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The Tuesday Birthday Problem

Posted by kdawson on from the if-it's-tuesday-it-must-be-a-girl dept.

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."

43 of 981 comments (clear)

  1. Re:Well? by Anonymous Coward · · Score: 3, Informative

    13/27

  2. Ordering and Convergence by eldavojohn · · Score: 5, Informative

    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.
    1. Re:Ordering and Convergence by mulvane · · Score: 4, Insightful

      The problem doesn't disallow twins as it doesn't give TIME of birth. Only day. A child born at 11:50PM on Tuesday and one born at 00:15 on Wednesday are still both twins. It also does not include if in that scenario is a single egg birth or whatnot so it could still be a boy and girl twin situation.

    2. Re:Ordering and Convergence by dakrin9 · · Score: 5, Informative

      This is incorrect - the question DOES not disallow the second child being a boy and born on Tuesday.

      Here's a reply to the article: (I haven't verified for mathematical correctness)

      "The "(and only one)" qualification suggested by Ralph Dratman is _not_ required. Indeed, in the first case of the analysis, "older child is a boy born on Tuesday", the possibility that the younger child is also a boy born on Tuesday is explicitly included and counted. The hypothesis for the second case does exclude the possibility of both being boys born on Tuesdays. The two cases are mutually exclusive and exhaustive.

      Note that if the puzzle had included the "(and only one)" qualification, then the possibility count would have been 13 (6 for boy and 7 for girl) in both cases, and the probability drops to 12/26."

    3. Re:Ordering and Convergence by Looce · · Score: 4, Informative

      The problem is stated thus:

      I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

      One of whom is not "exactly one of whom", so 'one' might be opposed to 'the other' or 'two'. All we know is that one of the problem poser's children is a boy born on a Tuesday. It states nothing about the relationship between the two children in time or space, so the probabilities are independent.

      Further, the problem doesn't ask about any probability related to the second boy's birthday. The problem doesn't ask, e.g., What is the probability that my other child is a boy not born on Tuesday?. That makes the birth weekday completely irrelevant.

    4. Re:Ordering and Convergence by joss · · Score: 4, Insightful

      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.

      --
      http://rareformnewmedia.com/
    5. Re:Ordering and Convergence by martin-boundary · · Score: 5, Insightful

      "it's more a trick of English converting to statistics than it is a true puzzle". And that is why I don't accept any mathematical answer to the riddle as 100% correct.

      Not necessarily. Any reading of an English sentence is an exercise in interpretation. We don't just read the words alone, we actually interpret them and use a mental model to help dismiss obviously incorrect ambiguities.

      Let's say the sentence has several interpretations. For each interpretation, we could solve for an interpreted probability answer. Then we could look at each answer and ask if it makes sense. If that answer doesn't make sense, we could dismiss that particular interpretation of the sentence. If a single answer remains after that, it would be THE answer (Sherlock Holmes style).

      In the example, here are some possible disambiguations:

      1) "I have two children, (exactly) one of whom is a boy born on a Tuesday."

      2) "I have two children, (at least) one of whom is a boy born on a Tuesday."

      3) "I have two children, one of whom is a boy born on a (particular) Tuesday."

      4) "I have two children, one of whom is a boy born on a (generic) Tuesday."

      There is also an ambiguity in the second sentence, which is only obvious to statisticians and probabilists:

      a) "What's the (Bayesian subjective) probability that my other child is a boy?"

      b) "What's the (objective) probability that my other child is a boy?"

      In case a), the problem is underspecified as it requires the full set of personal beliefs of the reader to be used for an answer (Bayesian subjectivists propose that a probability is merely a degree of personal belief, such that two people will not agree on the probability for the same event, because, being different people, they have different prior beliefs.)

      In case b), the problem can (should) be solved solely from the problem and general common knowledge of the world (which is still required to interpret the question).

    6. Re:Ordering and Convergence by bzipitidoo · · Score: 5, Insightful

      Yes, it is like the Monty Hall problem. What is the probability that 2 children are both boys? 25%. Knowing that one of the children is a boy does not change that probability as much as might be thought. The answer then is 33.3%, not 50%. This is because the additional information has cleverly NOT specified which child is the boy. If a particular child is picked out, eg. the first child is a boy, then it is 50% the other is a boy, because it always was 50% likely that a child is a boy. The bit about "born on Tuesday" does matter, because it comes close to specifying a particular child. The more improbable it is that both children fit some criteria, the closer the probability gets to 50%. If the info had been "one of whom is a boy born on Feb 29", the answer would be nearly 50%.

