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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."
Slashdot answer in 3..2..1..
Does this involve the Birthday problem and the Pigeonhole principle?
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.
Let's examine the question again: "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?". The answer is 0.5, because the second result is not influenced in any way by the first result (except in strictly biological and not mathematical circumstances).
Roulette is a similar game of probability - 18 red and 18 black numbers plus one (or two) zeroes. If you ignore the zeroes for the moment, then you could also say "I just played roulette and got a red 7. What the probability that I will get two reds?" Here the answer would be 0.5 again (if you ignored the zeroes). Of course in real life, the zero or zeroes exist and THAT is why the house always wins..
Never email donotemail@WeAreSpammers.com
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.
yea it would be 49% since women make up 51% of the population...
hear that men we are minorities... get your ObamaCheck at the door...
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.
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
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?
So the fact that they told us that the one kid was a boy born on a Tuesday is irrelevant.
The REAL issue is that we too easily miss the implication here: the OTHER kid *can't* be a boy born on Tuesday (or he'd have said that he has two such kids). Given that information, which rules out one of the possibilities, it is now not quite as much of a surprise that you get a chance not equal to 50%.
In the more general (mathematical) case, that is: I say: "my boy has [something with chance p of having]" if and only if I have a boy that has that 'something' and I have exactly two children and having 'something' is independent for each child. What is the chance of the other child being a boy?
Let's work it out, there are a few possibilities for the sexes:
M&M = 1/4
F&M = 1/4
M&F = 1/4
F&F = 1/4
The last is not going to work. There are four possibilities for 'having:'
H&H = p^2
N&H = p(1 - p)
H&N = p(1 - p)
N&N = (1 - p)^2
Now combine them we see that if we take together M&F and F&M (there is no inherent difference) we get a few possibilities:
M&M and H&H, N&H, H&N
M&F and H&H, H&N.
Compute the probabilities of being to say the statement is:
1/2*1/2*(p^2 + 2p(1 - p)) +
2*1/2*1/2*(p^2 + p(1 - p)) =
p-p^2/4
And the probability of there being another boy (independent of the statement) is:
1/2*1/2(p^2 + 2p(1 - p)) = p/2-p^2/4
And the probability of there being another boy is:
(p/2-p^2/4)/(p-p^2/4) = (p-2)/(p-4)
p = 1 and p = 1/7 gives the expected answer and as p -> 0, (p-2)/(p-4) -> 1/2 (because when p = 0 it is 1/2 and it is continuous on the interval [0, 1]). But p cannot be zero, or at least then the statement would never be said, so: (p - 2)(p - 4) is a strictly decreasing function from {0, 1] to [1/3, 1/2}.
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.
While it's true that the Tuesday remark adds something to the dataset, it doesn't however influence the probability of the sex of the next child.
Just like the sex of the first child doesn't affect the sex of the second child.
So in the end, it's going to be a 50% chance.
It's like asking, "I threw up a coin yesterday and it landed up heads. What is the chance it will land heads up today?"
I read the article and im wondering something. On the original question with answer 1/3 why is boy/boy only counted once it should be counted twice because with the given information there is 2 options. "Mentioned boy" / boy and boy / "mentioned boy" while they act as if the second boy is interchangeable with the mentioned boy which would be weird.
Because on the boy/girl option they do care if the mentioned boy is first or last.
Taking this in account you do get the intuitive answer of 1/2.
If anyone could tell why that would be great
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 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.
The condition "only one of which is boy born on Tuesday" places constraints on the sex of BOTH children.
If it were "only one of whom is a boy born on a Tuesday" then it would preclude the other child from being a boy born on a Tuesday but not a girl. The subset of events in which "only one of whom is a boy born on a Tuesday" thus contains fewer events in which both children are boys (because you can never have 2 boys both born on a Tuesday, but you can have a girl and a boy both born on a Tuesday) and thus the probability of the second child's being a boy is reduced.
This is irrelevant though because the statement does not specify "only" and thus the weekday on which the specified boy was born has no effect on the probability.
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 two children, one of whom is a boy born on a Tuesday.
What is the probability I am actually their father?
is not cleverness.
http://xkcd.com/169/
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.
> 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?"
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.
What if I say: "I have two pieces of fruit. This one is a banana". Do you feel this statement implies anything about the other piece of fruit? Can't that also be a banana?
