how knowing some information can affect the wave function.
There is still the possibility that the child is born a hermaphrodite and will be of both genders.
i.e. "Too many Cooks spoil the Broth"
If 1 cook makes a good Broth, then many cooks should make a better broth. But in reality the more cooks you add the worse the resulting Broth becomes.
Same with this question. The only relevant fact is what sex will the second child be. which is roughly 50/50. Well probably more close to 49.5 / 1 /49.5. (The 1% is to give leeway for dual-gender birth defects).
The more data you add to the question actually clouds the answer, which is counter-intuitive to the way most people are taught. (so more data does not equal a better result).
Laters Sol "Have you found the secrets of the universe? Asked Zebade "I'm sure I left them here somewhere"
I came up with a different answer, based on the summary's wording.
Firstly, the sex of the second child is not determined by the first. Whatever one child is, the other will always be 50% chance of being either.
What we can deduce from the wording is that his other child is not a son born on a tuesday.
We draw a two column, 7 row matrix. The rows are days of the week, and the columns are boy/girl. Write a tick in each cell if that is a valid sex and day for the child. We are left with 14 possibilities. 7 of those are girls (a girl can be born on any day), but only 6 are boys (as according to the wording, only ONE is a son born on tuesday...if the other is a son, it cannot be a tuesday, so we are left with 6 days if it's a boy. We give that probability to the girl column.
Thus we are left with 8 out of 14 chances being a girl, and 6 out of 14 being a boy. In decimal:
Girl: 0.57
Boy: 0.43
QED.
Sparks:Gadget:Beer Maker
The Gregorian calendar system repeats every 400 years, and the number of Tuesdays is not exactly one-seventh of the total number of days.
I am guessing that this is the key to the answer.
#include <stdlib.h> /* random() */ /* time() */ /* floor() */
/* 2**31 = RAND_MAX + 1 for random() */ /* number of simulations */
/* generate random number 0 <= x < 1 */
/* system init */
/* simulations loop */
/* Not both boys born on a Tue */ /* Is the other one a boy as well? */
/* output */
#include <stdio.h> /* printf() */
#include <time.h>
#include <math.h>
#define RANDMAX1 2147483648
#define SIMS 1000000000
#define SEXS 2
#define DAYS 7
enum Sex {GIRL, BOY};
enum Day {MON, TUE, WED, THU, FRI, SAT, SUN};
double drand(void);
int main (int argc, const char * argv[]) {
int sim, i, n, k, boy, hit, sex, day;
srandom((unsigned)time(NULL));
n = 0;
k = 0;
for (sim = 1; sim <= SIMS; ++sim) {
boy = 0;
hit = 0;
for (i = 0; i < 2; ++i) {
sex = (int)floor(drand() * SEXS);
day = (int)floor(drand() * DAYS);
boy += sex;
if (sex == BOY && day == TUE)
++hit;
}
if (hit == 1) {
++n;
if (boy == 2)
++k;
}
if (!(sim % 100000000))
printf("%i: %i / %i = %lf\n", sim, k, n, (double)k/(double)n);
}
return 0;
}
double drand(void) {
return ((double)random() / ((double)RANDMAX1));
}
Failed to read the article, hey?
You're absolutely correct IF you assume that the man is just describing one of his children. That is the logical interpretation of the problem. However, if someone went out and specifically selected a family with at least one boy, the probability of the other child being a girl is not 50% - the population in question has been artificially depleted of girls because the criteria excluded all two-girl families.
Shame shame shame...
The answer of course, is 42...
Silly birds off with you!
- Dan.
~ People that think they are better than anyone else for any reason are the cause of all the strife in the world.
What is the probability you will come across a dinosaur on the street today?
That depends on how you define dinosaur. By "dinosaur" do you refer to the clade of dinosaurs, which includes birds? Or are you using the paraphyletic definition, which excludes them? Do you include members of the U.S. Democratic Party who act more like Republicans, where dinosaur stands for "Democrat in name only, sorry-ass undercover Republican"?
This is similar to the confusion generated by the Monty Hall question. Why does he show you an empty door? If he would always show you the door immediately to the right of the one you chose, for example, and it just happened in this case to contain a goat, then there is no reason to switch. On the other hand, if he on purpose always shows an empty door to every contestant, then the usual reasoning applies and you should switch. (And if you imagine an 'evil' Monty Hall who shows a goat to a contestant if and only if that contestant originally chose the correct door, well then you should never switch.) It all depends on why you are being told this information, and what the general rule is about whether you are told or not.
-- Ed Avis ed@membled.com
There is a 50/50 chance each time a woman is pregnant. The ordering is irrelevant. The explanation in the linked article applies an implied need to suggest that ordering is necessary, but without making the case for it. Child A is a boy, this makes no impact on child B unless we consider that the pregnancy is specifically identical twin related. The day of the week and the order they're born on make no difference.
So to accurately answer the question based on the facts provided, we'd say that the chances of the second child being a boy is 50% plus an accepted percentage to account for identical twins derived from birth statistics which are not provided.
I agree with your reasoning assuming I were willing to take a leap of faith, however in a purely mathematical scenario where a specific answer is expected from a specific question, the data supplied does not allow for considering ordering in my opinion. Of course if the question was "What is the likelyhood that my youngest child is a boy" then the problem is adjusted to compensate for what appears to be the desired result.
Read the article. Whether the answer is 50% or not depends on the context, and that context is not specified in the problem.
If you meet the person with his son, and he tells you that that son is born on a Tuesday, then the chance of the other kid being a boy is 50%. If you see him as one of the thousands of parents who have two children, at least of which is a boy who's born on a Tuesday, the chance of the other kid being a boy is 13/27.
Three men eat at a restaurant. The bill comes to a total of $30. The waiter takes the money to the boss.
The boss tells the waiter that they're regular customers, give them back $5.
Waiter happily goes back, but on the way tries to figure out how to divide the $5 into three...
He decides to pocket $2, and give them each $1.
The men walk away happily, and comment on the great meals costing only $9.
3 x $9 = $27
The waiter took $2, totalling $29. Where's the missing $1 ?
(FF, FM, MF or MM)
FF is not possible
FM is not possible
Why is FM not possible? The M is a boy, isn't it?
(1/2)*(1/7)*6+(1/7)*(1/2)=1/2
that's my bet.
You were thinking
(1/2)*(1/7)*6+(1/7)*(1/4)=13/28
but this is wrong. You already know that it was a boy, so the 1/4th doesn't apply. Yes, there's 1/4th chance that there are two boys (or two girls). But since you already know it's a boy - you only go down the probability tree from there, and the other branch is discarded.
#
#\ @ ? Colonize Mars
#
Are one to assume that the father and mother of both kids are the same? If they are then the probability for the second kid to also be a boy is increased as statistically a male/female pair is more prone to produce more of the firstborns gender than the opposite.
Actually it's more like 49%, since there are slightly more girls than boys. I was surprised that I got this one so easy, this kind of puzzle usually discombobulates me.
Free Martian Whores!
According to the article, you have enumerated almost half of the possibilities. What if the second boy wasn't born on a Tuesday, then the first one must have been.
From the article:
Ah crap.. I get it now..
There are in fact only THREE possible combinations - HH, HT (or TH) and TT. I can't really draw a pie chart here so you will have to imagine it.
The probability of HH = 0.25
The probability of TT = 0.25
The probability of HT = 0.5 (i.e. 0.25 + 0.25)
Given that TT is not possible because we have at least one H, then we cut a quarter out of the pie chart so it looks a bit like a pacman.
The probability of HH = 0.33
The probability of HT = 0.67
So the probability of the second one being H (or a boy) is indeed one third.
I blame it on cognitive dissonance caused by insufficient caffeine.
Never email donotemail@WeAreSpammers.com
If "one" can mean at least one, with the possibility of more, then "two" can mean at least two, with the possibility of more.
It must be consistent for all values of X. Why is X == 1 a special case?
Confucius say, "Find worm in apple - bad. Find half a worm - worse."
Are the answers to the puzzles below equivalent? Are two equivalent? Which two?
The first is the original puzzle. Not being a native english speaker, I'd interpret the first as being equivalent to the second. And I might be completely wrong here, but if they are, then the information about tuesday is irrelevant, correct? Nothing would prevent the other child from being born on a tuesday if they could both be?
Isn't this much like the puzzle "two coins total 30 cents, one of them is not a quarter" (the other is).
If you have a bowl of fruit, specifying that one of them is a banana could be done by picking one up and saying: "This is a banana."
Assuming 50/50 chance of boy or girl and a 1/7 chance of each weekday, let's look at the general population of families with two childs:
B1B1 B1B2 B1B3 B1B4 B1B5 B1B6 B1B7
B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
B3B1 B3B2 B3B3 B3B4 B3B5 B3B6 B3B7
B4B1 B4B2 B4B3 B4B4 B4B5 B4B6 B4B7
B5B1 B5B2 B5B3 B5B4 B5B5 B5B6 B5B7
B6B1 B6B2 B6B3 B6B4 B6B5 B6B6 B6B7
B7B1 B7B2 B7B3 B7B4 B7B5 B7B6 B7B7
B1G1 B1G2 B1G3 B1G4 B1G5 B1G6 B1G7
B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7
B3G1 B3G2 B3G3 B3G4 B3G5 B3G6 B3G7
B4G1 B4G2 B4G3 B4G4 B4G5 B4G6 B4G7
B5G1 B5G2 B5G3 B5G4 B5G5 B5G6 B5G7
B6G1 B6G2 B6G3 B6G4 B6G5 B6G6 B6G7
B7G1 B7G2 B7G3 B7G4 B7G5 B7G6 B7G7
G1B1 G1B2 G1B3 G1B4 G1B5 G1B6 G1B7
G2B1 G2B2 G2B3 G2B4 G2B5 G2B6 G2B7
G3B1 G3B2 G3B3 G3B4 G3B5 G3B6 G3B7
G4B1 G4B2 G4B3 G4B4 G4B5 G4B6 G4B7
G5B1 G5B2 G5B3 G5B4 G5B5 G5B6 G5B7
G6B1 G6B2 G6B3 G6B4 G6B5 G6B6 G6B7
G7B1 G7B2 G7B3 G7B4 G7B5 G7B6 G7B7
G1G1 G1G2 G1G3 G1G4 G1G5 G1G6 G1G7
G2G1 G2G2 G2G3 G2G4 G2G5 G2G6 G2G7
G3G1 G3G2 G3G3 G3G4 G3G5 G3G6 G3G7
G4G1 G4G2 G4G3 G4G4 G4G5 G4G6 G4G7
G5G1 G5G2 G5G3 G5G4 G5G5 G5G6 G5G7
G6G1 G6G2 G6G3 G6G4 G6G5 G6G6 G6G7
G7G1 G7G2 G7G3 G7G4 G7G5 G7G6 G7G7
Each outcome has P = 1/(7*7*4) = 1/196
Let's only look at the families with (at least one) tuesday boy:
B1B2
B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
B3B2
B4B2
B5B2
B6B2
B7B2
B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7
G1B2
G2B2
G3B2
G4B2
G5B2
G6B2
G7B2
Of these 27 families, 13 have another boy. So P = 13/27.
This is just mathematician masturbation and it drives me nuts.
Order of birth, as is the exclusivity of the day of birth, are not given in this problem. Therefore boy,girl and girl,boy should not be included as separate items in the set, but rather treated simply as one item. Also why the whole Tuesday thing should be ignored as well. We're not told the other child was or was not born on a Tuesday. The information is irrelevant.
You're right! Pretty amazing, considering the article gives the correct answer and explains it pretty thoroughly. Even when you think: "But wait! Can't you look at it some other way?", the article does just that.
Really, this time it pays to RTFA.
If you want to be really pedantic, at birth, there are more boys than girls. Girls constitute a majority of the population because boys die more often, but at birth, the ratio is skewed in favor of boys. So even being a pedantic pain in the ass, you've made assumptions that invalidate your argument.
$_ = "wftedskaebjgdpjgidbsmnjgcdwatb"; tr/a-z/oh, turtleneck Phrase Jar!/; print
IANA mathematician, but: if you also list the probabilities for the Boy on Tuesday being the first child, you get 28 possibilities, of which 14 have two boys (giving you 1/2 again). However, then you've listed "Boy on Tuesday, Boy on Tuesday" twice, although there's no reason for it to be more likely than any of the other possibilities. So if you remove the duplicate, you get 13/27, as stated.
If, on the other hand, it stated that the younger child was a Boy born on a Tuesday, your list would apply, so the probability of two boys would be 1/2.
It is more like 51% since there are slightly more boys born than girls. The girls live longer which is why the female population is higher, but that is at the old women end of the spectrum.
You can read "one of whom" as "at least one of whom" or as "exactly one of whom", "one of whom" alone doesn't really tell you much about the other kid, as it only refers to the first.
...what that actually represents is both having one rock in your presence butt also...
I'm sure glad I don't have a rock present in my butt...
The order of birth is irrelevant. Child B in his list is the Tuesday born boy. That leaves only child A, a girl born on any day, or a boy born on any day.
When our name is on the back of your car, we're behind you all the way!
> He might be. The question does not say one way or the other.
What is the probability of that ?
I have nothing to lose but my bindings.
Just reading the comments gives me a horrible headache.
My older brother and I were both born on Tuesdays.
Nothing to see here, just a systemic anomaly. Move along, now.
When modding "Informative", please make sure it both has a source and IS actually informative.
As I was walking to Saint Ives I met a man with seven wives...
Free Martian Whores!
I've been a fan of MG since reading Dad's Scientific American, and apparently reprints elsewhere. I've always had a problem with these logic puzzles and I never knew why. So you have solved an old childhood mystery of mine. I had the same feeling while I was reading this article, and finally had the epiphany while reading your comment, so thank you for this.
I have been looking for real-world solutions, models or explanations, and the answers given to these questions never seem to reflect reality. The true puzzle here is, as you hinted, a semantic problem more than a math problem. It is worded simply enough, but it is phrased in the language of statistics. Just as words have specific meanings to lawyers and biologists and etymologists, the phrasing has a deeply specific meaning.
To solve the puzzle, you have to decipher the phrasing. The math is ancillary. That there are so many wrong answers should be a clue to this. The convergence given more specific criteria is one that has never been explained adequately, and does help explaining why the problem does not violate simple rules of chance.
Specific to this question, I will nitpick and say it is not clear whether "one of whom is a son born on a Tuesday" indicates that *only* one is a son born on a Tuesday. If you take this exclusionary interpretation the analysis is correct.
