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Math

The Tuesday Birthday Problem 981

Posted by kdawson
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."
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The Tuesday Birthday Problem

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  • by eldavojohn (898314) * <eldavojohn@gm[ ].com ['ail' in gap]> on Tuesday June 29, 2010 @05:20AM (#32727824) Journal
    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 [amazon.com].

    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?
    • by Looce (1062620) *

      What if the second child was born on Tuesday but was not a twin of the Tuesday boy?

      Say, the Tuesday 75 weeks later. It's still a Tuesday...

      • What if the second child was born on Tuesday but was not a twin of the Tuesday boy?

        Say, the Tuesday 75 weeks later. It's still a Tuesday...

        According to the quote that would have to be a girl born on Tuesday because the quote in question states that one and only one of the children is a boy born on Tuesday. It cannot be a boy born on Tuesday. The only information you get form that statement is that one of the two ordered children cannot be a boy born on Tuesday. We have two genders. We have seven days. That's fourteen permutations. We have two children that can possess those fourteen permutations but we know that one of those permutations

        • by dakrin9 (891909) on Tuesday June 29, 2010 @05:34AM (#32727906)
          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."
        • by Beale (676138)
          This isn't true, though - the question makes no statement about whether the other can also be a boy born on Tuesday.
          • by joss (1346) on Tuesday June 29, 2010 @06:50AM (#32728324) Homepage

            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.

        • i don't get it, are we taking that the second child was not a boy born on a Tuesday as implied? the statement doesn't really say one was a boy born on a tuesday and the other was not.
        • by Looce (1062620) * on Tuesday June 29, 2010 @05:50AM (#32728018) Journal

          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.

          • Re: (Score:3, Insightful)

            by mea37 (1201159)

            This problem (and related problems) are almost always poorly phrased. However, your assertion that failure to ask about the day-of-week of the 2nd child's birth makes the day-of-week of the 1st child's birth irrelevant is incorrect, if the problem is understood as it was intended. And actually my own phrasing there is a bit poor there, because the crux of the problem is that there is no ordering - no "1st child" and "2nd child".

            A better (or in any event less misleading) phrasing would be: "There are two c

        • by ignavus (213578)

          the quote in question states that one and only one of the children is a boy born on Tuesday.

          The quote is, and I 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?"

          Please point, click or otherwise indicate the word "only" in that quote.

    • by mulvane (692631) on Tuesday June 29, 2010 @05:28AM (#32727870)
      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.
    • Re: (Score:3, Informative)

      by Trepidity (597)

      Sounds vaguely similar to the same failure-of-intuition behind the Monty Hall problem [wikipedia.org]: you fix several quantities, then you reveal one of them, and then ask for a guess for the probabilities of the remaining, unrevealed quantities. Since the unrevealed quantities are completely independent of the revealed ones, it seems like the revealed information shouldn't matter, and your guess should still be that they're uniformly randomly distributed. But, it isn't.

      • Re: (Score:3, Informative)

        by AK Marc (707885)
        Since the unrevealed quantities are completely independent of the revealed ones,

        But they are related. When you eliminate one wrong answer, the probability of the remaining answer being right is 50%. When you eliminate one answer at random, the chance that the remaining answer is the right one is 33%. The chance of the one you chose being right is always 33%. When you reveal an answer known to be wrong, you change the probability of the ones that was chosen from. Since the door opened can't be the one
      • by bzipitidoo (647217) <bzipitidoo@yahoo.com> on Tuesday June 29, 2010 @07:13AM (#32728474) Journal

        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%.

    • by williamhb (758070) on Tuesday June 29, 2010 @05:36AM (#32727910) Journal

      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.

    • Re: (Score:3, Insightful)

      by radtea (464814)

      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.

      100% correct. All "paradoxes" of this form are very carefully, almost ritually, stated because they are almost entirely linguistic tricks designed to mislead and confuse rather than educate and enlighten.

      The very first thing anyone analyzing a problem of this type should do is restate it in several different ways, making as much implicit information explicit as possible. Such as:

      I have two children.
      I'm going to tell you some things about one of my children.
      I have not chosen that child randomly. For examp

  • by Shin-LaC (1333529) on Tuesday June 29, 2010 @05:22AM (#32727840)
    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.
    • by mrvan (973822)

      You miss one thing: if you have two children, there is a X% chance that they are twins. IF they are twins, they have a close to 100% probability of sharing a birthday, and a >50% probability of sharing gender (identical twins share gender, non-identical share gender in 50% of cases).

      Some googling says:
      P(identical twins) ~ .4%
      P(non-ident twins) ~ 2.6%

      The maths involved are left as an exercise to the reader :-)

      • by Shin-LaC (1333529)
        Yes, there is an assumption that the weekday of birth of a child is independent from that of its siblings, and uniformly distributed. The twin thing is only one of many factors that can violate one or both of these assumptions.
      • Probability (Score:5, Informative)

        by neoshroom (324937) on Tuesday June 29, 2010 @05:41AM (#32727938)

        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%."

