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The Tuesday Birthday Problem

kdawson posted more than 4 years ago | from the if-it's-tuesday-it-must-be-a-girl dept.

Math 981

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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Well? (0, Offtopic)

MistrX (1566617) | more than 4 years ago | (#32727792)

Slashdot answer in 3..2..1..

Re:Well? (0)

Anonymous Coward | more than 4 years ago | (#32727804)

49%

Re:Well? (0)

Anonymous Coward | more than 4 years ago | (#32727806)

Slashdot answer with a number of wrong answers in 3..2..1..

FTFY

Re:Well? (3, Informative)

Anonymous Coward | more than 4 years ago | (#32727810)

13/27

Re:Well? (2, Funny)

Anonymous Coward | more than 4 years ago | (#32728156)

Only if you believe in randomness. If the other child in fact is a boy the probablility for it is 1.

Pigeonhole principle (0)

Anonymous Coward | more than 4 years ago | (#32727812)

Does this involve the Birthday problem and the Pigeonhole principle?

Ordering and Convergence (5, Informative)

eldavojohn (898314) | more than 4 years ago | (#32727824)

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?

Re:Ordering and Convergence (1)

Looce (1062620) | more than 4 years ago | (#32727844)

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...

Re:Ordering and Convergence (1)

eldavojohn (898314) | more than 4 years ago | (#32727874)

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 is impossible so we have 27 permutations instead of 28. Now count how many of those available 27 permutations are boys and you'll find it's 13.

Re:Ordering and Convergence (5, Informative)

dakrin9 (891909) | more than 4 years ago | (#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."

Re:Ordering and Convergence (0)

Anonymous Coward | more than 4 years ago | (#32728094)

I have never seen "DOES not disallow" in my entire life. Normal humans never emphasize a positive immediately before a negative, it has no meaning. Unless it's a positive in front of a double negative like you did; which IMO is even less clear.

Thanks a ton; my brain hurts. Copy editor wants you to try "DOES allow" next time.

Re:Ordering and Convergence (1)

Beale (676138) | more than 4 years ago | (#32727912)

This isn't true, though - the question makes no statement about whether the other can also be a boy born on Tuesday.

Re:Ordering and Convergence (1)

kiddygrinder (605598) | more than 4 years ago | (#32727978)

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.

Re:Ordering and Convergence (4, Informative)

Looce (1062620) | more than 4 years ago | (#32728018)

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:Ordering and Convergence (1)

ignavus (213578) | more than 4 years ago | (#32728062)

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.

Re:Ordering and Convergence (4, Insightful)

mulvane (692631) | more than 4 years ago | (#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:Ordering and Convergence (2, Informative)

Trepidity (597) | more than 4 years ago | (#32727886)

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.

Shorter summary (5, Funny)

williamhb (758070) | more than 4 years ago | (#32727910)

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:Ordering and Convergence (-1, Troll)

martin-boundary (547041) | more than 4 years ago | (#32727942)

The interesting thing is that the answer to this comes down to 13/27.

That answer is WRONG (and yes, that means TFA's explanation is wrong too). And I'll tell you why it's wrong:

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. Now, the fraction 13/27 depends on the combinatorics of a 7-day-week, and therefore it is WRONG. Anybody who claims that answer as correct is implicitly implying that the 7 day week length convention has a preferred physical or natural significance.

This is NOT similar to the two children problem, where the number of possibilities (BB,BG,GB,GG) is fixed by nature and is not a human convention.

Re:Ordering and Convergence (1)

A Nun Must Cow Herd (963630) | more than 4 years ago | (#32728124)

Just because it's an arbitrary division of possibilities doesn't mean it isn't useful.

Saying one of the children is a boy who's favourite number between 1 and 7 (inclusive) is 3 would give the same probability of 13/27.

You're just narrowing down which boy you're talking about - the more specific you get the closer the probability tends towards one.

"The general rule is that for a set with N possibilities you would have (N*2 - 1)/(N*4 - 1)."
The wisdom of EldavoJohn [slashdot.org]

Re:Ordering and Convergence (1)

A Nun Must Cow Herd (963630) | more than 4 years ago | (#32728136)

I meant "the more specific you get the closer the probability tends towards one HALF."

Gah!

Re:Ordering and Convergence (1)

martin-boundary (547041) | more than 4 years ago | (#32728170)

Just because it's an arbitrary division of possibilities doesn't mean it isn't useful.

True, but if the answer encodes the arbitrary division you're given, then that answer simply cannot be right. And that means somewhere there is a mistake in the interpretation or the calculation or both.

"The general rule is that for a set with N possibilities you would have (N*2 - 1)/(N*4 - 1)."

Yes, but that assumes that the set truly has N possibilities. If it has X != N possibilities and you're still assuming it has N possibilities, then you'd be merely computing the wrong thing.

Sometimes you have to look at the putative answer to see if it is believable first, even if all the steps seem solid.

Re:Ordering and Convergence (0)

Anonymous Coward | more than 4 years ago | (#32727944)

...you can't have twins because only one was born on Tuesday. ...

What constrains twins to be born at that same time?

In fact, the twins case is irrelevant - one of them must be the older one.

Re:Ordering and Convergence (1)

aussieslovethecock (1840034) | more than 4 years ago | (#32727984)

ya this is pretty dumb.

So we're just missing the implication? (0, Informative)

Anonymous Coward | more than 4 years ago | (#32728008)

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%.

Re:So we're just missing the implication? (1)

CProgrammer98 (240351) | more than 4 years ago | (#32728168)

no, go read the article again. the solution DOES include the possiibility the the other child is also a boy born on a tuesday.