This is only true with the exclusionary interpretation. If you concede that the question as phrased does not exclude a second son born on a Tuesday (twins or not), then the answer is 1/2. And this is the key to the puzzle. Here's where I believe the analysis trips up:
This is a false dichotomy - a fundamental assumption which does not follow from the evidence. The analysis given is based on the assumption that one child either is or is not a boy born on a Tuesday. That assumption makes the analysis correct, but the question does not include the assumption, nor does it suggest it. The only way you can come up with that assumption is
What if I selected him because both were boys born on a Tuesday? We can't forget the lessons of other logic puzzles which are trick questions designed to mislead you. What if this is a trick question?
Of course it is all artificial anyway, which I always have a hard time reconciling. There are two answers for each of these questions - one statistician's answer, and one mathematician's answer. And I suppose a third answer, the wrong one, but depending on the person with whom you are arguing one of the first two will also be the wrong answer.
"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?"
As always, the challenge is the assumptions intentionally hidden in the problem statement.
"I" - was your family chosen at random, and if so, from what set?
"two children" - exactly or at least?
"one of whom" - exactly or at least?
"son" - was the sex to say chosen at random, or did you pick a child and announce his/her sex?
"Tuesday" - was the day chosen at random, or did you pick a child and announce his.her birthday?
"What is the probability..." - Some parent you are! Don't you know the sex of your own children?
Simply and honestly reveal the assumptions and the math is straightforward.
"Given a family, chosen at random from the set of all families that have exactly two children and have at least one son born on a Tuesday, what is the probability that both children are boys?"
To make the math easier, let's start with 196 families with two children, with the expected mix of boys and girls. 49 (25%) have two boys and 98 (50%) have a boy and a girl. Of the 98 boy-girl families, 84 do not have a Tuesday-Boy, leaving 14 that do. Of the 49 boy-boy families, 36 do not have a Tuesday-Boy, leaving 13 that do. That leaves a total of 27 families, of which 13 have at least one son born on a Tuesday.
So the probability is 13/27.
Reveal different assumptions, and the answer changes.
> The question is what is the chance that both kids have the same sex.
No, the question is what is the chance they're both boys ?
We know that one is a boy. Ordering and extraneous info is actually irrelevant. There are two possible sets -- boy boy and boy girl.
the chance is 1/2. Always. NO information is given about the second child, implicitly or explicitly. Anything else is mere trickery of the language -- "could have meant", etc. Logic has no place for ambiguity.
I have nothing to lose but my bindings.
After being so careful, I screwed up the bold part. The analysis chooses one child and asks "what if this child meets the criteria" followed by "what if this is child does not meet the criteria." In doing so, a third possibility is "What if both children qualify," which is neglected. That brings the answer back to 1/2.
Here is my reasoning. First, the boy could be one of an identical twin, in which case the other child is certainly a boy (maybe born on Tuesday, maybe not). The
probability of that is 0.004 (0.4%), according to this.
Of course, with probability 1.000 - 0.004 = 0.996 the children are not identical twins, in which case the probability of a boy is 0.512195 (the human live birth sex ratio).
So, the probability of a second boy is
0.004 + ( 0.996 * 0.512195) = 0.514146
Whatever the sex ratio, you gain 1/2 of the probability of identical twins if the other child is a boy (and you lose that amount if the other child is a girl, for the same reason).
I am assuming in all of this live births (i.e., that the children were not stillborn and did not miscarry), but the statement is "I have" not "I had," implying live births.
Order is Irrelevant. The question ask for the probability of the sex of the "other" child, not second child. It's like saying "One of my children's eyes are blue. He's a boy. What are the chances that a specific pregnant rural family in Russia will have a girl?"
I agree. Let's try this:
After having heard that Obama was born on a Friday, a parent remarked: "I have two children, one of whom is a boy born on a Friday. What's the probability that my other child is a boy?"
With the problem restated like this, the mathematical answer should be more acceptable.
> I threw a coin twice yesterday. On one of those throws it landed up heads. What is the chance that the other throw also landed up heads?
Correction -- The question would be, what is the chance that they both landed up heads, and if one is known heads, there are only two sets, heads heads and heads (what we will call headless for this discussion). Otherwise, you're correct, and I am appaled at the state of discussion and people willingly putting up with verbal diarrhea and ambiguities to MIS explain a definition of simplicity in the form of a puzzle.
I have nothing to lose but my bindings.
Please point out where this analysis goes astray.
Gladly. The available outcomes are actually:
Note that there are only 27 outcomes as both boys being born on a Tuesday is only one outcome, not two. There are 13 outcomes labelled BB, so the probability of having two boys is 13/27. Hope that's clear now.
This is a gambler's fallacy problem.
I agree. They have a problem and they need to admit it.
> Because there's only two potential outcomes for the other child: boy or girl
It's not quite a fair coin toss, though. The ratio of males to females is a big higher than 1:1 among children (and much lower among the elderly, though that's less relevant to the question). Of course, this is assuming that the speaker is human...
Cut that out, or I will ship you to Norilsk in a box.
Exactly...
One thing I'm wondering, is why those would be solvers do not take into account the possibility of hermaphrodite then ? Other "common sense" deviations ?
In the boundaries of the puzzle (2 genders, nothing known or assumed about other of the set of 2), so worded, the answer is and forever will be 1/2.
I have nothing to lose but my bindings.
This is where it gets so ambiguous and well... stupid really. At one point in the article they mention that the numbers change depending on why the son born on tuesday was mentioned, as in, "I'm incredibly stupid and live my life based on a set of arbitrary rules, and I'll only mention the gender of one of my children if it is a boy that was born on a tuesday, but otherwise wouldn't tell you anything other than the number of children."
So in that instance, the biological probability isn't the only thing taken into account, but the motivation to tell you the gender and day are taken into account as well.
A "mathmatics" article with stuff like "(approximately) " and "This analysis ignores minor differences like the..." is rather amusing I think.
- http://www.milkme.co.uk
The order isn't irrelevant for the answer, but it is relevant when enumerating all the possibilities. RTFA, it does a very good job of explaining it.
Why does it matter what order they are in. You only have two. If you have two sticks of different color and you put one in a box before the other and shake the box, it doesn't change the outcome of pulling out the red stick.
Every time I start to have faith in humanity, I ruin it by driving to work between 7 and 8 am.
Assuming 50/50 chance of boy or girl and a 1/7 chance of each weekday, let's look at the general population of families with two childs:
B1B1 B1B2 B1B3 B1B4 B1B5 B1B6 B1B7
B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
B3B1 B3B2 B3B3 B3B4 B3B5 B3B6 B3B7
B4B1 B4B2 B4B3 B4B4 B4B5 B4B6 B4B7
B5B1 B5B2 B5B3 B5B4 B5B5 B5B6 B5B7
B6B1 B6B2 B6B3 B6B4 B6B5 B6B6 B6B7
B7B1 B7B2 B7B3 B7B4 B7B5 B7B6 B7B7
B1G1 B1G2 B1G3 B1G4 B1G5 B1G6 B1G7
B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7
B3G1 B3G2 B3G3 B3G4 B3G5 B3G6 B3G7
B4G1 B4G2 B4G3 B4G4 B4G5 B4G6 B4G7
B5G1 B5G2 B5G3 B5G4 B5G5 B5G6 B5G7
B6G1 B6G2 B6G3 B6G4 B6G5 B6G6 B6G7
B7G1 B7G2 B7G3 B7G4 B7G5 B7G6 B7G7
G1B1 G1B2 G1B3 G1B4 G1B5 G1B6 G1B7
G2B1 G2B2 G2B3 G2B4 G2B5 G2B6 G2B7
G3B1 G3B2 G3B3 G3B4 G3B5 G3B6 G3B7
G4B1 G4B2 G4B3 G4B4 G4B5 G4B6 G4B7
G5B1 G5B2 G5B3 G5B4 G5B5 G5B6 G5B7
G6B1 G6B2 G6B3 G6B4 G6B5 G6B6 G6B7
G7B1 G7B2 G7B3 G7B4 G7B5 G7B6 G7B7
G1G1 G1G2 G1G3 G1G4 G1G5 G1G6 G1G7
G2G1 G2G2 G2G3 G2G4 G2G5 G2G6 G2G7
G3G1 G3G2 G3G3 G3G4 G3G5 G3G6 G3G7
G4G1 G4G2 G4G3 G4G4 G4G5 G4G6 G4G7
G5G1 G5G2 G5G3 G5G4 G5G5 G5G6 G5G7
G6G1 G6G2 G6G3 G6G4 G6G5 G6G6 G6G7
G7G1 G7G2 G7G3 G7G4 G7G5 G7G6 G7G7
Why are you assuming that order matters, but boys are interchangable? If we label the child that was mentioned in article as A and the other one as B then B2B2 splits into two cases when A has an older brother and when he has a yonger brother, therefore we get:
B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
B1B2
B2B2
B3B2
B4B2
B5B2
B6B2
B7B2
B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7
G1B2
G2B2
G3B2
G4B2
G5B2
G6B2
G7B2
Of these 28 families 14 have another boy, so P=14/28=1/2
Because he didn't say, "Take two of my children."
He said, "I have two children."
Every time I start to have faith in humanity, I ruin it by driving to work between 7 and 8 am.
See my other reply - I formalized the problem in a different, yet straightforwarde way, and lo and behold, my answer is also 1/2.
:P
http://slashdot.org/comments.pl?sid=1701394&cid=32729366
The point is here, most improtantly, reducing the problem in a graceful manner. Nothing is known about the OTHER (not first or second, other) child. The state of the first child does not influence the state of un known child. Therefore, there are two possible sets of children -- boy boy and boy girl. Of the question "what is the probability of them both being boys", the correct answer will always be 1/2. Give or take a cat
Note that I will not change my mind or see the supposed light in this matter, because there is nothing to see, but the amount of delusion.
Note, if you do not allow them both to be born on Tuesday and them both being boys, you are doing it wrong.
I have nothing to lose but my bindings.
However, if someone went out and specifically selected a family with at least one boy, the probability of the other child being a girl is not 50% - the population in question has been artificially depleted of girls because the criteria excluded all two-girl families.
This is the crux of the problem: the choice of the population and the information provided to you by the twit who thinks this is an interesting way to spend time.
The cannonical form of this problem is, "If at least one of my children is a boy, what is the probability the other is a girl?" The answer is not necessarily 50%, depending on information that the person stating the problem has not provided and you have no way of knowing.
The usual answer that is claimed to be "correct" is 66%, since you have four cases: bb, bg, gb, gg, but the last case is excluded by the formulation of the question.
But suppose that the question was formulated by flipping a coin and picking from these cases randomly, and then posing the question in terms of boy or girl on that basis. One way of doing that would be to split the above list in half:
bb, bg | gb gg
and ask "At least one of my children is a boy..." in the first case and "At least one of my children is a girl..." in the second case.
Mathematicians will complain that this isn't "fair", and they would be right.
But then I have to ask: why is this problem always stated in terms of the child of known sex being a boy? Do mathematicians always and only have male children? In that case the probability of the other child being a girl would be zero!
Blasphemy is a human right. Blasphemophobia kills.
The more tangents you throw at it, the closer you get to .5 (50%), while never reaching it. This is the limit, why? Because there's only two potential outcomes for the other child: boy or girl.
Though as others have pointed out, the probability of having a boy is actually slightly higher than that, around 51%. The "two outcomes equals 50% chance" only applies to the special case of each outcome having equal probability, which is why problems that are being rigorous will always say a "fair" coin or die.
I could roll a die 99 times, and get 6, the probability of getting 100 6's when I've already got 99 6's is still 1 out of 6, not 6^100.
Yes, but human biology is more complicated than a fair die roll. Specifically, some men tend to produce more sperm of one sex than another, or have more vigorous sperm of one sex than the other. Henry the VIII famously had this problem of only producing viable female sperm, and he cut his wives' heads off for it.
So there's a conditional probability based on having N children of one sex that the father has such a condition, and in that case the probability of a boy would be different. The result would like P(boy given no condition)*(1 - P(condition given one boy)) + P(boy given condition)*P(condition given one boy).
I don't have the information to actually calculate the probability. But it's not 50%, and "extraneous" information like "I've already had a boy" is not actually extraneous and not an example of the Gambler's Fallacy.
The enemies of Democracy are
Indeed, is everyone drunk or something...
Second, the probability that the other child is a boy is either 1 or 0, it's something that has already occurred...
...snip...
The probability of guessing correctly by saying the second child is a boy would therefore be 1/2(6), or 3, divided by 6 and a half, which gives you 6 out of 12 and 1/2 odds.
Using this logic, what's the probability that I won the lottery last week? 50/50?
If you have two sticks of different color and you put one in a box before the other and shake the box, it doesn't change the outcome of pulling out the red stick.
Probability is all about what you know and what you don't.
Of course, if you already have the info that both sticks are of a different color, then the probability of pulling out another red stick (boy) after you've already got a red one is exactly 0.
However, if you only knew that:
The key here however is how you find out about 3. If it is by pulling out one stick, you'd have established an order (the order with which you pull sticks out), so the probability for the other one to be red would only be 1/2.
However, if the (trustworthy) game-show host would tell you that there is (at least) one red stick in the box, this wouldn't establish an order among sticks, and so the probability of their being 2 red sticks would be only 1/3.
It just goes to show you how people can manipulate numbers to mean what they want... then fervently defend that position as the right answer.
Every time I start to have faith in humanity, I ruin it by driving to work between 7 and 8 am.
It's not that the order in time is especially relevant. It's that the identity of the child matters. You could call them child A and child B, and list the permutations, and the answer would still be 13/27. Child A being born on a Tueday are NOT the same set of permutations as child B being born on a tuesday. They have to be listed separately so that duplicate permutations can be removed from the list. That's what makes the difference.
Did I win the lottery last week? The unknown only has two possible outcomes: I won or I lost.
Therefore, based on your math, my odds are 50-50%.
University of Phoenix online wants their diploma back. :-)
Yes, I read through the article. Yes, I understand the "basics" of probability. Yes, I know how to do statistics.
And Tuesday IS irrelevant.
Anyone trying to say otherwise, quipping that "it depends" is doing the problem wrong and making a classic mistake of correlating a statistic that doesn't affect the outcome. Those who go through the mental gymanstics of making it matter are forcing an outcome that truly is not associated with the problem.
The real problem is if you have two kids, can remember the sex of only one of them and have to resort to statistics to know whether the other kid is a boy or a girl.
Nae king! Nae laird! Nae yurrupiean pressedent! We willna be fooled again!
I think a lot of people are missing what's actually going on, here.
First off, there's no "only one child/boy was born on Tuesday" constraint.
If someone says, "I have two children. One of them is a boy." The odds of the other one being a boy is 1/3.