Another variant: Let's say I call up a car salesman and ask if he has a green car. If he says yes, does he have exactly one green car? What if I ask if he has 'one' green car. I'd imagine he'd say yes if he has at least one green car. And not because all car salespeople are lying scum either.
English is not my native tongue, but I used to think I understood this sort of subtlety. Am I wrong?
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.
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
I met my girlfriend on Thursday. She is receptionist. What is the probability of my second girlfriend being supermodel?
Zero if the current girlfriend hears you talking about "second girlfriend".
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.
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.
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));
}
I've actually seen this before as "child at the window"-problem in a german C++-forum: http://www.c-plusplus.de/forum/viewtopic-var-t-is-115631-and-start-is-0-and-postdays-is-0-and-postorder-is-asc-and-highlight-is-.html
The thread got 88 pages, probably one of the longest threads ever in that forum. And by the way... the answer is simply 1/2, also for the tuesday-problem, because the probabilities are independent. But it seems very easy to get it wrong at some point.
If I flipped a coin on Friday that showed "heads", what are the chances of the next coin flip being heads on the next coin flip?
It's Sunday, September 5th, 2014, and I pulled a "Diamond" suited card out of a full deck of cards... What are the chances of the next card being a "Diamond"?
Wow, how fun is this?
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
How do we know that the perceived boy is not actually a girl? Should we not factor this into the probability as well?
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.
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 ?
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.
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 you drop the superfluous notion that one boy could not have been born on Tuesday from the method given in TFA, you get 28 cases, 14 of which are boys and 14 of which are girls. 14/28=1/2.
I'll take who gives a fuck for $10.
My other sig is a knife wound.
The chance is 50% since
Boygirl = girlboy = 50%
Boyboy = 50%.
Now if he HAD stated that the _older_ is male or the second child is NOT born on a tuesday, that would be an entirely different thing.
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).
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.
It depends on how exactly the information was obtained.
Case A:
You: I have exactly two kids
Me: Is at least one of them a boy who was born on a Tuesday?
You: Yes
The odds of the other being a boy are 13/27
Case B:
You: I have exactly two kids
Me: Is at least one of them a boy?
You: Yes
Me: Was at least one boy of yours born on a Tuesday?
You: Yes
The odds here are now 1/3
Case C:
You: I have exactly two kids
Me: Tell me the gender and day of birth of one of them
You: At least one is a boy born who was born on A Tuesday
The odds here are now 1/2
Case D:
You: I have exactly two kids, at least one of which is a boy who was born on Tuesday
It depends on what you WOULD HAVE told me had there not been a boy born on Tuesday. If you would have told me the gender and day of birth of a different child, the odds are 1/2. If you would have only told me "None are boys who were born on Tuesday", the odds are 13/27. If you would have told me "None are boys", the odds are 1/3.
I'm amazed by how many trolls are getting modded +5 or higher today. 4chan must be having fun.
Given that it's actually 49/51, here's the deal. The mentioned boy is irrelevant, absolutely. The other child's outcome was not or would not have been influenced by the mentioned boy, short of all the twins, conjoining, BS exceptions. End of story. Unless the story problem says that the mentioned has an effect on the unmentioned, the other is unaffected. "Oh, but I was so close to three cherries on that last slot machine pull - I'm so close to winning!"
"I have two stories, one of whom is posted on a Tuesday. What's the probability that my other story is a dupe?"
his daughter couldn't solve it as well...
sorry, i haven't even read the first reply, but or the linked story. tl;dr.
but the answer is obvious to anyone who's done a bit of maths in uni, if you can put aside the irrelevant facts from the question.
ie it's 50%. or close enough to it. depending on the population stats for the area. i'm assuming the population has as many boys born on average as girls which is probably not true.
sex of one child (given) is not connected to sex of other child.
they are disconnected events.
maybe statistically identical twins are more likely or less likely than twins from two sperm, but I doubt that is the point behind the story.
so my simple answer would be it doesn't matter what the ho tells us about one of her kids she said she's got two and it's approx 50/50 that each is a boy or girl. unless you're in india or something where boys might be considered more valuable and girls might be terminated beforebeing born - but again, i doubt that's the point of the story.
it's like the goat prize puzzle.
you are in a tv competition with three doors. behind one is a good prize but behind the other two is a goat (ie a dud or shitty prize).
the host of the tv show asks you to pick a door and tells you you'll get to have whatever is behind the door you pick.
So you pick a door.