      --
      Intellectual Property is a monopolistic, selfish, and defective concept. It is "tyranny over the mind of man"
    7. Re:Ordering and Convergence by radtea · · Score: 5, Insightful

      These kind of problems require competence in both English and Maths which is why so few people get them right.

      Mostly they require competency in psychology, so you can figure out how the twit posing the problem is deliberately trying to mislead you by using ambiguous English and claiming on the basis of their poor communication skills to be clever.

      --
      Blasphemy is a human right. Blasphemophobia kills.
  3. Let's try it without reading TFA by Shin-LaC · · Score: 5, Informative

    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.

    1. Re:Let's try it without reading TFA by cjnichol · · Score: 3, Informative

      He has already had the children and he is only telling you that at least one was a boy. He didn't say which.

    2. Re:Let's try it without reading TFA by jimicus · · Score: 5, Interesting

      It's playing games with words and attaching significance to two sets that in any practical case I can think of would be considered one.

      The argument is that if you were to consider it as a set, there are four possible ways for your children to be distributed:

      1. (Boy, Boy)
      2. (Boy, Girl)
      3. (Girl, Boy)
      4. (Girl, Girl)

      We already know that your children can't possibly fall into the fourth set, and so looking at the sets it appears that the probability should be 1/3. But this misses one minor point - you've added an extra set which only makes sense if you wish to attach significance to the order in which the children were born (Sets 2 and 3). But as soon as you do attach that significance, the information you are given in order to establish the probability of any particular outcome (eg. the boy is older) allows you to eliminate two sets rather than just one.

  4. The difference between a man and a woman by dimethylxanthine · · Score: 5, Funny

    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.

  5. Shorter summary by williamhb · · Score: 5, Funny

    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.

  6. Re:Rubbish by guyminuslife · · Score: 4, Informative

    "I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?" .. this is the same as saying "I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?

    No, it isn't. It's the same as saying, "I have just tossed a quarter twice, and given that it came up heads at least once, what's the chance it that it came up head both times?"

    We can do an analogue, but not in the way you're saying. I.e., "I have just tossed a coin and it came up heads, what's the chance it will come up heads if I toss it again" == "I have just had a kid and it's a boy, if I have another kid, what's the chance it will be a boy" == 50%

    If you don't understand the problem, that's fine. It's counterintuitive, these things usually take a second look. If you don't understand the problem and you want to claim the solution is incorrect because you don't get it, well, that's something else entirely.

    --
    I don't believe in time. It's a grand conspiracy designed to sell watches.
  7. Probability by neoshroom · · Score: 4, Informative

    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.
  8. Re:Rubbish by Joce640k · · Score: 4, Informative

    Nope.

    "I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?"

    Is *not* the same as:

    "I just tossed two coins and one of them was heads, what is the probability that the other one was heads as well?"

    --
    No sig today...
  9. Principles of Restricted Choice by tangent3 · · Score: 3, Insightful

    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.

  10. Johnny's Mom Has 3 Kids... by mim · · Score: 5, Funny

    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.

  11. I love mathematicians... by itsdapead · · Score: 5, Funny

    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.
    1. Re:I love mathematicians... by Asmor · · Score: 5, Funny

      In other news: you can't actually build a hotel with an infinite number of rooms - you'd run out of bricks

      Don't let the venture capitalist who's funding me know that. I duped him into paying me an infinite amount of money! He paid me $1000 for the first room, $500 for the second, $250 for the third, etc. I'll be rich! Rich, I tells ya! I've been eying this nice $2000 watch, and I should have enough for it any day now...

  12. Re:Summary misstates the problem by tomtomtom777 · · Score: 4, Insightful

    "I have two children, one of whom is a boy. What's the probability that my other child is a boy?" ... it is given that the FIRST child is a boy.

    I must admit that English is not my native tongue but I fail to see how this gives that the FIRST child is a boy. Doesn't "one of whom" implies that it can be either the first or the second?

  13. Re:What's counterintuitive about it? by X0563511 · · Score: 3, Insightful

    This is a question written in purposely misleading English.