If someone says, "This is little John. I have another child." The odds of the other one being a boy is 1/2.
Saying, "One of my children is a boy born on Tuesday," is a lot like saying, "This is John," except the possibility that there's another boy born on Tuesday slightly skews the odds away from 1/2. "Boy born on Tuesday" almost, but not completely, identifies the child.
You're right the question posed in the topic is not really the question at all. I hope someone gets fired over that blunder!
I'm applying the sticks in the above equation to the children. (and I admit, I did say they were a different color which I should retract.)
We know there are two sticks. Let's assume (because using color can equate to millions of results) that you know the only colors that exist are red and blue.
The question can be rephrase to:
I have two sticks of which one is red that I pulled out on Tuesday. What's the probability that the other stick is red?
It is in fact, 50%. It's either red or blue. You don't know that they are different colors.
There were two sticks in the box. The possible combination pairs would be:
red - red
blue - blue
red - blue
By pulling out a blue stick, you eliminate red-red leaving only two possibilities.
The one solution gets everything wrong by deducing order.
ie:
girl - girl (can't exist)
boy - boy
boy - girl
girl - boy (irrelevant, we already cover this situation above... by ordering them you've put weight on there being different sexes.) [this was my original point]
So there are only two outcomes. You have two boys or a boy and a girl.
Every time I start to have faith in humanity, I ruin it by driving to work between 7 and 8 am.
World population is split closer to 50.25% male, 49.75% female. Therefore, it's 0.5025 probability of being male. 201/400
http://wiki.answers.com/Q/What_is_the_ratio_of_women_to_men_in_the_world Abstract thought ftw!
Finally had enough. Come see us over at https://soylentnews.org/
Out of all fathers of two kids, 1/2 have a girl as the first one, and the other 1/2 - a boy. Same for the second one.
This makes 4 groups of fathers that have two kids:
1. boy, boy - 1/4
2. boy, girl - 1/4
3. girl, boy - 1/4
4. girl, girl - 1/4
That's something I don't think you'd doubt.
If I tell you my first child is a boy, this means I'm either from group 1, or from group 2. Which means that the chance that the second child is a boy is 1/2.
If I tell you that one of my child is a boy, I'm either from group 1, 2 or 3. If you do the counting, you'd find that only 1/3 of the parents from those groups have their second child being a boy, too!
Well I seem to be blocked, so that's out. Is it a US only site or something?
To have a right to do a thing is not at all the same as to be right in doing it
If order doesn't matter, then consider the basic question, where no information is offered: IfI have two children, what's the probability that I have two boys? Let's list the possibilities without taking order into account:
Boy, Boy
Boy, Girl
Girl, Girl
So it's 1/3? Well no, there's overwhelming acceptance that the probability is 1/4, because order does matter. Note that ordering doesn't have to be by the age of child - that's just a convient way of ordering. We could equally order then by their name alphabetically, or whatever. But in probability, some notion of ordering is important. Consider the same question with randomly selected blue and green balls - the probability of selecting two blue balls is still 1/4, not 1/3, because the ordering matters, even though neither ball has an "age".
That's not to say I've wrapped my head round the paradoxes presented by these problems. But it's not as simple as saying ordering doesn't matter - if we ignore ordering, we get into far more obvious problems.
You're misreading the question. If you exclude the Tuesday part (which I still don't get, and I still doubt), the question is not what's the chance that the second child is a boy, which would be 50%. The question is what is the chance that both kids have the same sex. Since the fact that one of them is a boy doesn't bring any new information to the table about they being the same sex or not, the probability remains the same -- 1/3.
Actually, the probabilty changes and is therefore 33% (not quite 1/3). If we do not know the sex of either child, the odds of both of them being boys is 25.8%, the odds of them being a girl and a boy is 52.2%.
If we were to change the problem to one is a girl, the odds that the second one would be a girl would be 30%.
The issue with this problem is that the gender of children is not truly random.
The truth is that all men having power ought to be mistrusted. James Madison
The way the question is almost always worded (including in this case), there is at the very least a strong implication that you're talking to a random person who is telling you about his kids, in which case the sex of the other child is independent of the sex of the child he's telling you about.
You have to jump through verbal hoops to actually state the situation unambiguously in the way the usual answer assumes. Someone has to go out and specifically select a person who has at least one child of a particular, predetermined sex, and introduce that person to you. Once you add all that detail to the question it's pretty clear something is going on, which takes all the fun out of it for people who like silly puzzles, but does turn it into a meaningful lesson about selection bias.
Wow. There must be 100 posts before someone clearly and simply states the correct analysis.
Intron: the portion of DNA which expresses nothing useful.
But if you split one of those cases in two, each has only half the chance of occuring in the first place.
The chances are identical then, when considering any day of the week, not just tuesday, and also in the case of a girl born on tuesday (or any day). The fallacy comes from stipulating that one of your children is a boy, not that it occurs on a particular day.
Is it just my observation, or are there way too many stupid people in the world?
This is slashdot. If you're asking a logic problem like this, you are probably lying and have never reproduced.
There, go ahead and count. There are 27 possible families when where the man has (at least) one son who's born on a Tuesday. You can then easily see that there are 13 cases where the man has two sons.
My birthday is today, and it's a Tuesday.
What are the odds this post will be modded up?
The society for a thought-free internet welcomes you.
One way of doing that would be to split the above list in half:
bb, bg | gb gg
and ask "At least one of my children is a boy..." in the first case and "At least one of my children is a girl..." in the second case.
I don't see how you think you've changed the problem. If you are grouping 'bg|gb' as one group on par with bb and gg, first that's 3/4ths of the group count and the middle group is twice the size of the other groups. If you are saying that bg, bg is one group, and gb gg is another group, then "at least one of my children is a boy' would describe the entire first group and half of the second, making the grouping meaningless. If you are saying the goal is to pick which half of the | describes it, then using the unordered information provided, the probability of guessing the correct side can be 2/3rds.
You haven't changed the problem in any way whatsoever..
XML is like violence. If it doesn't solve the problem, use more.
From the article, you also have to take into account that it was the older boy that was born on Tuesday.
I would RTFA but the site has been broken all morning - I think they pulled the article.
No. I live in Europe and had no problem reading it.
seems like the monty hall problem, we start off with four possibilities: girl-girl girl-boy boy-girl boy-boy then we are given that girl-girl is incorrect since one child is a boy this leaves: girl boy boy girl boy boy 2 of those solutions have a girl, so the odds of a girl child are 2/3
It's either false dichotomies, or the terrorists win, you decide.
I'm still able to reach the site. Here's the entire text:
When intuition and math probably look wrong
A twist on the Two Children Problem shows how information can steer what looks probable.
By Julie Rehmeyer
I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?
Gary Foshee, a puzzle designer from Issaquah, Wash., posed this puzzle during his talk this past March at Gathering 4 Gardner, a convention of mathematicians, magicians and puzzle enthusiasts held biannually in Atlanta. The convention is inspired by Martin Gardner, the recreational mathematician, expositor and philosopher who died May 22 at age 95. Foshee’s riddle is a beautiful example of the kind of simple, surprising and sometimes controversial bits of mathematics that Gardner prized and shared with others.
“The first thing you think is ‘What has Tuesday got to do with it?’” said Foshee after posing his problem during his talk. “Well, it has everything to do with it.”
Even in that mathematician-filled audience, people laughed and shook their heads in astonishment.
When mathematician Keith Devlin of Stanford University later heard about the puzzle, he too initially thought the information about Tuesday should be irrelevant. But hearing that its provenance was the Gathering 4 Gardner conference, he studied it more carefully. He started first by recalling a simpler version of the question called the Two Children Problem, which Gardner himself posed in a Scientific American column in 1959. It leaves out the information about Tuesday entirely: Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?
Intuition would suggest that the answer should be 1/2, since the sex of one child is independent of the sex of the other. And indeed, had he been told which child was a boy (say, the younger one), this reasoning would be sufficient. But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth:
Boy, girl
Boy, boy
Girl, boy
Since one child is a boy, we know that girl, girl isn’t a possibility. Of the three approximately equally likely possibilities, one has two boys and two have a girl and a boy — so the probability of two boys is 1/3, not 1/2, Devlin concluded.
He used this same method on the Tuesday birthday puzzle, enumerating the equally likely possibilities for the sex and birth day of each child and then counting them up.
If the older child is a boy born on Tuesday, there are 14 equally likely possibilities for the sex and birth day of his younger sibling: a girl born on any of the seven days of the week or a boy born on any of the seven days of the week. (This analysis ignores minor differences like the fact that slightly more babies are born on weekdays than on weekend days.)
Now suppose that the older child isn’t a boy born on Tuesday. The younger child then must be, of course. Now we count up the possibilities for the sex and birth day of the older child. If she’s a girl, she might have been born on any day of the week, generating seven more possibilities. If he’s a boy, he could have been born any day except Tuesday. (Otherwise this case would already have been counted in the first scenario: the older child a boy born on Tuesday). This second scenario generates just six, rather than seven, more possibilities.
Since each of these cases is (approximately) equally likely, we can compute the probability by dividing the number of cases in which there are two boys by the total number of cases. The total number of cases is 27: 14 if the older child is a boy born on Tuesday and 13 if the older child isn’t. In 13 of those cases both children are boys (7 if the older child is a boy born on Tuesday and 6 if he isn’t), yielding a probability of 13/27.
Devlin w
You are neglecting to treat the two children as independent non-exchangeable objects. It is easier to think about if you consider that you have two pets a dog and a cat, each of which can be male or female. Then the enumeration of possibilities:
Male Dog, Male Cat
Male Dog, Female Cat
Female Dog, Male Cat
Female Dog, Female Cat
If you select a family where (at least) one of them is a Male that leaves the following options:
Male Dog, Male Cat
Male Dog, Female Cat
Female Dog, Male Cat
So the chances of both being male are 1/3.
Now moving onto the date question. You have selected a family from families known to have one male pet born on Tuesday (and one dog and cat each). The options are:
Male Cat born Tuesday, Male Dog born Monday
Male Cat born Tuesday, Male Dog born Tuesday *
Male Cat born Tuesday, Male Dog born Wednesday
Male Cat born Tuesday, Male Dog born Thursday
Male Cat born Tuesday, Male Dog born Friday
Male Cat born Tuesday, Male Dog born Saturday
Male Cat born Tuesday, Male Dog born Sunday
Male Cat born Tuesday, Female Dog born Monday
Male Cat born Tuesday, Female Dog born Tuesday
Male Cat born Tuesday, Female Dog born Wednesday
Male Cat born Tuesday, Female Dog born Thursday
Male Cat born Tuesday, Female Dog born Friday
Male Cat born Tuesday, Female Dog born Saturday
Male Cat born Tuesday, Female Dog born Sunday
Male Dog born Tuesday, Male Cat born Monday
Male Dog born Tuesday, Male Cat born Tuesday *
Male Dog born Tuesday, Male Cat born Wednesday
Male Dog born Tuesday, Male Cat born Thursday
Male Dog born Tuesday, Male Cat born Friday
Male Dog born Tuesday, Male Cat born Saturday
Male Dog born Tuesday, Male Cat born Sunday
Male Dog born Tuesday, Female Cat born Monday
Male Dog born Tuesday, Female Cat born Tuesday
Male Dog born Tuesday, Female Cat born Wednesday
Male Dog born Tuesday, Female Cat born Thursday
Male Dog born Tuesday, Female Cat born Friday
Male Dog born Tuesday, Female Cat born Saturday
Male Dog born Tuesday, Female Cat born Sunday
* Note that I enumerated the case where both are males born on Tuesday twice. These are redundant and one must be discarded else I will double count that situation. After doing so there are 13/27 cases where both are males.
Notice that if you ignored the fact that one was a dog and the other was a cat you would have merged the two lists, ending up with your original list, and double counting the case where both are boys born on Tuesday.
In other words your mistake is that you assumed you had been given the sex and birth date of child A, and enumerated the sex and birth date of child B. However, you don't know the sex and birth date of child A or B, just that one of child A or B have that sex and birthdate. That is a subtly different problem.
So the order that they are born in is irrelevant, but keeping track of the fact that they are the two unique items while enumerating the cases is vital (and older and younger is a simple label to use while doing so).
It goes astray where you are assuming the 2nd child is the one born on a Tuesday. What your list should look like is:
1 Girl on Monday, Boy on Tuesday
2 Girl on Tuesday, Boy on Tuesday
3 Girl on Wednesday Boy on Tuesday
4 Girl on Thursday, Boy on Tuesday
5 Girl on Friday, Boy on Tuesday
6 Girl on Saturday, Boy on Tuesday
7 Girl on Sunday, Boy on Tuesday
8 Boy on Monday, Boy on Tuesday
9 Boy on Tuesday, Boy on Tuesday
10 Boy on Wednesday Boy on Tuesday
11 Boy on Thursday, Boy on Tuesday
12 Boy on Friday, Boy on Tuesday
13 Boy on Saturday, Boy on Tuesday
14 Boy on Sunday, Boy on Tuesday
15 Boy on Tuesday, Girl on Monday
16 Boy on Tuesday, Girl on Tuesday
17 Boy on Tuesday, Girl on Wednesday
18 Boy on Tuesday, Girl on Thursday
19 Boy on Tuesday, Girl on Friday
20 Boy on Tuesday, Girl on Saturday
21 Boy on Tuesday, Girl on Sunday
22 Boy on Tuesady, Boy on Monday
23 Boy on Tuesday, Boy on Wednesday
24 Boy on Tuesday, Boy on Thursday
25 Boy on Tuesday, Boy on Friday
26 Boy on Tuesday, Boy on Saturday
27 Boy on Tuesday, Boy on Sunday
27 possibilities, 13 of them are 2 boys.
As lots of people have said, the wording/situation is ambiguous, but this is the correct analysis for the statement: Assuming subsequent children have independent birthdates and genders, and assuming the probability of being a boy is 1/2 and the probability of being born on a Tuesday is 1/7, what is the probability of a 2 child family having 2 boys given that they have at least one boy born on a Tuesday.
The order is totally irrelevant for even enumerating the possibilities. The author of the article is a dimwit. The question is: What is the probability of the sex of the other child? That will always be 50/50. What if the first child was listed as an unknown sex? Does that suddenly magically make the probability of the other child's sex more 50/50? The question was worded incorrectly to reach the answer in the article. You're looking at [b,b][b,g][g,b][g,g], while failing to realize that since order is irrelevant, [b,g] and [g,b] are the same thing. Your choices are between [b,g] and [b,b]
well did you or didnt you?
I know I'm replying to my own post, but I wanted to point out that the result would be 13/27 if you *asked* the parent if they had a son born on Tuesday. In this case, the question is posed by the parent, meaning they made the conscious choice (assumed 50%) to pick Tuesday over the day of the week of their other son in cases with 2 boys born on different days.