But then the host say 'hey, actually I'm going to open one of the doors with a goat behind it and then i'll give you another chance - stick to the door you've already chosen, or change your choice to the other door (the only one that is still closed).
The question of the puzzle, is should you stick with your original door or change to the other one.
The plain and simple answer is that you should ALWAYS change.
it doesn't matter what door the host shows.
Quite simply, you had 1/3 chance of being right with your first choice, and 2/3 chance of being wrong.
So if you change to the other door (remember there's only two after the host opens a goat door), THAT DOOR has 2/3 chance of being RIGHT.
google it.
it's on wikipedia.
you can argue with your mum about this one for weeks and they will still maintain it's 50/50.
Just reading the comments gives me a horrible headache.
As I was walking to Saint Ives I met a man with seven wives...
Free Martian Whores!
The chance of the second child being a boy is 1/2. Period. Any thought going beyond that is brain power wasted on semantics.
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.
And the answer only works if you're pedantic enough to assume that your sample space was self-selected to include only those questioners with two children, one of whom was a boy born on Tuesday.
That seems... silly.
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.
Occums razor.
You want to know the likelyness that something is male - as long as its a "normal" reproduction -> we presume it is as its not mentioned to any more likely one way or the other.
So, 50%.
- http://www.milkme.co.uk
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.
The chances of a boy already born being a boy is 100%, chances of a girl already born being a boy is 0%. Best course of action is to ask a follow-up question: is that other child a boy? Usually parents tend to know this.
Is it just me, or isn't it obvious why the probability changes? All you are doing is adding more parameters to what you are trying to actually determine. The most basic question you can do (not in this question, but overall) is simply: What is the odds of a child being a girl or boy. That's theoretically 50/50. Now, each other question is subtly altering the grouping of what you are trying to determine the probability. It's just that english (or probably any language) obscures the fact that you are making those other elements important. The real trick of these is the fact that language is vague, and most people are missing that you are making necessary additions to the group you are trying to determine information about. You are in fact not just asking about the sex of the other child, but you are building a situation that makes the information about the date, or the order of the children, important. If the importance of those other elements disappears, then they are not bounding boxes to the situation and the results fall back to 50/50.
Webmaster of the webcomic 'Stupid and Insane Defenders Against Chaos' at http://www.onezumi.com
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.
"This is not the puzzle you thought it was"...
Here's the part I don't get with this Tues puzzle and the article. When order was considered in the analysis (ie allowing boy/girl, girl/boy), the puzzle was changed and there is a missing option.
There are really 5 possibilities when considering order:
1. Boy(Tues), Boy
2. Boy, Boy(Tues)
3. Girl,Boy
4. Boy,Girl
5. Girl,Girl
Eliminate #5, and you have 2/4 or 50%.
If you don't consider order, there are only 3 options:
1. Boy, Girl
2. Boy, Boy
3. Girl, Girl
When you eliminate #3, it is still 50%.
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
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.
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.
Quote:"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?"
He has 2 kids:
1 Male
1 Unknown
So possible outcomes: (0=Male 1=Female)
0,0
0,1
(Note 1,0 is the same as 0,1)
So probability is 50%
Case closed...
Laters Sol "Have you found the secrets of the universe? Asked Zebade "I'm sure I left them here somewhere"
A = chance of first child being a boy
B = chance of second child being a boy
P(B|A) = P(A and B)/P(A)
but both events are independent so
P(A and B) = P(A) P(B)
= 1/2 * 1/2 = 1/4
so
P(B|A) = 1/4 / 1/2 /thread
= 1/2
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/
When you flip a coin, there is a very high probability that it will come up either heads or tails. There is, however, a teeny little chance it could land on edge and stay that way.
This is also the case with the children. There is a very high probability that it will be either male or female. But the probability is not 100%, as most of you seem to assume. There are various medical conditions that make the sex of the baby ambiguous. Hermaphrodites, for example. Also there are people born with male chromosomes and female genitalia.
So to answer this question you need to define your terms as far as "boy" and "girl" more carefully. If your only looking at genitalia, then of course hermaphrodites cause a problem for your percentages. If you're looking at chromosomes then a parent may think they have a daughter while genetically it is a son.
So yeah, the question is more complicated than you all are saying.
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.
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.
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).
This is better than an Order of Operations Troll on 4chan
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
From the original article
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.