    • Nice work! I read the article and it agrees with your math. 13/27 is exactly what the author concludes.

    • Re: (Score:2, Insightful)

      by cjnichol (1349831)
      I'm still not convinced about the 1/3 probability that the second is a boy in the original problem (ie. without the day of the week). If the order matters when the other child is a girl, why doesn't it matter when the other child is a boy? That is, if we know one child is a boy, let's call him Ba, why don't we have the following possibilities with non-zero probabilities: Ba + Bb, Bb + Ba, Ba +Ga, Ga + Ba
      • by jimicus (737525) on Tuesday June 29, 2010 @06:54AM (#32728342)

        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.

    • by Trepidity (597)

      Shouldn't ordering matter in the first part? You're told that the first child was a boy, so of the four ordered possibilities (boy, boy), (boy, girl), (girl, boy), (girl, girl), you can eliminate both (girl, boy) and (girl, girl).

      • Re: (Score:3, Informative)

        by cjnichol (1349831)
        He has already had the children and he is only telling you that at least one was a boy. He didn't say which.
        • by Trepidity (597)

          Hmm, somehow I misread that several times in the problem statement. I could've sworn it said, "he has two children, the first of which was a boy born on a Tuesday". But then clearly the problem would be much less paradoxical, so makes more sense now.

    • by Shin-LaC (1333529)

      So yes, the weekday of birth is significant.

      To clarify: I mean, of course, that the fact that the weekday of birth is specified is significant. It doesn't matter whether that day is Tuesday, Friday, or even Monday. (Assuming uniform probability of childbirth across weekdays, etc.)

    • Re: (Score:2, Insightful)

      You are good:) I have a question.

      I met my girlfriend on Thursday. She is receptionist. What is the probability of my second girlfriend being supermodel?

    • by Zelos (1050172)

      If it is known that I have a boy, there is 1/3 probability that the other is also a boy.

      I'm confused. Say we took 900 families with two children, one of whom is a boy. Are you saying that we would expect 300 of them to be (boy,boy) and 600 would be (boy, girl)?

  • 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.

  • 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?
    • Re: (Score:3, Informative)

      by Maddog Batty (112434)

      Compare these two questions:

      "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?"

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

      Most people would think that the being born on a Tuesday bit was irrelevant and would make no difference to the answer. In fact it makes a big difference to the answer.

    • by BoberFett (127537)

      Don't try to think too hard about it, statisticians think about these kinds things in order to convince themselves that they're more intelligent than people who actually do something worthwhile with their lives. From the original problem:

      "Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?"

      They then go on to list options of:

      Boy, girl
      Boy, boy
      Girl, boy

      when birth order has nothing to do with the original question. Options one and three are the

  • 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,
  • Other problems (Score:3, Insightful)

    by Exitar (809068) on Tuesday June 29, 2010 @05:48AM (#32728002)

    "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?

    • Yes.

      It's P=(4x-1)/(2x-1) where x is the number of options for the extra information.

      So first day of year, x=365, (ignoring leap years),
      born in January x=12,
      winter x=4.
  • by tangent3 (449222) on Tuesday June 29, 2010 @05:50AM (#32728014)

    This is related to the Principle of Restricted Choice [wikipedia.org] 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.

  • by mim (535591) on Tuesday June 29, 2010 @05:54AM (#32728044)
    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 questio
    • Re:OK I'm stupid (Score:4, Informative)

      by ceoyoyo (59147) on Tuesday June 29, 2010 @08:40AM (#32729154)

      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%.

  • by itsdapead (734413) on Tuesday June 29, 2010 @06:06AM (#32728108)

    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!

    • by Asmor (775910) on Tuesday June 29, 2010 @07:16AM (#32728490) Homepage

      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...

    • Re: (Score:3, Insightful)

      by halcyon1234 (834388)

      "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?"

      "I don't know - you have not given me enough information"

      So the correct answer is "That depends. Is your other child a boy?"

  • by houghi (78078) on Tuesday June 29, 2010 @06:40AM (#32728262)

    If the dad is Schrödinger the other kid is both born and unborn at the same time.

  • by hellop2 (1271166) on Tuesday June 29, 2010 @07:04AM (#32728412)
    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.
  • by natbrooks (465129) on Tuesday June 29, 2010 @09:24AM (#32729686)

    "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.

  • by RevWaldo (1186281) on Tuesday June 29, 2010 @11:25AM (#32731628)
    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)


    .
  • by liam193 (571414) * on Tuesday June 29, 2010 @12:39PM (#32732646)

    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%

  • Dear /. (Score:3, Funny)

    by geekoid (135745) <dadinportland AT yahoo DOT com> on Tuesday June 29, 2010 @01:48PM (#32733750) Homepage Journal

    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.

"It's like deja vu all over again." -- Yogi Berra

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