The real riddle isn't the idea that you need to interpret a set of data, that's (relatively) easy. It's actually identifying which set of data you have presented to you.
the dad of B2B2 is twice as likely to appear in your test group than the other dads, which makes it 14/28 = 1/2
I think another way to see the word game here is that they are enumerating genders and not objects. Real objects are not interchangeable... if there are two boys they could be born in two different orders:
Boy1 Boy2
Boy2 Boy1
Girl1 Boy1
Boy1 Girl1
=1/2
Or to put it another way - In the real world given that "at least one is a boy" we know that the boy was born either first or second... so write it as two trees and average them together to get 1/2.
Possibility 1 - boy born first:
Boy Girl
Boy Boy
=1/2
Possibility 2 - boy born second:
Girl Boy
Boy Boy
=1/2
Both are equally likely possibilities so:
Possibility 1 + Possibility 2 = 1/2.
Pat
Normal English strongly implies it does. If you say to someone I have several pieces of fruit and one of them is a banana, when in fact two of them are bananas, most people would call that lying. One could argue that strictly speaking the statement was true: you did have one banana, you just also had an additional banana, but that level of honesty is only tolerated in politicians. If you had said "at least one of which is a banana" that would be fine, otherwise the statement is deliberately misleading.
By the same standards, stating that one child was born on a tuesday also implies the other child was not born on a tuesday, regardless of gender.
You cant twist it both ways to state that it implied "the other child was not born on tuesday unless it was a girl".
Consistancy is key.
As such I would say the odds remain unchanged (assumed 50/50).
"lt;dr" is the correct response to most of my posts.
It's the 1970's. Two math professors, old friends who both live in London, are on the phone discussing an upcoming conference in Edinburgh they'll both be attending.
- Hey, we could fly over together if you'd like.
- Thanks, but I'll be driving.
- All that way? It'd take you most of the day! Whatever for?
- Well, I recently made a study of the statistics of bombs being smuggled on board passenger planes. And while the odds of it occurring on any particular flight are high, the possibility still makes me uncomfortable with flying.
- Well, suit yourself. I'm going to take the plane.
A short time later, the one professor is boarding her flight out to the conference, and who should be sitting in the adjacent seat but her old friend! They're both pleasantly surprised, and the first professor settles into her seat. She leans in and quietly asks her friend -
- So what about that whole probability issue? Was your math off, or did you just work up the nerve?
- Wrong on both counts! I did have a breakthrough, however.
- Really? How do you mean?
- Well, I went over the statistics again, and worked out the odds of two bombs being separately smuggled on board the same flight.
- High?
- Astronomical! You've a better chance of being struck by lightning!
- So how does knowing that make you more comfortable with flying?
- (singsongs) Guess what I've got in the briefcase...(pats the case on her lap)
.
Prisencolinensinainciusol. Ol Rait!
Two six-sided dice.
Chance of a sum of seven?
Are there 36 possible outcomes? Really? Why? We're only interested in the sum. Addition is commutative. We don't care about order.
So again are there really 36 possible outcomes?
If you're attempting to count possible sets and [3,4] is the same as [4,3] to you, then maybe not. But if you're attempting to calculate probabilities of rolling a seven, you'd better be clear that there are 36 possible outcomes and you must count [3,4] separate from [4,3] as you calculate your probabilities.
Or... chuckle... come on over here and let's start some serious gambling...
You're right, and now that I read the article again, this is actually explained there pretty well... But your point goes only half way, because the selection of telling of a boy affects this exactly in the same way as you explained for the weekday, so that the result in the end is P=1/2. Just as intuition would suggest. I think.
And I buy four lottery tickets a week. Woo-hoo, I won twice!
Your ad here. Ask me how!
I don't get it. He tells us that he has a boy. So that leaves two possibilities. He has a boy-boy or a boy-girl. That 1/2.
Don't take life so seriously. No one makes it out alive.
How is that not the same? Either way, once you know that one coin is heads, the only two options left are HH or HT.
Don't take life so seriously. No one makes it out alive.
...in one little sentence.
Before you mod me down in trolling rage, listen:
This sentence is perfectly clear and unambiguous to everyone but mathematicians.
In normal social interaction, there simply is no doubt about what the “one of” belongs to.
Only mathematicians with their surreal alien thought processes and social inferiority (Because of their work. So no absolute judgment here. Only relative to what we call normal.), would ever struggle with that.
It feels like their minds completely miss any kind of default assumptions of a society. But not only that. It even is seen as a taboo and as something unacceptably bad to them. (Because of course in mathematics and programming, it is.)
But the real world is based on them, and they work really well.
Sure there are wrong assumptions here and there, but they can’t remotely outweigh the benefits.
Mind you that I was very close to also becoming that strict, a couple of years ago.
But nowadays, after having been out much, I have become unable to stand more of 2-3 lines of conversation with a mathematician.
Because for every shit they don’t accept it, if there isn’t a rule and definition for every shit.
I’m just standing there, yelling “You know *exactly* what I mean, but you’re too much of a micromanaging dick to simply make that assumption!”.
It’s sad, because mathematicians don’t have to be that way all the time.
I can perfectly see the point of it, when doing mathematics. It’s good.
But leave it there, and stay human for the rest. Or you’ll find yourself all alone at home, having tinkered about shit like the Tuesday birthday “problem” for months at a stretch.
Any sufficiently advanced intelligence is indistinguishable from stupidity.
He makes the assumption that only one was a boy born on Tuesday. The question would be much different if it said "Only one was a boy born on tuesday."
Here is a less ambiguous problem that shows the same effects.
Take two decks of cards. Shuffle each deck. Deal a card from deck 1 and another card from deck 2. If one of the cards is a spade, stop. If no card is a space, put the cards back into their original decks, shuffle again and repeat. Continue repeating until one of the cards dealt is a space.
Question: At the time you stop, what is the probability that the other card is black?
The problem "solution space" overlooked the first thing that I thought of that would skew answers - twins. So the odds in the "solution" are wrong.
Current odds of twins 1/33
Identical twins 1/250.
And if you're OctoMom ...
The AC you responded to has it right, both cases of "two Tuesday boys" need to be counted. You are also correct that, given two boys born on Tuesday, there is equal probability that a specific one is older or younger than the other. This second fact, however, is not a good reason for excluding one of the cases. In fact, it's the precise reason why we need to count both.
Here are the two "two Tuesday boy" cases as I see them:
(Boy I've met is older, boy I haven't met is born on Tuesday)
(Boy I've met is younger, boy I haven't met is born on Tuesday)
Excluding the second of these instances is equivalent to saying that if I'd known the boy I met was the younger then I'd also know that his brother couldn't have been born on Tuesday. This is clearly not a true claim.
Neither would it be reasonable to exclude the first one, for the same reason: knowing that the boy I've met is older than his brother doesn't mean that his brother couldn't be born on Tuesday, too.
The probability of the "brother I haven't met" being older or younger must be uniform, which requires two equally likely cases to be counted.
Don't forget intersex, hermaphrodites and other non-boy/non-girl sexes!
Exactly what I'm talking about!
Because out of the 14 BG/GB combinations, 50% of them would be telling you about their *girl*
That pulls 7 out of the total set, and leaves 7/14 being the probability for 2 boys.
I cheated though, because I've already read this blog:
http://blog.tanyakhovanova.com/?p=221
It's from a couple months ago and summarizes several solutions also seen in this thread. But it's easier to get people to pick up the concept that it's just a trick question designed to get you to determine your dataset.
There are three options: HT, TH, and HH, each with equal probability. All three can be described as "one of them was heads".
It's easier to visualize if you make them different coins. The stated problem is equivalent to, "I just tossed a quarter and a dime. One of them was heads; what is the probability the other was heads as well?"
With no information, there are four equally-probable scenarios: QH/DH, QH/DT, QT/DH, QT/DT. The information provided (one of them was heads) eliminates the QT/DT possibility. So there are three equally-probable scenarios, only one of which fits the description of "the other is heads".
Article currently reads:
"Error. You are unable to view this section .."
cd
Ok, the 1/3 was making the most sense to me and the Tuesday thing bugged me. So I coded the possibilities and did some grepping. storm@octopus:~$ perl BGB.pl | grep -v G[1-7]G | grep B2 | grep -c -v G 13 # -v is the not operator so number of boy-boy options storm@octopus:~$ perl BGB.pl | grep -v G[1-7]G | grep B2 | grep -c G 14 # Expected from article. storm@octopus:~$ perl BGB.pl | grep -v G[1-7]G | grep B[1-3] | grep -c G 42 storm@octopus:~$ perl BGB.pl | grep -v G[1-7]G | grep B[1-3] | grep -c -v G 33 # but if he said mon tue or wednesday, the numbers move toward 1/3 storm@octopus:~$ Storm
(Assuming 50/50 change of boy/girl and no twins.)
There is only one source; "I have two children, one of whom is a boy born on a Tuesday."
The answer is 50%. (excluding complexities on how to interpret the line mathematically.)
As soon as you add a second source the answer changes;
Foo says: Bob has two children.
Bar says: Bob has a son.
Now the answer to Bob having two boys becomes 33 1/3%
The stick or switch problem (can't think of the more common name) has two sources, first the random choice and the second caused by elimination action after this.
Devlin started by listing the children’s sexes in the order of their birth:
Boy, girl
Boy, boy
Girl, boy
Since one child is a boy, we know that girl, girl isn’t a possibility. Of the three approximately equally likely possibilities, one has two boys and two have a girl and a boy — so the probability of two boys is 1/3, not 1/2, Devlin concluded.
All this analysis appears to be over-complicating the problem. But I'll see if I can play the game, too.
While it's true that g/g is obviously eliminated, the boy MUST be either the younger or the older. Since he's either one or the other, we can infer that either b/g or b/g is also impossible. If he's older, then g/b is not possible, and if he's younger, then b/g is not possible. One of them is not possible, even though we don't know which one it is.
So we have:
Boy > boy
OR
Girl > boy | boy > girl.
The woman cannot have girl > boy if the boy is older or boy > girl if the boy is younger.
I conclude intuition is correct, 1/2.
I believe the reasoning here is completely flawed:
The list should be:
_Boy_, girl
_Boy_, boy
Boy, _boy_
Girl, _boy_
Where the precondition is one of the underlined ones. Thus the probability become 1/2 that the other kid is a boy.
The thing is, if the "girl, boy" event differs from "boy, girl", then "_boy_, boy" must differ from "boy, _boy_".
"Civis Europaeus sum!"
I have two slashdot readers and one of them did not read the article before they posted a reply on a Tuesday. What is the probability that the other didn't read the article as well?
100%
The "boy is born on Tuesday" problem is really the same as an old card puzzler. Deal out13 cards (i.e. a Bridge hand). In some cases the player says "I have an ace." In other cases the player says "I have the ace of spades." In which case is the player more likely to have 2 or more Aces? :-) by counting the number of hands w/ at least one ace, counting the number of hands with the ace of spades, and then seeing the percentage of each of those sets with more than one ace.
The answer is of course, when he says he has the Ace of spades. You can brute-force it easily
And don't post that I'm wrong. I'm not, neither is Martin Gardner, and neither is Marylin vos Savant.
https://app.box.com/WitthoftResume Code: https://github.com/cellocgw
Why 6/13? Clearly it's either a boy born on one of the seven days, or a girl born on one of the seven days. Or are you falling into the trap that just because one of the boys is born on Tuesday that the other can't be born on Tuesday as well? It's a riddle. If both boys were born on Tuesday, the statement that one of them was born on Tuesday is true. It's the same as the two coin/30cents/not a nickel riddle.
When our name is on the back of your car, we're behind you all the way!
Ok, here's my thought after 10s of thinking:
> "I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?"
Well, that it's Tuesday is irrelevant unless you're gonna go data mining into actual births mapped to days of the week.
But that "a boy was born on a particular day of the week" is important. Certain births are multiple births, possibly twins in this case. Hence one could expect that, knowing precisely 2 were born, that there's a slightly increased chance the other child was born on the same day, being a twin.
So that's my 10s of thought, written over about two minutes.
(-1: Post disagrees with my already-settled worldview) is not a valid mod option.
So... all we know is that there is a son that was born on a Tuesday. The other child was born either before or after him. Maybe even on the same day. The possibilites are: He has an older brother. He has a younger brother. He has an older sister. He has a younger sister. The specific weekday doesn't matter at all. The answer is 2/4.
Actually, the basic two children problem isn't so bad
Sorry, retract that, just read TFA again... That's the one that's ambiguous... My head hurts.
In a survey of 100 programmers, 111111 thought that duck-typing was a good idea.
The thing is, if the "girl, boy" event differs from "boy, girl", then "_boy_, boy" must differ from "boy, _boy_".
This is incorrect; you can distinguish boy < girl from girl < boy by comparing the ages. Underlining one or the other doesn't alter the number of states for the system, boy < boy counts only once.
Yes, but you didn't address the question.
There is ambiguity in the language, and this ambiguity *allows* for other constraints (i.e. how was the family selected, is only one child male, does order matter etc...).
So you try to solve the problem by anticipating every possible unmentioned constraint that could have been implied. You come to an answer. Then the person who posed the problem says "Aha! You're an idiot! What about constraints X,Y and Z!". Since the question as posed didn't rule out X,Y and Z, you feel stupid and the person who posed the question pats themselves on the back.
The problem is that it's not rational.
In the absence of any explicitly supplied constraints, the only rational course of action is to assume that there are no other constraints and work only with the information given. This is how math works. Have you ever seen a math test where you answer the question with the information given and then the tester adds extra information after the fact and tells you that you're wrong? No? Me either, but that's exactly what's happening here.
The only information given is that this person has two children, at least one of whom is a male born on a Tuesday. The fact that the one child is born on a Tuesday is *only* relevant if you start assuming unmentioned constraints (e.g. maybe the family wasn't selected at random, maybe the child was selected because of his birthday, maybe the man actually has 3 children... etc...).
So, without trying to anticipate each and every *possible* constraint allowed by the ill-posed question, the only certain information you have, and thus the only information it is rational to work with, is: two children, one's a boy.
The answer in this case is simple.
Adding additional constraints after you have already solved the problem with the information provided and then claiming you're wrong is simply a dirty trick philosophers use to make themselves feel smart. If relevant constraints are not provided before hand, they are not part of the problem.
P.S. I should qualify, I don't really have anything against philosophers, I really like a lot of philosophy - but philosophy is the only field I have seen where supplying constraints that affect the outcome of a problem after the fact is a valid way to win an argument.