But this makes no sense whatsoever. For simplicity, I will assume that the human sex ratio is 1:1, and ignore the possibility of identical twins. I will also ignore the possibility of the "Monty Hall effect" - i.e., that there is other information that skews reporting of the first child. (Without this, anything can go. Suppose that I belong to a religion that believes that the birth of two boys in a row is deeply shameful and should never be mentioned. If you know that, saying that I have one boy make the probability of a girl 100%. Any other probability you might want is also possible, and so this has to be ruled out.)
To make the English clearer, suppose that I am on a beach with equal numbers of blue (B) and green (G) pebbles, and I pick two up, carefully noting the color of each.
So, an observer who just knows I have picked up two pebbles (but not the color of either) has the following probability table
BG - 25 %
BB - 25 %
GB - 25 %
GG - 25 %
Ok, I continue to walk along and meet up with the observer, and as I do so one pebble drops at random out of my bag. It is Blue (B). The observer sees this, and modifies his probability table as follows :
New probability = old probabiity x probability of having a Blue pebble fall out (renormalized to sum to unity, if necessary). This yields
BG - 25 %
BB - 50 %
GB - 25 %
GG - 0 %
So, the problem with the original "Birthday Problem" analysis is that, while there are indeed three choices, one with a second boy, and two with a girl, the probability of these choices are not equal ! With the correct probabilities, the chance that the second child is a boy is 50%
I would argue that, in the universe of these problems (where you are not supposed to need additional, unstated, information beyond basic things like people want to minimize jail time and maximize revenue), this is the correct analysis of the original "Birthday Problem," and the one presented in the original article is wrong.
The original "Tuesday Birthday Problem" analysis is likewise wrong, in the same fashion.
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!
"I have two children..."
The probability that a given child is a boy is either 100% (the child is indeed a boy) or 0% (the child is a girl). The child either is a boy or isn't. Once the child is born, there is really no probability. It would be rather disturbing to walk up to the non-Tuesday child and say "there is a 33% probability that you are a boy".
...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?
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.
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. The second case needs to be counted as well, which brings the count to:
28 possible events
14 of which have 2 boys
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.
just read about this puzzle last night in a book called "born on a blue day"
The children could just as easily be adopted and/or born from two different mothers. Sorry to bring you into the 21st century.
He's got the right idea.
Article currently reads:
"Error. You are unable to view this section .."
cd
(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.
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
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.
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
I think he knew that...
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
As stated, the fact that one child was born on a Tuesday makes no difference (I can always make a statement of that form, substituting "girl" for "boy" and whatever day of the week works). The probability that both children are the same sex remains 1/4.
On the other hand, if we decide ahead of time that we're specifically looking for people with two children, one of whom is a boy born on a Tuesday, that's a different story. We're not looking at a random pair of children any more. We're looking at pairs, one of whom has an unusual property. And the likelihood that the other child is also a boy is therefore, as near as makes no difference, 1/2 (it's not *quite* 1/2 because the case where both children are boys and share the property distorts the calculation very slightly - and the more unlikely that is, the closer the probability gets to 1/2).
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.
...that gender is by no means binary. You have the possibility of transgendered children, mutated children, etc. Which basically makes this whole question pointless. The nerdiest among us manage to get very detailed decimal answers while failing to notice one of the most basic things that they (universally) take for granted. Shame, all you so-called expert problem solvers, shame!!! Just say 50% anyways, and go outside and enjoy nature. Spending any more time on this is pretty ridiculous.
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 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.
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
I was thinking various scenarios that can happen in this case :
(I mean what are different possibilities of 2 kids can born to a woman) :
1. Woman gave birth to 2 kids at the same time – and both boys > BB
2. Woman gave birth to 2 kids at the same time – and both GALs> GG
3. Woman gave birth to 2 kids at the same time – and 1 boy 1 gal> GB it is at same time so u can say GB og BG
4. Woman gave birth to 2 kids– First Boy then Gal > B- G
5. Woman gave birth to 2 kids– First Boy then Boy > B- B
6. Woman gave birth to 2 kids– First GAL then GAL > G- G
7. Woman gave birth to 2 kids– First GAL then Boy > G- B
Now let’s listen the puzzle. I have 2 kids and one is boy.
Now lets get only those scenarios which can satisfy the statement at this time :
Number 2 and 6 are out. We have only five different options now(2 options are removed).
Now what the prob of having 2 boys :
= 2 / 5 = 0.4
This is based on consideration that 2 kids can take birth at the same time (exact same time )
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.
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...