    This, in other words, is a shit question.

    --
    For large sets, this will be our guide even unto death, for the LORD will work for each type of data it is applied to...
  14. The last name is even MORE important by houghi · · Score: 4, Funny

    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.
  15. Man this question pisses me off. by hellop2 · · Score: 5, Insightful

    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?
  16. Re:Science and Intuition defeating Fun Math by snowgirl · · Score: 3, Informative

    No joke, you break out the SQL.

    I did so.

    I generated a bunch of SQL records, into a table with two relevant fields, both binary. (Male was indicated by false, female was indicated by true).

    Then I counted how many SQL records were returned WHERE gender1 = male OR gender2 = male
    Then I counted how many SQL records were returned WHERE gender1 = male AND gender2 = male.

    The results? Out of a population of 100,000 records: 24940 male-male, 74893 at least one male. Yielding 1/3 to the third decimal place.

    The mistake you made (and it is a mistake, because you claim your experiment would show a result that it empirically does not) is that you failed to count properly. There will always be approximately twice as many records with one boy as there are two boy records.

    It's not the BIRTH ORDER that turns this problem into something interesting.

    As a side note, in the future, actually run your experiments if it is actually feasible to do so.

    --
    WARNING! This girl exceeds the MAXIMUM SAFE standards established by the FDA for BRATTINESS
  17. Re:The other problem posed in TFA by A+Nun+Must+Cow+Herd · · Score: 3, Insightful

    That's incorrect - you've just skewed the population!

    In 1000 pairs of children you'll have 250 girl/girl, 250 boy/boy, 500 girl/boy.

    Of those, the ones that have at least one boy are the 250 boy/boy and 500 girl/boy pairs. So there's a 33% chance it's boy/boy if you know one is a boy.

    The whole point is you could be talking about either of the boys in the 250 boy/boy pairs - it doesn't increase the probability that it's boy/boy instead of girl/boy (you're still twice as likely to have a girl/boy pair relative to a boy/boy pair). If you specify more about the boy you're talking about - for example (ironically) saying his name is Peter - then the boys are no longer interchangable and the probability tends towards 1/2.

    It is tricky ;-)

  18. Re:Well? by suso · · Score: 3, Insightful

    My older brother and I were both born on Tuesdays.

  19. Re:Well? by ticklemeozmo · · Score: 4, Informative

    This is a gambler's fallacy problem. The more tangents you throw at it, the closer you get to .5 (50%), while never reaching it. This is the limit, why? Because there's only two potential outcomes for the other child: boy or girl.

    What you (or the website you copied and pasted the ratio from) fail to take into account (and why it's a Gambler's fallacy problem) is that when involving chance, anything that happened in the past is completely irrelevant to future probables. I could roll a die 99 times, and get 6, the probability of getting 100 6's when I've already got 99 6's is still 1 out of 6, not 6^100.

    The reason the chi square doesn't come into play here is because it doesn't MATTER the order. Has she said "What is the probability my SECOND-BORN was a boy?" it would be perfectly logical to write the square because the boy who was born on Tuesday could be either the first born or the second born, she never stipulated.

    We can say that the boy, who was born on a tuesday, was also a Gemini. Does this change the ratio? No, the probability of having two boys is still 50-50%, because the unknown only has two possible outcomes: boy or girl.

    --
    When modding "Informative", please make sure it both has a source and IS actually informative.
  20. Re:The other problem posed in TFA by Shin-LaC · · Score: 4, Informative

    No, you're wrong. Look at this. If Mr. Smith has two children, at least one of whom is a boy, it is two times as likely for him to have a boy and a girl than it is for him to have a boy.

    Your mistake is in believing that, by virtue of naming one of the boys Peter, the probabilities are magically equalized. They're not. The correct probabilties for your table are:

    Peter, Boy = 1/6
    Boy, Peter = 1/6
    Peter, Girl = 1/3
    Girl, Peter = 1/3

  21. Re:Well? by ceoyoyo · · Score: 4, Informative

    Failed to read the article, hey?

    You're absolutely correct IF you assume that the man is just describing one of his children. That is the logical interpretation of the problem. However, if someone went out and specifically selected a family with at least one boy, the probability of the other child being a girl is not 50% - the population in question has been artificially depleted of girls because the criteria excluded all two-girl families.