Automatic Assumption #1: "The day of the week is irrelevant. The probability is 1/2. Dice and (theoretical) Wombs don't have memory."
Automatic Assumption #2: "This is one of those loaded math problems, which means my first automatic assumption is almost certainly false and I need to apply further examination to work out why."
Please note that. . .
Intuition had exactly NOTHING to do with that process. Both assumptions were based on fast pattern recognition and old survival tactics. Old monkey/reptile brain stuff. You can program all of that into a machine. Real intuition cannot be programmed.
REAL Intuition works from the soul level and taps into a knowledge base which includes data from past life experiences and the "infinite universal hard drive". (Which, of course, is still a form of pattern recognition and response, but it allows for more information than is actually available in the physical realm.)
Here's what My REAL intuition told me: "There's something fishy about this. The mathematicians working on this are making assumptions which are actually debatable. This isn't as cut and dried as the Monty Hall problem; you cannot apply this to the real world. Also. . , most of Slashdot is going to take this personally and get all huffy."
Real intuition, if people followed it, would see them switch doors when Monty offers the option. Their guts would tell them to because their guts have access to the universal probability calculator. Unless of course the prize was behind the first door.
Jedi don't lose at Monty Hall. But then, of course, Jedi don't go on game shows either. On the way to becoming a Jedi, you realize that money and fame and glittery distraction for the masses belongs to a different realm. -A realm where automatic behavior is confused with intuition.
-FL
I could change the answer by saying there a 52 Tuesdays in the year and using that as what the Tuesday means instead of just a day of the week.
In fact, the question assumes they where born in the same week in order to get their 'answer'.
The Kruger Dunning explains most post on
It's an interesting problem, but a slightly trickier probability puzzle in the Snooker Table of Doom:
http://www.skytopia.com/project/imath/imath.html#13
Why OpalCalc is the best Windows calc
Here's the problem I'm seeing with everyone's "it's not 50/50" result - comparing females vs. males alive today is not the same as females vs. males births. Also, I'll cite my reference:
http://www.scientificamerican.com/article.cfm?id=why-is-life-expectancy-lo
The assumption is that it's current day, world ratio - obviously defining time or region will adjust the numbers, but the question is vague and we haven't been counting all births since humanity began.
(Note 1,0 is the same as 0,1)
Nope, it isn't. If you had addition information of whether or not it was the younger child that was born on Tuesday, then it would be eliminated
You got nearly the right answer, but for all the wrong reasons.
The Kruger Dunning explains most post on
Can you please create a flag for everyone that posted a wrong conclusion and then filter them out of all my views in the futures.
I wish to do this because it will eliminate people who don't read the articles and people who can't do math.
Thank you.
The Kruger Dunning explains most post on
The answer of course, is 42...
Well spotted... strange how so many people seem to think that was just goofy British humour and miss the important point about making sure you know what the question means before trying to calculate an answer.
In a survey of 100 programmers, 111111 thought that duck-typing was a good idea.
The output of the following is:
663152 out of 1377790 = 0.48131573026368313
663863 out of 1377323 = 0.4819951456557394
That's pretty close to 13/27 (0.481481...).
I even sliced it a couple of different ways.
public class TuesdaySon {
public static void main(String[] args) {
TuesdaySon ts = new TuesdaySon();
ts.go();
ts.go2();
}
public void go() {
int hits = 0;
int criteriaMet = 0;
int max = 10000000;
for (int i = 0; i < max; i++) {
Child c1 = new Child();
Child c2 = new Child();
if ((c1.isBoy && c1.dayOfWeek == 2) ||
(c2.isBoy && c2.dayOfWeek == 2)) {
criteriaMet++;
if (c1.isBoy && c2.isBoy) {
hits++;
}
}
}
System.out.println(hits + " out of " + criteriaMet + " = " + (double) hits / (double) criteriaMet);
}
public void go2() { // We have at least 1 boy // We have a boy born on Tuesday // The other child's a male
int hits = 0;
int criteriaMet = 0;
int max = 10000000;
for (int i = 0; i < max; i++) {
Child c1 = new Child();
Child c2 = new Child();
if (c1.isBoy) {
if (c1.dayOfWeek == 2) {
criteriaMet++;
if (c2.isBoy) {
hits++;
}
Tiller's Rule: Never use a word in written form that you've only heard and never read. You will end up looking foolish.
Bah. It's a 50:50 chance - we all know this. Clearly what this demonstrates is merely that the laws of mathematics are wrong!
Aide-toi, le Ciel t'aidera - Jeanne D'Arc.
Nope... it's still 50% The statement "I've tossed a coin twice, at least one of which was heads, what's the probability that both were heads?" is still 50%, given that one of them was heads. The false premise here is that there are 3 possibilities :
H H
H T
T H
when in fact there are 4 :
H h
h H
T h
h T
The lowercase h represents the one that's guaranteed heads. The false assumption here is that equal weight is given to H / h, when in fact it represents 2 possible choices - the result being discussed where "at least one of which was heads" could have been the first toss or second toss.
If the first toss was the heads being referred to, then the 2nd toss could be either heads or tails (h H or h T). If the second toss was the heads being referred to, then the first toss could be either heads or tails (H h or T h). Either way you look at it, it's still 50%
Global warming and other natural disasters are a direct effect of the shrinking number of pirates - Gospel of the FSM
I have two sisters. They each have a sister and a brother, but I don't.
I am glad you mentioned the existence of the alternative bayesian approach. I would like to argue in addition that both a) and b) need prior information, which are the demographics of the country in question.
Whatever interpretation of "probability" is used (frequentist or bayesian), a realistic answer has to depend on the specific prior information about both actual proportions of boys and girls in the specific country, and details on their weekday of births.
Assuming that the percentages are 50% boys, and that there are born equal numbers of boys per weekday is a non-problem for me - it's just an étude that does great disservice for the science of probability and statistics. I don't think health insurance works in real life by estimating the risks of, say, heart attack due to bad nutrition, starting with the premises of equal probabilities for males and females and equal probabilities for them eating healthy and junk food.
If I were asked to provide an answer, I'd look up the demographics and make a pure 'classical' frequentist estimate, based on the actual frequencies of American boys' births and their respective dates of birth. For a different county, the results are altogether different.
This reminds me of another riddle whose catch involves taking words too literally in exactly this way:
I hold in my hand precisely two coins of current US denomination whose total value is 55 cents. One of them is NOT a 50 cent piece. What are the respective values of the two coins?
(For those not in the US, the current US denominations of coin are 1, 5, 10, 25, 50, and 100 cents).
The answer is a 5 cent piece (a "nickel" as we say colloquially) and a 50 cent piece.
See, only ONE of them is not a 50 cent piece. The OTHER ONE is. The one that isn't a 50 cent piece is a 5 cent instead.
-Forrest Cameranesi, Geek of all Trades
"I am Sam. Sam I am. I do not like trolls, flames, or spam."
Comment removed based on user account deletion
So the point of this is to judge probability solely by the interpretive qualities of implied language? There's little value in this, and misdirecting. That's not a puzzle; it's a con.
Actually, I'm going to give a less glib answer now. Somebody tell me if it's wrong. The whole thing hinges on firstly, the other child being 50:50 boy or girl (we'll ignore twins, triplets, etc. and that society generally has a slight gender imbalance) and secondly, this business about the possible ordering of the children: (boy,girl),(boy,boy),(girl,boy). The thing is, that whilst the first probability is relevant, the second one is entirely contingent on the first and has no bearing of itself. The correct possibilities are:
Other child a boy (one sub-possibility: boy-boy, at 100%)
Other child a girl (two sub-possibilities: girl-boy and boy-girl, at 50%/50%)
So the probabilities for the girl-boy and the boy-girl are actually 25% each, adding up to 50%, so it's still 50:50 whether it's a boy or a girl. These people seem to have just taken three permutations and said that they have equal probabilities with no foundation for saying so.
Aide-toi, le Ciel t'aidera - Jeanne D'Arc.
You're absolutely correct IF you assume that the man is just describing one of his children. That is the logical interpretation of the problem. However, if someone went out and specifically selected a family with at least one boy...
I hate to be a smart ass on this one... but the question was posed by a man who was just describing (at least) one of his children.
When modding "Informative", please make sure it both has a source and IS actually informative.
Following assumptions/criteria:
Exactly 2 children.
At least one is a Boy with Blonde hair born on Tuesday
While obviously not even close to true, for the sake of probabilistic analysis, let's assume that possible hair colors are evenly distributed (1/4) between blonde, brown, red and black.
What is the probability of the 2nd child being a boy?
Answer is 55/111
111 total possibilities with at least one Blonde Boy born on Tuesday: Question is not what day of the week the other child is born or what hair color he/she has, therefore normally 4 possible answers: boy/boy, boy/girl, girl/boy, girl/girl. Additional information takes these possibilities times 7 (days of the week) and times 4 (hair colors). 4x4x7 = 112. Here, each possibility allows for EITHER the first or the 2nd child child to have any gender/day of week/hair color. However, with the restriction that one MUST be a Blonde Boy born on Tuesday we lose one possible combination (you cannot count Blonde Boy born on Tuesday for BOTH the first and the 2nd child - that's like rolling dice and counting "double 1" or "double 2" twice in the possible dice combinations). This leaves 111 possible combinations for the other child - of these 55 are boy ("1st child=Brown Boy on Thursday" OR "2nd child=Brown Boy on Thursday" OR "1st child=Black Boy born on Sunday" OR "2nd child=Black Boy born on Sunday" OR "another Blonde Boy born on Tuesday", etc) and 56 are girl.
Well this thread burned an otherwise boring hour and a half at work for me. Thanks everybody.
But, the question more properly is "I have just tossed a 10 pence coin twice and it came up heads once on a Tuesday, what is the probability that the coin toss will came up heads?" And you might also ask where did it land!
> While it's true that g/g is obviously eliminated, the boy MUST be either the younger or the older.
Can't RTFA, but does the problem explicitly rule out twins?
The article doesn't mention C-Section wasn't a possibility, therefore it is possible for both to be the same age.
A mathematician poses a problem about a father with two children, mentioning that one child is known to be of a certain age and born on a certain day, and asks what the probability is that the other child is the same gender.
The answer to this question (assuming 50:50 birth rates) is 50%.
This is because I'm accumulating probability over sets of questions & fathers & children without omitting any cases, much as the non-50% answer accumulates probability over sets of children and omits some cases.
Long version:
If you try to understand probability without trying to count over large sets of examples (i.e. statistics), you will almost surely make mistakes.
Let's step back to just the Two Children Problem and, I'm going to try to point out what I hope is a slightly deeper intuition about the two answers.
The naive answer of 1/2 just assumes the child gender is independent. Woohoo, rock on.
The "clever" answer imagines that we have thousands of fathers with two children. Some of those fathers are the fathers of two girls. The rest are the fathers of at least one boy. Only the latter fathers can pose this problem; therefore, if we randomly select a father from that set of fathers who can pose this problem, we have only a 1/3rd chance of drawing a father who is the father of two boys.
But we're linking the fathers to the problem in an overly specific way. Suppose we draw a father randomly from the set of all fathers with two children. That father may have two boys, two girls, or one of each. If we draw a father with at least one boy, we can proffer him to the reader and ask what the odds are his other child is a boy. If we draw a father with at least one girl, we can proffer him to the reader and ask what the odds are his other child is a girl.
So, if you actually had to do this--pick one person, ask the question, and you're done--what would you do? Well, it's dumb to draw a random father and then go "oh crap, we can't actually ask a question, this whole thing was a waste of time". So you'd probably ask the "at least one boy" question if they had two boys, the "at least one girl" question if they had two girls, or... one or the other question in the other case. Maybe you always ask the boys question. Maybe you randomly pick. If your tie-breaker is to just always ask "at least one boy" if it's possible, then the answer to "at least one boy" is 1/3 chance the other is a boy, but the answer to "at least one girl" is a 100% chance the other is a girl!
If you choose "fairly", then when you draw a father with both a boy and a girl, then half the time you ask "at least one boy" and half the time you ask "at least one girl".
If you choose that way, and happen to draw a father with the "at least one boy" question, then the probability the other child of his is a boy is 50%. (That is, given 100 fathers, the three cases are 2 girls: 25; 2 boys 25; 1 each 50; you split the 1 each case between the 2 questions, so you have 50 cases where you ask "at least one girl"--25 of those are they have 2 girls, 25 of those they have one of each; and 50 cases where you ask "at least one boy" with the same odds.)
Thus a complete analysis of all the cases under a more plausible situation for where you might be asking this question returns to the naive solution: the other child will be a boy 50% of the time.
So, as other people have said it depends on context. If the context is that it's a mathematician posing the question once about a hypothetical non-real situation, and the mathematician is sexist or otherwise favors mentioning boys, then maybe it's 1/3rd. But if he's not--if he chose the question randomly based on the gender choices available to him--then it's 50%.
Moreover, even if you choose a sexist bias for how you split the middle cases, if you measure over asking ALL questions, both "at least one boy" and "at least one girl", no matter how you split up the cases, the average odds that the other c
For this to work you're not allowed to know *which* coin is heads.
If you open one of your hands and look at one of them the probability function collapses back to 50%.
The trick is to get somebody else to look at them for you...
No sig today...
I'm just trying to show that the over-analysis was incomplete.
Truly, probability doesn't change even when you add a ton of irrelevant data. Having seven boys does not one bit affect the sex of the next child. If it doesn't affect the sex, then it doesn't change real probability.
Statistics is the art of crafting convincing lies.
Oops, sorry. "drew". English is not my first language.
This is long, but please read it for an ACTUAL EXPLANATION of what is going on.
-----
You are invited to a party by an old friend. When you show up, he opens the door and welcomes you.
-----
Hello and welcome to my home.
As we get the party started, I would like to play a little game that I think you will enjoy. All my guests today have played this game and I'd love for you to play too.
First I would like to introduce you to my family.
I have two children. I'd like to see if you can guess whether they are a boy or girl before you meet them.
Let us bring one out shall we? Have you made your guess? What do you think the odds are? I assure you, we're a perfectly normal family.
Here's Chris, my son. Did you guess right? The odds of being correct were 1 in 2, of course.
Next, I'd like you to meet my other child. Would you care to guess again? What are the odds this time?
Two children can of course be:
Boy Boy @ 1/4
Boy Girl @ 1/2
Girl Girl @ 1/4
Given that Chris is my son, what are the odds that my other child is a boy?
Simple intuition may suggest that the odds are 1 in 2. But let's try math, shall we?
We all know that P(Y|X) = P(XY)/P(X).