Oops, sorry. "drew". English is not my first language.
All of you are missing the real question here. Not seeing the trees(baby’s) for the numbers.
As the question is phrased, we are talking about human birth here, not a true abstract math problem.
The reality of human biology is the root and the answer.
It is always 51/49 boy/girl birth ratio without knowing of any individual family history that would show a predisposition to either conceiving or carrying to term one sex over the other.
So, the answer is 51%, because that is the percentage of boys born to humans.
No math ‘tricks’ applied.
The ratio is not 50/50. And ‘Monty Hall’ type probability does not apply as we are not talking about a random gamble but biological fact.
The fact is 51% of the time a baby is born alive it will be a boy, 49% of the time it will be a girl.
Now We could go deeper into demographics and look at survival rates past some age as the question would leave open room for looking at the chance that both children are still baby’s and then look at the survival rate at the most probable age range, but you get my point.
We are not talking about a simple data set with 50/50 chances (not really) and then distilling down
Boy|Girl
Boy|Boy
Boy|Hermaphrodite (forgot about that one?)
We are looking at the real (biological) result of a complex system with a known probability of outcome.
51% chance the second child is a boy.
Now, if we want to infer from the truth stated that one child is a boy that the parent in question is capable of either siring or carrying to term a boy then we would need to look deeper at the medical research to determine how much that knowledge influences the chances of the other child being a boy. In fact, having a boy (or boys) without any girls does indeed slightly reduce the chance of the next child being a girl, but the way the question is phrased we don’t know if the boy was born first.
The reason(s) boys are born more often are complex, one theory is that the Y sperm is lighter and faster than the X and has a higher chance of reaching the egg first. Also the smaller size of the DNA package might make the actual process of penetrating the egg slightly easier.
Other factors with more complex controls would be evolutionary pressures leaning to boys because of the higher mortality rate of human men, the need for familial men to protect females and children for survival of the group (and the groups genes), and long standing and widespread cultural preferences for women to give birth to male offspring.
Still, without more information:
51% chance the second child is a boy.
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.
> 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.
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.
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.
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.
FFS, the probability of both children being boys is 1/2.
I have nothing to lose but my bindings.
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.
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.
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.
When it's just stupid word games. This isn't math. This is someone twisting words that a majority of people would understand one way to an obscure meaning that any normal person would disagree with.
It's not news, it's idiocy.
1. The day when the children are born is not relevant. As far as I know, aside from whatever superstition mumbo-jumbo might say, days of the week do not affect the gender of the child; even if they do affect it, in some minimal way, it would have an even smaller effect on the gender of any future or past children. The information about Tuesday is irrelevant to the question.
2. When the man says that one of his children is a son, that makes the odds of having a second son 50% (ignoring any differences in gender ratios). There is a DIFFERENCE between the odds of having two sons and the odds of having a son. The odds of having a son is 50% (two fair options); the probability of having two sons is 25% (.5*.5). The question, however, "What's the probability that my other child is a boy?", is only asking what the probability is that his son is a boy, not the probability of an outcome with two sons.
Answer = 50%.
It'd be the same thing as saying, "A person of indeterminable gender walks up to me and says, 'One of my children is a son. What is the probability that my other child is a boy?" and acting as though the gender of the parent affects the probabilities.
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?
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
The probability that your other child is a boy is: zero if your other child is a girl, and one if your other child is a boy. Honestly, just ask the doctor if you're not sure.
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.
There are only four options:
Boy(Tuesday), Boy (day irrelevant)
Boy(Tuesday), Girl (day irrelevant)
(day irrelevant)Boy, Boy(Tuesday)
(day irrelevant)Girl, Boy(Tuesday)
Two options have Boy, Boy and two options have Boy, Girl simple 50%
Nobody has come up with the correct answer so far. There are two children, therefore in addition to one boy born on Tuesday, there is one other child, who cannot be also a boy born on Tuesday, because then there would then be two boys born on Tuesday instead of one.
So what are the possibilities for the other child?
* Girl born on any day of the week = 7 possibilities
* Boy born on any day except Tuesday = 6 possibilities
All possibilities are equally likely, therefore the probability of the other child being a boy is 6/13.
All the stuff about first child or second child is irrelevant. Even the guy who did SQL solved the wrong problem. Even the original article gives the wrong answer. Sheesh!
At this point they should know the sex of all there children!
This may be one of the few problems where 42 is not the right answer.
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.
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.
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
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.
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.