  22. Re:Well? by mcvos · · Score: 4, Informative

    Read the article. Whether the answer is 50% or not depends on the context, and that context is not specified in the problem.

    If you meet the person with his son, and he tells you that that son is born on a Tuesday, then the chance of the other kid being a boy is 50%. If you see him as one of the thousands of parents who have two children, at least of which is a boy who's born on a Tuesday, the chance of the other kid being a boy is 13/27.

  23. Re:OK I'm stupid by ceoyoyo · · Score: 3, Informative

    Close. The situation with coins is this:

    I flip a coin twice and record the answer. I repeat this many times. I discard all the pairs where both coins came up tails. I then select a pair where at least one toss came up heads. What is the probability that the other is also heads?

    The selection criteria screw up the probabilities. If instead I flip a coin, see that it comes up heads, and ask what the probability of it coming up heads again is, the answer is 50%.

  24. Re:Well? by prionic6 · · Score: 5, Informative

    Assuming 50/50 chance of boy or girl and a 1/7 chance of each weekday, let's look at the general population of families with two childs:

    B1B1 B1B2 B1B3 B1B4 B1B5 B1B6 B1B7
    B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
    B3B1 B3B2 B3B3 B3B4 B3B5 B3B6 B3B7
    B4B1 B4B2 B4B3 B4B4 B4B5 B4B6 B4B7
    B5B1 B5B2 B5B3 B5B4 B5B5 B5B6 B5B7
    B6B1 B6B2 B6B3 B6B4 B6B5 B6B6 B6B7
    B7B1 B7B2 B7B3 B7B4 B7B5 B7B6 B7B7

    B1G1 B1G2 B1G3 B1G4 B1G5 B1G6 B1G7
    B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7
    B3G1 B3G2 B3G3 B3G4 B3G5 B3G6 B3G7
    B4G1 B4G2 B4G3 B4G4 B4G5 B4G6 B4G7
    B5G1 B5G2 B5G3 B5G4 B5G5 B5G6 B5G7
    B6G1 B6G2 B6G3 B6G4 B6G5 B6G6 B6G7
    B7G1 B7G2 B7G3 B7G4 B7G5 B7G6 B7G7

    G1B1 G1B2 G1B3 G1B4 G1B5 G1B6 G1B7
    G2B1 G2B2 G2B3 G2B4 G2B5 G2B6 G2B7
    G3B1 G3B2 G3B3 G3B4 G3B5 G3B6 G3B7
    G4B1 G4B2 G4B3 G4B4 G4B5 G4B6 G4B7
    G5B1 G5B2 G5B3 G5B4 G5B5 G5B6 G5B7
    G6B1 G6B2 G6B3 G6B4 G6B5 G6B6 G6B7
    G7B1 G7B2 G7B3 G7B4 G7B5 G7B6 G7B7

    G1G1 G1G2 G1G3 G1G4 G1G5 G1G6 G1G7
    G2G1 G2G2 G2G3 G2G4 G2G5 G2G6 G2G7
    G3G1 G3G2 G3G3 G3G4 G3G5 G3G6 G3G7
    G4G1 G4G2 G4G3 G4G4 G4G5 G4G6 G4G7
    G5G1 G5G2 G5G3 G5G4 G5G5 G5G6 G5G7
    G6G1 G6G2 G6G3 G6G4 G6G5 G6G6 G6G7
    G7G1 G7G2 G7G3 G7G4 G7G5 G7G6 G7G7

    Each outcome has P = 1/(7*7*4) = 1/196

    Let's only look at the families with (at least one) tuesday boy:

         B1B2
    B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
         B3B2
         B4B2
         B5B2
         B6B2
         B7B2

    B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7

         G1B2
         G2B2
         G3B2
         G4B2
         G5B2
         G6B2
         G7B2

    Of these 27 families, 13 have another boy. So P = 13/27.

  25. Re:Well? by mcvos · · Score: 4, Informative

    You're right! Pretty amazing, considering the article gives the correct answer and explains it pretty thoroughly. Even when you think: "But wait! Can't you look at it some other way?", the article does just that.

    Really, this time it pays to RTFA.

  26. Re:Well? by bsDaemon · · Score: 3, Funny

    ...what that actually represents is both having one rock in your presence butt also...