If we let
X = That I have at least 1 boy
Y = That both are boys
Then
P(X) = 3/4
P(XY) = P(Y) = 1/4
P(Y|X) = 1/4 / 3/4 = 1/3
Aha! We see that the odds are indeed 1 in 3 that my other child is also a boy. This also works out intuitively. I assure you I brought out my children in a random order. Given that one of my two children is a boy, we know that there are 3 possible outcomes, and all are equally likely. Only one of those outcomes is that both are boys, and as such, the odds are 1 in 3!
Now, as you make your guess shall we put a wager on the outcome?
If you wager $100 that my other child is a girl, I'll pay you $60 if you are right.
Given the odds, you're expected winnings are (2/3 * 60) - (1/3 * 100) = $7.66 .
You can of course, take the other bet, that my other child is a boy.
If you wager $100 that my other child is a boy, I'll pay you $200 if you are right.
Given the odds, you're expected winnings are (1/3 * 200) - (2/3 * 100) = $0.00 .
Would you care to make the wager? Which one?
Ah, here's my wife now with my other child, Alex.
As you can see, she's due any day now. Then we'll know the result of our wager!
-----
Your host thinks he's pretty clever. After you agree to take the first wager, he reveals that all the other guests have done the same thing, and that he stands to make quite a decent amount of money off of tonight's party, since he has invited 100 guests.
He's trying to trick us. We're guests at his party, and he's giving us two wagers. With the odds he suggests, the smart guests will take the first wager, that his other child is a girl, because it has a higher expected payout.
When he reveals that his wife is still pregnant, and that the sex is unknown, our intuition kicks in again. Strongly. You KNOW that there is a 50/50 chance on the sex of that child. Thus, the odds he gave you in the wager are skewed. The the first wager is skewed in his favor (you expect to lose $20), while the second wager is skewed in your favor (you expect to gain $50).
The man is offering the wager to all the guests at the party, which we now know to be a baby shower.
But where did the reasoning he offered, and that you agreed with, go wrong? The fact that the sex of the unborn child is not known makes no difference, as the man assured us he brought forth his children randomly. Indeed, he could have brought out his pregnant wife and asked you to wager if his other child would be of the same or opposite gender. He could have walked through the same reasoning, stating "If this child is born a boy, then..." and "If this child is born a girl, then...", with both cases yielding the same 1 in 3 outcome as he presented to you initially.
The problem here is that there are not four differ
You are making the assumption that the probability of having a boy is the same of having a girl. This is not true.
But, assuming it is you still have a problem with identical twins. You can break the set of all families into two sets, One those without identical twins and another those with identical twins.
Your tables are valid for the set of families who do not have identical twins. The probabilities for them is granted to be equal for each instance of your set of possibilities.
With the case identical twins there are seven groups of BB and GG, only one of which occurs on a tuesday. So you have a p = 1/2 for those families with identical twins.
You then need to modify your total result to include both probability cases based on the probability of identical twins occurring in your set of families.
Then you will need to adjust things for the case of identical twins having different birthdays because they were born around midnight.
Then you need to adjust these results to take into account the probabilities of births which are performed by inducing labor or performed by surgery. This will alter the assumption that births occur with equal frequency on each day of the week.
And, we are of course assuming the children being mentioned are the currently living children. It would be common to not mention those who have died. The probabilities for deaths of children are probably not evenly distributed for each sex, and probably vary according to age.
You then need to apply the distribution of these death probabilities by sex and age to the distribution of possible child ages for the set of families with non identical twins.
And, we haven't even mentioned transgendered children.
You are wrong.
The chance is 50%, regardless.
Look at the non-day-of-week version.
A person with two children has either:
B B
B G
G B
G G
Given that they have at least one boy, what are the odds of both of them being boys? 1 in 3.
This makes perfect sense, and works out mathematically.
P(AB) / P(B) = P(A|B)
1/4 / 3/4 = 1/3
The trick is that revealing a child is not the same as stating a given condition. And no, it has nothing to do with order or whether or not the reveal is random.
In a reveal situation:
B B 1 (both boys, revealed boy 1)
B B 2
B G 1
B G 2
G B 1
G B 2
G G 1
G G 2
The question (without the day-of-week stipulation) is asking: Given that I have 2 children, and that I have shown you a boy, what are the odds that my other child is a boy (and thus both children are boys)?
Count em up. Do the math.
2 children, 1 reveal = 8 states, listed above.
4/8 states = showing a boy.
2/8 states = both children are boys.
Count. 2 out of 4 states satisfy the rules.
Try math.
A = Both boys. 2 out of 8 states
B = Shown a boy. 4 out of 8 states
P(AB) = P(A) (If A is true, B MUST be true). 2/8
P(AB) / P(B) = P(A|B)
2/8 / 4/8 = 2/4 = 1/2
Right, it's 14/28 ((7 days * 2 kids * 1 sex) / (7 days * 2 kids * 2 sexes ) or 1/2), but because boy 1 on tuesday and boy 2 on tuesday is counted twice in this instance, you lose one from both numerator and denominator.
"If a nation expects to be ignorant and free in a state of civilization, it expects what never was and never will be."
that word is a horror motif
you can be born/ not born, which is the proper schrodinger scenario you are referring to
but its some sort of cthulhu ritual to contemplate the process of being UNborn
i guess its like being undead: not alive, but still moving
unborn: you never were born, but you still exist! MUAHAHAHAHAHAHA
intellectual property law is philosophically incoherent. it is your moral duty to ignore it or sabotage it
The fact that someone told someone else the outcome of a random event has no effect on probability of other independent random events -- even if they are similar.
Contrary to the popular belief, there indeed is no God.
Well, I do not agree that boy - girl and girl - boy are not equivalent.
Rephrased without double negatives: I think there are only 2 groups, boy - boy and boy-girl. Importance of ordering is not at all discussed in the question, leaving us with only 2 sets.
Sorry, but I am not a lawyer - I interpret the question in the most simple way possible. Maybe this is where I am supposedly "wrong", but I do not care :P
I have nothing to lose but my bindings.
> What we can deduce from the wording is that his other child is not a son born on a tuesday.
No we can not - nowhere it is shown, in the wording of the question, to be the case.
In fact, NOTHING is mentioned, or is known about the other child.
http://slashdot.org/comments.pl?sid=1701394&cid=32729366
http://slashdot.org/comments.pl?sid=1701394&cid=32729694
http://slashdot.org/comments.pl?sid=1701394&cid=32729826
I have nothing to lose but my bindings.
This is what I was trying to impart to deluded ones, it is so painful to watch people jumping off the cliff in lemming-like droves.
I wrote a fair share of well - reasoned comments, just because this article struck a deep nerve in me - about certain kind of people who always seem to conveniently (to themselves) misinterpret things.
Well, guess what, 1/2 is correct in all closed-universe cases with 1/2 gender normality and 2 genders.
PERIOD.
Glad to see I am not the only one not drawn in by this baroque nonsense :P
I have nothing to lose but my bindings.
No, that's my interpretation of the question too. The article describes how the originator didn't think of the straightforward interpretation, or didn't realize it implied different probabilities.
The order of the children doesn't come into it though - it's purely a matter of selection bias.
You are correct. Both the article and the slashdot posting are somewhat misleading (almost intentionally).
I'd expect a disproportionately low number of slashdotters to be fooled by this kind of nonsense.
This is akin to saying that a house could fall on my head today or not. Since there are only two possibilities, there's a 50/50 chance that a house is going to fall on my head today.
That's clearly intuitively wrong, and it's mathematically wrong as well. Similar math teasers (like proving that 1 = 2) have existed for eons. This plays on a common probabilistic mistake, taking the likelihood of some compound situation (two boys) and applying those odds to a situation in which you already have some information determined.
The child you met on the street (boy born on Tuesday) is a coin flip that you've already seen (heads). The child you haven't met is a coin that you're going to flip in the future. The coin flip is 50/50.
Read my other comments on this thread, where I try to expose these faux-mathematicians for the lawyers that they are -- as per normal, common sense wording (english and all) of the question, 1/2 is always the answer, because the other child is absolutely non-dependent on existing given constraints of the question. I suspect these particular mathematicians, lawyery as they are, seem to not really grasp proper use of language operators. I suspect they would make bad programmers. Mind you, I met some nice mathematicians who share no such flaws, hence the partial defense on my part :P
I have nothing to lose but my bindings.
...lies, damned lies, and statistics.
You failed to see that the independent events are conditionally dependent on having a particular outcome. Since you like the dice example:
Say I have a cup with 2 fair 6 sided dice. I roll them and turn over the cup so you can't see them. I peek under the cup, look at both dice, and without changing either die I slide out one of the dice to reveal that is a 6. Do you think the odds of the other dice being a 6 is still 1/6? No, it's 1/11 and here's the breakdown:
Odds of not rolling any 6s: 25/36 (I can't reveal a 6, so these cases are impossible as I showed you at least one 6)
Odds of rolling exactly 1 6: 10/36 (I reveal the 6 and the other die is not a 6)
Odds of rolling exactly 2 6s: 1/36 (I reveal either 6 and the other die is a 6)
It's like a mini Monty Hall problem. The key is that I have advanced knowledge of the outcome all the events and selectively reveal information. The only way the die I revealed is a 6 and the in the cup is a 6 is if I rolled double 6s to begin with (odds 1/36), and by revealing one of the a dice as a 6 I eliminate 25 of the possible cases to make the remaining odds of double 6s 1/11, which is still worse than 1/6 on an independent roll.
Heya :P
Boy this struck a nerve with me - you can read my comments in this story, by far my most prolific so far, to see my attempts to show some light to lemmingy people who assume that non-existing language opperators in the given question are the same as existing. Well, there is no "ONLY one of two kids". I don't see it.
The answer is always 1/2, because nowhere in question, if you read as per normal rules of grammar, is there any indication that the second child's gender depends on ANY outside parameters given in the question.
Thus there are only 2 possible solution sets -- boy boy and boy girl, which gives answer to the question and exact probability of 1/2, if we don't take into account random gender variations and possibility of more genders than 2.
And a cat :P
I have nothing to lose but my bindings.
You gave a good summary of the article's logic, but TFA's author is wrong to exclude the 7th younger boy possibility. The author assumes that any case where two boys are born on Tuesday is identical, and so duplicates should be removed; this is not true.
It's simply not the case that there are duplicates. There is just one event there are two boys who are born on a Tuesday.
<eldest boy born on Tuesday, youngest boy born on Tuesday>
There are two events where there are two boys and one is born on a Monday and the other is born on a Tuesday.
<eldest boy born on Monday, youngest boy born on Tuesday>
<eldest boy born on Tuesday, youngest boy born on Monday>
Your confusion arises from introducing a new random variable (whether you've met X or not) and enumerating over this, but only for the both-born-on-a-Tuesday event. You would need to do this for all the other events as well; this would have the effect of needlessly multiplying events but wouldn't change the relative frequencies of the original events.
No.
Extended set is
bb
bg
gb
bb
BB is two times because you do not know which is the first mentioned boy, as with the girl case. Do you see the light now ? gb bg constricts to bg for stat purposes, and bb bb constrict to bb for stat purposes. Tuesday, as per spoken language of kings, Yon English, is irrelevant.
1/2.
I have nothing to lose but my bindings.
Here's the way to look at it:
Suppose you took all of the people in the world who have 2 children and started tallying the genders of the older child and the younger child. You would get these results, in roughly equal proportions:
1) Older=boy, younger=boy
2) Older=boy, younger=girl
3) Older=girl, younger=boy
4) Older=girl, younger=girl
1/4 of these families have two boys, 1/4 have two girls, 1/4 of them have an older boy and a younger girl, and 1/4 of them have an older girl and a younger boy. So 1/2 of families with two children have one boy and one girl. This, I think, makes sense to everyone. If you don't specify that at least one of the children is a boy, then the chance of them having 2 boys is 1/4.
If you do specify that one of the children is a boy, then you can eliminate the girl-girl scenario. So then you're looking at the following possible outcomes:
1) Older=boy, younger=boy
2) Older=boy, younger=girl
3) Older=girl, younger=boy
So really, instead of having only two possibilities (boy-girl or boy-boy), there are 3 possibilities---boy-girl, girl-boy, and boy-boy. Each of them being equally likely. So the chances of them having 2 boys is actually 1/3.
The reason everyone thinks that the answer is 1/2 is because they think that the two children are distinguishable. They're not, in the problem given; you're not stipulating that Child A is a boy, you're saying that at least one of Child A or B is a boy. If you distinguish the children by saying something like, "My older child is a boy" then the gender of the younger child becomes independent, and is exactly 1/2.
That's what makes the Tuesday birthday problem so interesting. Usually, if we want to distinguish between the children, we use their ages, because this gives them perfect distinguishability. (You don't have 2 oldest children, for instance.) By saying that one of them is a boy born on Tuesday, you're giving that child more specific attributes, also making them more distinguishable.
If you say, "I have two children, one of whom is a boy named Bartholomew, who won the lottery in 2009, and was born on the day the Berlin Wall came down," then you've made an almost perfect distinction between the children, and the chances of the other child being a boy are very, very close to 0.5.
I don't believe in time. It's a grand conspiracy designed to sell watches.
You are incorrect. This isn't semantics, or improper reasoning. If you actually look at real two children families with at least one boy, only one third (or something very close) of them are two boy households. That is simply fact, not one possible mathematical interpretation. I shall explain below, further below I will show you one situation where your reasoning does work. Hopefully you will be able to see the difference.
Of all two children households, their oldest is either a boy or a girl, with 50% chance (simplifying slightly). Their youngest is a boy of a girl, with 50% chance. Therefor of all 2 children households:
50% have one boy and one girl
25% have two girls
25% have two boys
But, we are removing the families with two girls, because we are only looking at households with at least one boy. Therefor:
2/3 have one boy and one girl
1/3 have two boys
So you see, given our set the odds of "the other child being a boy" or, put better, the odds that the household is a two boy household, is 1/3. This isn't semantics, this is how actual household data would be distributed in reality.
Now to clarify why we get this unintuitive result I will give another situation, one we would encounter day-to-day. I meet my neighbor Scott. Scott has his son with him. I say, "Hey I knew you had two kids but I didn't know one of them was a boy!". Scott says, "Yes, indeed, now to test if you have a real understanding of statistics I have a question for you: what are the odds my other kid is a boy?"