    I'm sure glad I don't have a rock present in my butt...

  27. As always, ambiguous language by natbrooks · · Score: 3, Insightful

    "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.

  28. Re:Well? by Anonymous Coward · · Score: 3, Informative

    Assuming 50/50 chance of boy or girl and a 1/7 chance of each weekday, let's look at the general population of families with two childs:

    B1B1 B1B2 B1B3 B1B4 B1B5 B1B6 B1B7
    B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7
    B3B1 B3B2 B3B3 B3B4 B3B5 B3B6 B3B7
    B4B1 B4B2 B4B3 B4B4 B4B5 B4B6 B4B7
    B5B1 B5B2 B5B3 B5B4 B5B5 B5B6 B5B7
    B6B1 B6B2 B6B3 B6B4 B6B5 B6B6 B6B7
    B7B1 B7B2 B7B3 B7B4 B7B5 B7B6 B7B7

    B1G1 B1G2 B1G3 B1G4 B1G5 B1G6 B1G7
    B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7
    B3G1 B3G2 B3G3 B3G4 B3G5 B3G6 B3G7
    B4G1 B4G2 B4G3 B4G4 B4G5 B4G6 B4G7
    B5G1 B5G2 B5G3 B5G4 B5G5 B5G6 B5G7
    B6G1 B6G2 B6G3 B6G4 B6G5 B6G6 B6G7
    B7G1 B7G2 B7G3 B7G4 B7G5 B7G6 B7G7

    G1B1 G1B2 G1B3 G1B4 G1B5 G1B6 G1B7
    G2B1 G2B2 G2B3 G2B4 G2B5 G2B6 G2B7
    G3B1 G3B2 G3B3 G3B4 G3B5 G3B6 G3B7
    G4B1 G4B2 G4B3 G4B4 G4B5 G4B6 G4B7
    G5B1 G5B2 G5B3 G5B4 G5B5 G5B6 G5B7
    G6B1 G6B2 G6B3 G6B4 G6B5 G6B6 G6B7
    G7B1 G7B2 G7B3 G7B4 G7B5 G7B6 G7B7

    G1G1 G1G2 G1G3 G1G4 G1G5 G1G6 G1G7
    G2G1 G2G2 G2G3 G2G4 G2G5 G2G6 G2G7
    G3G1 G3G2 G3G3 G3G4 G3G5 G3G6 G3G7
    G4G1 G4G2 G4G3 G4G4 G4G5 G4G6 G4G7
    G5G1 G5G2 G5G3 G5G4 G5G5 G5G6 G5G7
    G6G1 G6G2 G6G3 G6G4 G6G5 G6G6 G6G7
    G7G1 G7G2 G7G3 G7G4 G7G5 G7G6 G7G7

    Why are you assuming that order matters, but boys are interchangable? If we label the child that was mentioned in article as A and the other one as B then B2B2 splits into two cases when A has an older brother and when he has a yonger brother, therefore we get:


    B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7

              B1B2
              B2B2
              B3B2
              B4B2
              B5B2
              B6B2
              B7B2

    B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7

              G1B2
              G2B2
              G3B2
              G4B2
              G5B2
              G6B2
              G7B2

    Of these 28 families 14 have another boy, so P=14/28=1/2

  29. Re:Well? by Chapter80 · · Score: 3, Funny

    We can say that the boy, who was born on a tuesday, was also a Gemini. Does this change the ratio? No, the probability of having two boys is still 50-50%, because the unknown only has two possible outcomes: boy or girl.

    Did I win the lottery last week? The unknown only has two possible outcomes: I won or I lost.
    Therefore, based on your math, my odds are 50-50%.

    University of Phoenix online wants their diploma back. :-)

  30. Re:MOD PARENT UP by prionic6 · · Score: 4, Insightful

    In my experience, many non-intutive probabilty results are easier to understand if you spell out the full population. For example, I coudn't understand http://en.wikipedia.org/wiki/Berkson's_paradox until I drawed it up graphically.

  31. Order is irrellevent but uniqueness is not. by pavon · · Score: 3, Informative

    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).

  32. Reminds me of an old joke by RevWaldo · · Score: 5, Funny

    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!
  33. Two Slashdot Readers by liam193 · · Score: 3, Funny

    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%