I know Scott's household is a two family household, and I know that the two girls option is off the list. So once again we are at this statistical point:
2/3 chance of one boy and one girl
1/3 chance of have two boys
But this time I can't stop there. Because I have one further piece of information, he has one of his kids with him and that is how I know he is a boy. One kid has been selected. Now, I'm going to assume his selection was more or less random in regards to gender. That is to say, I am assuming Scott doesn't prefer to hang out with boy children more than girl children, and that he selected which kid was with him with 50/50 randomness. So our new breakdown is thus:
1/3 chance of one boy and one girl, with the boy being introduced
1/3 chance of one boy and one girl, with the girl being introduced
1/3 chance of have two boys, with one of the boys being introduced
Now, we know the theoretical girl was not introduced, so we can scratch the second option of the list, it isn't part of our set. After doing that, you can see that the more intuitive answer of 50% has been reached. But don't confuse this example with the one above. In the original, there would be factual evidence that of two kid households with at least one boy, only 1/3 will have two boys.
Bends the mind around a little, doesn't it? A seemingly statistically insignificant fact like circumstance actually affects the hard numerical realities.
I'm glad somebody brought up the biological angle, although it can be more nuanced than the AC states. First, the ratio of live male births to live female births is (at least in the industrialized world) 1.05:1.00, although studies report there is a long term falling trend on male births, such that 1:1 appears to be coming.
That said, since many men (most, perhaps?) have a predilection toward fathering either boys or girls, the fact that one of the children is known to be a boy may influence the likelihood away from 50/50. I knew an anecdote is not a statistic, but enough of them can look a lot like a statistic, so here are a few:
-My dad fathered three sons, no daughters.
-I fathered four daughters, no sons.
-My next door neighbors has three sons, no daughters.
-The majority of the kids my daughters go to school with come from families where the siblings are predominantly one gender, and in many of them, the siblings are all of the same gender.
So, while I'm not a mathematician and I may be talking out my arse statistically, it seems to me that knowing one of the children in question is a boy raises the chance that the other is also a boy to some (possibly indeterminable) number well above fifty percent.
FFS, the probability of both children being boys is 1/2.
I have nothing to lose but my bindings.
And that is 1/2.
Fboy boy 1/4
boy Fboy 1/4
Fboy girl 1/4
girl Fboy 1/4
Group to Fboy boy 1/2 and Fboy girl 1/2.
1/2.
I have nothing to lose but my bindings.
A helpful analogy from the comments is:
-You come across a guy on the street who happens to have one of his two children with him, and it's a boy. That has no impact whatsoever on the gender of his other child, it's still 50/50.
-You call a guy up on the phone (he has 2 kids) and ask "Is either one of your kids a boy?" He answers yes. What does that mean for the gender of the other child?
Firstborn|Secondborn
Boy Boy (possible, the other child is a boy)
Boy Girl (possible, the other child is a girl)
Girl Boy (possible, the other child is a girl)
Girl Girl (Impossible)
So in 2/3 cases the other child is a girl.
Another insight from the comments:
The tuesday thing is a constraint on the boy. It's unlikely that any given boy is born on a tuesday. So a boy-boy pair is much more likely to have a tuesday-born boy than a less boy-populated boy-girl pair.
Again though, it's a question of whether you happen to come across a guy on the street with a random one of his boys, or if you specifically ask him if any member of his kids is a tuesday-born boy.
Having the person volunteering this information, as if "tuesday born boyness" is a desirable quality, and if none of his children had it he wouldn't be bragging about it, makes it sort of fall into the second category.
Wrong.
How did this wall of babbling nonsense earn an Interesting mod point?
And the are 250 Tboy boy AND 250 boy Tboy pairs.
Also 250 Tboy girl and 250 girl Tboy pairs.
The question a priori rules out there being possible a pair of girl girl.
Gees, now I talk like a lawyer.
Really, it is easy. There are others. Join us, live a meaningful life.
Before applying a solution, first, read the problem. Then, construct the solution.
1/2, yo :P
I have nothing to lose but my bindings.
Tboy boy 1/4
boy Tboy 1/4
Tboy girl 1/4
girl Tboy 1/4
Relevant added probabilities Tboy boy 1/2 Tboy girl 1/2.
1/2.
Tuesday - irrelevant pseudo logical garbage, shoving contempt for language operators.
Really, this is one of those rare cases where the article is somehow, excruciatingly, painfully wrong. This thread made my ears steam.
I have nothing to lose but my bindings.
If you already have an H, TH is not longer an option for you. You're left with HH and HT.
Don't take life so seriously. No one makes it out alive.
Age doesn't matter to the question. You either have a boy and a girl or a boy and a boy. In this case, BG and GB are the same outcome.
Don't take life so seriously. No one makes it out alive.
what does this have to do with the task at hand ?
girl girl pairing is ruled out by the stating of the problem itself. You keep forgetting this, and add gg prior to considering the true scope of the problem, which has no gg. This is what I think, and I think it correct. Seemingly, I am not the only one in this thread. The truth is, you cannot assign mothers here, it is inconvenient and irrelevant to the ACTUAL problem.
Again, you keep forgetting that one of the boys, Tboy (Tuesday-born boy), is unique, and persistent.
There are 4 groups, as per problem:
Tboy boy
boy Tboy
Tboy girl
girl Tboy
each 1/4 prob which adds up to 1/2 for Tboy boy (ordering is irrelevant, only the sum matters). And Tuesday, is too, not relevant. Except as a convenient naming measure for the unique boy, Tboy.
I have nothing to lose but my bindings.
Two children can of course be:
Boy Boy @ 1/4
Boy Girl @ 1/2
Girl Girl @ 1/4
This is where your first mistake is. We already met one boy.
And here is where mathematics and the real world depart. If I roll a die 99 times, and it comes up '6' all 99 times, I would no longer assume that the die is fair and would expect the die to come up with '6' almost surely the next time.
The dad of B2B2 is also twice as rare as B1B2 for example. That takes it back to 13/27.
>>The real confusion occurs due to the use of odd numbers...
Lol at this from the gp.
This problem has not been stated sufficiently clearly in the summary:
"I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?"
The way I read it, the answer is 50%, if we assume that the distribution of sexes is even - with the first sentence only there to mislead.
The problem should have been more clearly stated - for example:
"If I have two children, what is the probability that both are boys if one is a boy born on a Tuesday?"
The original question does not tie the first sentence into the probability question correctly. You might have well asked
"I have two children, one of whom is a boy with red hair. What's the probability that my other child is a boy?"
I am not generally picky about grammar, but in this case, where the question is obviously designed to be some sort of "trick question", the submitter has to be a much more careful in how the question is stated.
I like to look at it as coins. Take a 10 cent coin and a 5 cent coin with distinct heads or tails. There are the following possible combinations:
10 | 5
H | H
H | T
T | H
T | T
They are very distinct in the sense that if we toss the coins, we have a 25% chance of any combination appearing. If we're told that in a given toss, there was one that was heads and are now asked what is the probability of the other being heads, it is one in 3 since we've eliminated the third option.
The point is, the boy/girl combinations really are distinct and unique outcomes that do factor into the overall probability.
Strictly parsed, you are correct.
However, when making a statement to a friend or coworker, it is generally considered dishonest to be ambiguous. In normal context, the statement in the original question would imply that the other child is not a boy born on a Tuesday. If you were to insist on your interpretation, you may be considered pedantic, picky, or impolite. Only lawyers and logicians would respect you.
You wouldn't ask "do you have a green car" you would ask "do you have any green cars?"
This is related to the phenomenon of why the phrase and/or exists. Most English speakers hear "or" and assume XOR, not logical or. As in, do you want to go to the movies or dinner? The polite answer is "movies" or "dinner", not "yes" or "no".
Bottom line: the correct interpretation of a math problem is not necessarily the same as the idiomatic interpretation.
I forgot an important punctuation mark. I should have stated "Do you want to go to the movies, or dinner?" the pause emphasizes that it is a choice between two mutually exclusive options. When stated as a run-on, "yes" would be a fine answer.
It's wrong. Suppose that you randomly selected 100 families with two children. We're ignoring all the things that you mention above and assuming that gender distribution 50:50. Statistically speaking, our 100 families would consist of 25 boy/boy, 25 boy/girl, 25 girl/boy and 25 girl/girl. We assign a number from 1 to 100 to each family, write that number on a series of cards, and you randomly select a card and hand it to me. I look at it, see the number, and announce that the family has at least one boy. What are the odds that they have two boys? What are the odds that they have a girl and a boy? There are 25 girl/girl cards, but you know you did not pick one of them, so they can be ignored. So you know you picked one of 75 cards. Of those 75 cards, 25 are boy/boy, 25 are boy/girl and 25 are girl/boy. So there is a 25/75, or 1 in 3 chance that you picked a boy/boy card. There's a 50/75 or 2 in 3 chance that you picked a card with a boy and girl combination.
"The legitimate powers of government extend only to such acts as are injurious to others." Thomas Jefferson.
funny how there's controversy over whether "one" means "at least one" or "exactly one," but "two children" means "exactly two children." to do this properly, you need to know how likely it is that the person has three children, four children, five children, six children,..., 6.7 billion children. after all, this person may consider the all humans his/her children.
Nope. Wrong. Age does matter. Not because it correlates with gender or anything, but because it makes the two children distinguishable. (E.g., one of them must be the older child, one must be the younger.)
You don't have to distinguish them with ages. You could distinguish them with height, e.g., "the taller child is a boy" or anything else.
Probability in this case is about how much information you have. The more information you have, the more you can refine your results. If I know that a parent has two children, then I can give the probability of both children being boys depending on how much information I have:
no further information -> .25 .333 .5
one child is a boy (but we don't know which one) ->
one child is a boy (and we do know which one) ->
one child is a boy, and so is the other child -> 1.0
one child is not a boy -> 0.0
TFA does a pretty good job of explaining why people have such a hard time understanding this. The reason is that they assume that the parent is picking a child at random and then stating their gender. That's not the process. If they choose one of their children at random and she happens to be a girl, they skip over her and then state the gender of the other child.
I don't believe in time. It's a grand conspiracy designed to sell watches.
This is similar example of the gamblers fallacy. This is when you have a really random event such as gender of a child. Knowledge you have of the past or existing events have no effect on future or unknown events. So a balanced die will come up 1 1/6th of the time. If you roll it and 1 comes up 5 times in a row it doesn't mean anything for the next roll the probability of the next roll coming up 1 is 1/6. Now if before you started you asked what are the odds that the die will come up 1 6 times in a row it is 1/46656. The main thing is asking if the event is random.
I know this because I worked as a manufacturing engineer and used statistical process control. The basis for this is that you don't have to measure every part if your process is in control and all variations are random. You can then randomly select parts and measure them to prove the rest of the parts are good with a high level (6 sigma) of probability. The critical assumption is that the process is in control and there are tests that are used to determine if you are in control (The variations are random).
I love Jesus, except for his foreign policy.
This is why we don't let English majors write statistics texts.
I have two children. One is retarded and creates mathematical puzzles based on ambiguously loosely worded assumptions. What is the probability that the second child is goatse?
The problem with that is that the distribution of sexes is not 50-50 to begin with.
There's an article at wikipedia about the simpler problem (no Tuesday confusion). http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
What is most annoying in all of these discussions is the (seemingly automatic) passing over of the possibility of identical twins.
Stephan
http://stephan.sugarmotor.org
Sorry, my bad. I wasn't aware that names of days were unique throughout the entire history of the world.
Confucius say, "Find worm in apple - bad. Find half a worm - worse."
According to your distribution list, the odds that a family would have a boy born on Tuesday, and another boy born on Tuesday are 1/196 while the odds that a family would have a boy born on Monday, and another boy born on Wednesday are 2/196. If the birth date of the two children were independent the odds would be the same. So your assumption that P=1/196 for each outcome may be flawed.
As another poster pointed out, your list actually has 28 boys born on Tuesday and 14 of them have a brother. So if the gender and birth date are independent the odds are 14/28=1/2.
in case mentioned boy is older:
B, B p = 1/3
B, G p = 2/3
in case mentioned boy is younger:
B, B p = 1/3
G, B p = 2/3
Why is it in both cases 1/3 vs. 2/3, and not, as you claim, 1/2. You have to look at the probability distribution of the underlying set. It is:
B, B = 1/4
B, G = 1/4
G, B = 1/4
G, G = 1/4
So all have the same likelyhood. In our breakdown into the two paths "younger and older" above notice how "B, B" is mentioned in both of them. Since "B, B" will occur on average just as often as "B, G" or "G, B", all the "B, B" cases will be split between the "boy younger" and "boy older" path, and it's probability of occurring in either path is thus halved when compared to it's alternative
When all Probabilitys are added up, the end result for the probability that you get B, B is the same.
p = 1/2 * p[path 1] + 1/2 * p[path 2]
p = 1/2 * 1/3 + 1/2 * 1/3 = 1/3
Incorrect! TH is only not an option if the *first* one is necessarily H. If you flip both coins and say "one of them is heads", you're not specifying which is heads.
(On the other hand, if you flipped two coins, picked one arbitrarily, and said whether it was heads or tails, it wouldn't influence the probability of the other.)
At this point they should know the sex of all there children!
Then the question is incorrectly phrased in two ways. Firstly, it omits the fact that there are 100 families all with two children only and that the parent concerned is a parent in one of these families. Secondly, the question is about whether or not one has guessed correctly, not what the child actually is.
Aide-toi, le Ciel t'aidera - Jeanne D'Arc.
The problem with that is that the distribution of sexes is not 50-50 to begin with.
Parent specifically included the assumption that "the distribution of sexes is even," so yes it is 50-50 by definition. Of course the distribution based on the data of actual births might not be, but to point that out would be impertinent.
Better to be despised for too anxious apprehensions, than ruined by too confident a security. --Edmund Burke
Can't RTFA,
Yes you can. I just posted it as your GP.
In other words, there will be a 0.2% bias towards a second boy, as opposed to what you would expect without the "twin effect".
Devlin started by listing the children's sexes in the order of their birth:
Boy, girl
Boy, boy
Girl, boy
This is incomplete. The identified boy is marked in capitals:
... the relative ages of the children does not matter
BOY, girl
BOY, boy
girl, BOY
boy, BOY
You're right, the relative ages don't matter and that's why Devlin's list is complete, and we need not distinguish between BOY, boy and boy, BOY.
Assuming male/female births is a coin-toss, any mother having a child has a 50% chance of having a boy, and a 50% chance of having a girl, for both the first and second children. Which gives us 4 possible combinations of 1st and 2nd children: (girl, girl); (girl, boy); (boy, boy) and (boy, girl). Since we are told that this is not a (girl, girl) family, this leaves the 3 possibilities listed by Devlin, and with 2 chances in 3, that the other sibling is a girl.
Better to be despised for too anxious apprehensions, than ruined by too confident a security. --Edmund Burke
the boy MUST be either the younger or the older. Since he's either one or the other, we can infer that either b/g or b/g is also impossible.
No you can't, because you don't have that information. He can be either the older or the younger, or they can even be the same age. As long as you're not talking about a specifically identified boy, the chance is 1/3.
If he's older, then g/b is not possible, and if he's younger, then b/g is not possible.
But because we don't know, they're still both possible.
The odds of having an Fboy *at all* are higher if there are two boys than if there are a boy and a girl. That's the clue of the problem: you cannot ignore the a-priori probabilities in that case.
Excellent explanation! Especially the last bit.
The question is incompletely phrased, which is entirely what TFA is about. Read it.
The simple point is this:
If you have selected a child, independent of gender, and that child happens to be a boy, then the gender of the other child is independent of this, and therefore has 50% change of being a boy.
If, on the other hand, you select a child specifically for being a boy and part of a 2-child family, then the gender of the other child is not independent.
In fact, it's even more complicated than that: do you select a 2-child families that has at least one boy, then 1/3 of those families will have 2 boys.
But if you take all 2-child families, and randomly pick a boy from all those children, then there's 50% change he's from a 2-boy family. Because they have twice as many boys, and therefore a bigger chance of being selected.
The famous 2-buy problem is about the middle option. TFA discusses the difference between the first and the second option, neglecting the third option. But a lot of people here seem to be focusing on the first and the third, and don't want to admit the legitimacy of the second option.
The gambler's fallacy is irrelevant. The dice have already been rolled. What matters is how you select the dice that you get to see.
This may be one of the few problems where 42 is not the right answer.
Bah and peh ! :P I can and I will.
Cos I am a pirate :P
I have nothing to lose but my bindings.
Yes, this finally makes sense to me and I now understand where my analysis differs from that in the article. Still though, I do not believe their analysis. They say there are four possibilities, bb, bg, gb, gg, then you drop the last, etc.
So here is the second thing that needs clearing up, if order does not matter then why are bg and gb different outcomes? If order does matter arent the outcomes bb, bb, bg, gb, gg, gg... which would still lead to a 50% chance its a boy. The whole saying order matters to generate the outcomes then later saying order doesn't matter to get rid of the other bb just doesn't sit well with me.
so... stupid people are less likely to have boys then?
So here is the second thing that needs clearing up, if order does not matter then why are bg and gb different outcomes? If order does matter arent the outcomes bb, bb, bg, gb, gg, gg... which would still lead to a 50% chance its a boy.
Just look at big numbers. If there are hundred families with two kids, 25% of those will have two boys, 50% will have a boy and a girl, 25% will have two girls. There's no need to try to second guess that.
If you have trouble dealing with that, then you need an introductory course on probabilities and statistics. Highschool level.
Yes, I discovered this distinction somewhere else in this discussion. It matters whether you select a family first, before randomly selecting a boy from that family (leads to a chance of 1/3), compared to directly selecting a boy from a variety of families that have at least one boy (leads to a chance of 1/2, because families with 2 boys are twice as likely to be selected).
The way the problem is presented, however, I believe the family is selected first.
Don't forget to read up on the Monty Hall problem. It's very much related to what you're saying.
It should be:
If I had two children and if one of the children was both a boy and born on a Tuesday then what is the probability that the other child is also a boy.
Any chance the original question was a translation from another language?
correct, it is just about english assumptions in the article.
had he stated
"I have two children, only one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?"
Then the problem would have been obvious. The whole article assumes that the word "only" is optional in that sentence with the same meaning with or without it. Then confuses the issue by then assuming "I have two children, only/b> one of whom is a boy. This time omitting this clarification he can assume that because the solution would be obviously One boy, one Girl that what we must really mean is that instead of assuming "only" we are assuming "at least" instead.
Birth order matters.
...
...
You could also have:
Boy on Tuesday, Girl on Monday
Boy on Tuesday, Girl on Tuesday
Boy on Tuesday, Boy on Monday
SKIP Boy on Tuesday, Boy on Tuesday
Boy on Tuesday, Boy on Wednesday
Giving 13/27.
--
JimFive
Please stop using the word theory when you mean hypothesis.
You're an idiot.
That's the overall case for two children.
Meeting the first one removes the girl girl possibility, as the party host explains (and erroneously leads you down the 1/3 reasoning).
Actually Ive taken at least 4 like that, it never sticks though. Heres what the probability tree looks like:
.5 B <
.5 G <
.5 B
.5 G
<
.5 B
.5 G
Its obvious from that why 25% will have two boys, etc. But If you already know one is a boy, regardless of order, its 1 in two chance the other child is a boy as well...
B/G and G/B are both possible, but mutually exclusive. If the boy is older, then G/B is not possible. You know that because he's a boy you cannot have G/G. You don't know his age, or the age of his sibling in question, but that data is not undetermined, it's merely undiscovered.
Listing their sexes in order of their birth:
B/G
B/B
G/B
Since you're qualifying order of birth as part of this calculation, you must take into consideration whether Tuesday is the younger or the older. You don't know which one he is, that data is undiscovered, but it's not undetermined. He is either older, or younger, and so you can KNOW that G/B and B/G are mutually exclusive to the order of birth, having only one boy that is Tuesday.
Another way to express it, since you're talking about order of birth, is to actually define who Tuesday is in the sequence. Obviously Tuesday would be the boy in the B/G and G/B sequence, but who is he in the B/B sequence? He could be either. Since the question is who is the OTHER, to define the other so clearly in B/G and G/B but not define who the other one is in the B/B leaves the whole thing lacking that detail: is he older or younger than the other boy?
T/G
T/B
B/T
T/G
So you have 2 chances for the other being a boy and 2 chances for the other being a girl.
Where T is the Tuesday boy:
G/G - impossible
T/G -
Okay, so then I re-flip and a freak meteor falls out of the sky and vaporizes the coin before it hits the ground. It also vaporized you.
Big apple, new Yorik, undig it, something's unrotting in Edenmark.
#!/usr/bin/env python
: :
: : ...
#If you think the "statistics" is a lie, or the problem is
#being "over-analysed", why not actually try doing it and
#see what you get.
import sys, random
SIZE=100000
def population (size)
"Generate a population of random kid pairs"
def birth () : return random.choice(('girl','boy'))
while size
size -= 149898
yield (birth(), birth())
counter = {'girl' : 0, 'boy' : 0}
for pair in population(SIZE)
if 'boy' in pair
#one of the kids is a boy
counter[pair[not pair.index('boy')]] += 1
#the other is
sys.stdout.write('girls: %(girl)d\nboys: %(boy)d\n' % counter)
sys.stdout.write('ratio of boys: %f\n' % (
float(counter['boy']) / (counter['girl'] + counter['boy'])))
#example output
#girls: 49898
#boys: 24975
#ratio of boys: 0.333565 #<--Please explain this result
Better to be despised for too anxious apprehensions, than ruined by too confident a security. --Edmund Burke
The line "size -= 149898" should in populate(), of course, read "size -= 1"
Better to be despised for too anxious apprehensions, than ruined by too confident a security. --Edmund Burke
Its obvious from that why 25% will have two boys, etc. But If you already know one is a boy, regardless of order, its 1 in two chance the other child is a boy as well...
Only if you know which one is the boy. If you don't, then there's two options that have a boy and a girl, and one that has two boys, therefore 1/3 chance that the other is a boy.
I've handled this issue in several other comments. Read this one, for example.
The probability of any kid being a boy is 0.5. The gender and age of any of your other kids is irrelevant. You can keep playing number and word games if you like, but the question is "What's the probability that my other child is a boy?" It's 0.5.
"TFA does a pretty good job of explaining why people have such a hard time understanding this."
Yes, he admits that people have a hard time because it's bullshit: "The difficulty of these problems is rooted in their artificiality: In real life, we almost always know why the information was selected, whereas these problems have been devised to eliminate that knowledge."
Don't take life so seriously. No one makes it out alive.
Yes, that's right, if you take a random sampling where either the older (column 1) or the younger (column 2) child is a boy, there will be approximately 2 girls for every boy in the other column.
But divide these results into 2 different sets. One set where the child in column 1 is a boy and the second set where the child in column 2 is a boy. If Tuesday boy is the older child, his sibling belong to the first set of data. If he's the younger child, his sibling is in the second set of data.
I did your test in SQL for 1000 families, results are 260 boys, 495 girls. However, when applying an identity to Tuesday boy, the results are:
Tuesday is the older child in a family with 1 or more boys: 260 boys, 252 girls.
Tuesday is the younger child in a family with 1 or more boys: 260 boys, 243 girls.
Tuesday is a member of both sets (all boys are), but only as a younger child or an older child. But his sibling can belong to only one of these sets, if a girl. And that data shows approximately 50/50.
If, on the other hand, you select a child specifically for being a boy and part of a 2-child family, then the gender of the other child is not independent.
Intuition would say that it is independent. TFA is all about how intuition is wrong. Just because your data (and theirs) shows that it is not independent doesn't mean it isn't. You may have your data wrong. Intuition may be right.
If you've selected a child specifically for being a boy, he is specifically either older or younger than his sibling in question, and so the set of data which he belongs to is not the group with either a younger boy or an older boy, but to one of two sets of data which are each subsets of that data.
Just because you don't know if he is older or younger doesn't mean you discard that data, but you form two sets of data to represent the two cases. Average out those two sets of data to determine probability, which comes to approximately 50/50.
See here http://slashdot.org/comments.pl?sid=1701394&cid=32752678
Comment removed based on user account deletion
"Gardner himself tripped up on his simpler Two Children Problem. Initially, he gave the answer as 1/3, but he later realized that the problem is ambiguous in the same way that Peres argues that the Tuesday Birthday Problem is. Suppose that you already knew that Mr. Smith had two children, and then you meet him on the street with a boy he introduces as his son. In that case, the probability the other child is a son would be 1/2, just as intuition suggests."
is the difference between taking "one is a boy" to mean "at least one boy, can't tell you which" and "a specified boy." I'm kind of surprised it ever took Gardner any time to realize that distinction existed.
However, that's not the same ambiguity as the one that determines whether or not the information about the birth day can be informative.
The information about the birth day can only lead to the 13/27 answer if the speaker only would have told you the birth day if it were Tuesday. If the speaker might have told you the birth day was Wednesday if that were the case, the answer is no longer 13/27. To get an answer of 13/27, you have to imagine the speaker answered the question "Was at least one child a boy born on a Tuesday?" or something similar. This much is clear from the article.
In contrast, "at least one is a boy" can be informative without knowing what prompted the speaker to say it. It is still informative even if the speaker might have said under other circumstances "At least one is a girl."
That means that this paragraph is, at best, unclear:
"Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy. If I specifically selected him because he was a boy born on Tuesday (and if I would have kept quiet had neither of my children qualified), then the 13/27 probability is correct. But if I randomly chose one of my two children to describe and then reported the child’s sex and birthday, and he just happened to be a boy born on Tuesday, then intuition prevails: The probability that the other child will be a boy will indeed be 1/2."
Specifying "if I randomly chose one of my two children" does indeed necessitate that the probability the other one is a boy is 1/2, but it misses a possibility. That possibility is that the speaker is reporting not on a specific child, but reporting the gender of at least one child. In that case, the probability is 1/3, unless the the speaker also responding to a prompt like "is at least one of your children a boy born on a Tuesday?" The information about the birth day and the information about gender do not have the same epistemological status in the problem. The informativeness of the day of birth is totally dependent on what prompted the utterance. The information about the gender of at least one child is not. (I have code that demonstrates this in a simulation; I wasn't confident on the point until I had proved it to myself.)
Whether one can accept an answer of 1/3 does depend on an additional assumption necessary for full specification of the problem: that the speaker would be equally likely to report "at least one of my children is an X" in all cases it's true.
What's interesting is that you're claiming that the math behind this is a "number game" and yet you're arguing about numbers. With that approach, you could claim that the probability is 150%, and of course, since you apparently disdain mathematics, you wouldn't accept any proof otherwise.
So my alternative theory to TFA is that you have a hard time understanding this because you simply don't want to.
I don't believe in time. It's a grand conspiracy designed to sell watches.
Intuition would say that it is independent. TFA is all about how intuition is wrong. Just because your data (and theirs) shows that it is not independent doesn't mean it isn't. You may have your data wrong. Intuition may be right.
Since when does intuition trump data?
See here http://slashdot.org/comments.pl?sid=1701394&cid=32752678
Nice! Experimental data, even.
Can we agree that that settles the case that the chance is indeed 1/3 if you select by family?
Before I was given the proof I could immediately understand ( http://science.slashdot.org/comments.pl?sid=1701394&cid=32729242) I started a little OpenOffice database to give me some experimental evidence. This OpenOffice Base file unfortunately has one error which keeps me from checking bigger datasets but at least it did show me that 100,000 families is not enough. The probability for a brother ranges between 0.42 and 0.51 in my runs.
Since when does intuition trump data?
It's not the data, it's the analysis.
Nice! Experimental data, even.
Can we agree that that settles the case that the chance is indeed 1/3 if you select by family?
No. Read my reply to that comment.
The reply where you start by agreeing that 1/3 is correct, and then propose some meddling in order to prove your hurt intuition right again?
You're just trying to justify your intuition, and beat the data into a shape that feels right to you. The analysis of the data in the article (and by me) is correct, though. Unfortunately for your intuition.
The reply where you start by agreeing that 1/3 is correct, and then propose some meddling in order to prove your hurt intuition right again?
You're just trying to justify your intuition, and beat the data into a shape that feels right to you. The analysis of the data in the article (and by me) is correct, though. Unfortunately for your intuition.
No, I agreed that the data they provided showed 1/3, but that data is incorrect.
Explain how the data I beat "into a shape that feels right" to me is wrong.
To beat a dead horse, having girl/boy and boy/girl are mutually exclusive events and cannot both be counted in statistics. Tuesday boy is a subject of the study, not part of the statistics. He can be either younger or older, but not both. Although that data is undiscovered it is not indeterminate, so the data must be split into two sets to avoid counting mutually exclusive events in the statistics.
Essentially, in statistics the concept of something being mutually exclusive serves to prevent it from being counted more than once in the overall tally and has less to do with it being true or false over something else (although it is always preferable to count only true data).
Although that data is undiscovered it is not indeterminate, so the data must be split into two sets to avoid counting mutually exclusive events in the statistics.
No it mustn't. It only must in order to satisfy your intuition that's unable to grasp the simple math behind this problem.
What's your opinion on the Monty Hall problem? It's very much the same. Counter-intuitive, yet true, if you work it out.
Ok, I get where I was wrong. I was selecting a boy at random, not a family at random, as some of the comments I was responding to incorrectly said, if you're a boy in a 2-child family you have a 1/3 chance of your sibling being a boy and 2/3 chance of your sibling being a girl.