Re:Ordering and Convergence (1, Funny)

Anonymous Coward | more than 4 years ago | (#32728034)

Yeah, but the chance of it being a girl is more than 1 in 2, it is about 51 percent, and then you have to allow for which country the kid was born in (to cater for female infanticide)... if you are being pedantic.

Re:Ordering and Convergence (0)

Anonymous Coward | more than 4 years ago | (#32728140)

you can't have twins because only one was born on Tuesday.

The question, as stated, does not say that. It merely states that one of the children is a boy born on a Tuesday, but makes no statement at all about the other; in particular, it does not claim that the other was not born on a Tuesday.

If the English language worked like that - if "one was born on a Tuesday" meant "precisely one", then the probability that the other child is a boy would actually be zero, since we'd already have stated that precisely one was a boy.

Re:Ordering and Convergence (1)

RichiH (749257) | more than 4 years ago | (#32728202)

They can be twins if one was born at 23:58 and the other at 00:02 or similar.

As you said, the basic problem is that "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. That being said, it's still fun, in a way.

Lies, dam'd lies... (1)

c0lo (1497653) | more than 4 years ago | (#32727836)

and statistics (umm... probability juggling based on hind-sights).

Let's try it without reading TFA (5, Informative)

Shin-LaC (1333529) | more than 4 years ago | (#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.

Re:Let's try it without reading TFA (1)

mrvan (973822) | more than 4 years ago | (#32727890)

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 :-)

Re:Let's try it without reading TFA (1)

Shin-LaC (1333529) | more than 4 years ago | (#32727926)

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 (4, Informative)

neoshroom (324937) | more than 4 years ago | (#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%."

Good job. (1)

neoshroom (324937) | more than 4 years ago | (#32727892)

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

Re:Let's try it without reading TFA (2, Insightful)

cjnichol (1349831) | more than 4 years ago | (#32727894)

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

Re:Let's try it without reading TFA (0)

Anonymous Coward | more than 4 years ago | (#32727902)

Yes it is, because "Entropy". Information theory gives a perfect explanation for this kind of puzzles.

Re:Let's try it without reading TFA (0)

Anonymous Coward | more than 4 years ago | (#32727908)

if I have two children, it is equally probable that they are born as boy+boy, boy+girl, girl+boy, girl+girl.

Except that , in real life, it is not equally probable. Ratio is 51/49. If you had a boy, chances are you'll have another.
http://www.in-gender.com/XYU/Odds/Gender_Odds.aspx [in-gender.com]

Re:Let's try it without reading TFA (1)

Trepidity (597) | more than 4 years ago | (#32727914)

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:Let's try it without reading TFA (3, Informative)

cjnichol (1349831) | more than 4 years ago | (#32727920)

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

Re:Let's try it without reading TFA (1)

Trepidity (597) | more than 4 years ago | (#32727934)

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.

Re:Let's try it without reading TFA (1)

cjnichol (1349831) | more than 4 years ago | (#32727998)

I still think that ordering should matter in the first part though (I posted right before you it seems). Why does the order matter if the other child is a girl but not if it is a boy?

Re:Let's try it without reading TFA (1)

((hristopher _-*-_-* (956823) | more than 4 years ago | (#32727950)

I'd go with 1/2.

I'm no statistician, but why compare the odds of a series?

If I flip a coin, does the previous 10 flips impact the result of particular flip? No.

Even using the odd's of a series, if you put in boy(tuesday)+girl, and girl+boy(tuesday), then why don't you use boy(tuesday)+boy and boy+boy(tuesday) also?

Re:Let's try it without reading TFA (1)

Shin-LaC (1333529) | more than 4 years ago | (#32727962)

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:Let's try it without reading TFA (0)

Anonymous Coward | more than 4 years ago | (#32727970)

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

There is a mistake in your opening argument, if you don't know the sex of any of the two children (child A and child B) the probability is equally A=boy+B=boy, A=boy+B=girl, A=girl+B=boy or A=girl+B=girl. If we say that child A is a boy that rules out A=girl,B=girl and A=girl, B=boy which just leaves us with boy, boy and boy girl. a porbility of 1/2 not 1/3

If you've already specified one child as being a boy the other one can only be a boy or a girl. It can't be a boy, a boy or a girl.

Re:Let's try it without reading TFA (2, Insightful)

William Robinson (875390) | more than 4 years ago | (#32728042)

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?

Re:Let's try it without reading TFA (0)

Anonymous Coward | more than 4 years ago | (#32728066)

Of course it is. The addition of the born on Tuesday part makes the problem *less* interesting. Suppose I said, "I have two children, one of whom is Jesus". Now, this precondition is very strong, and its probability is very low. The probability of both children being Jesus is zero or virtually zero. So the other child will not be Jesus and we don't have to consider it - answer degenerates to about 0.5.

The problem "I have two children, one of whom is a boy" is more interesting because the pre-condition appears to skew the probability for the other child. But that's only because both precondition and final test relate to the same thing - the population of boys in the family.

Adding the Tuesday or Jesus conditions tends to separate the precondition from the final test by making the precondition probability (very) small compared to the (prior) test probability, simplifying the calculation and getting us back to (about) 0.5.

Re:Let's try it without reading TFA (0)

Anonymous Coward | more than 4 years ago | (#32728086)

The probability of one boy being born on a Tuesday given two boys? I figure there are two ways to figure this out.

The way you did it, there is a 1/7th chance of the first boy landing on a Tuesday. Then you add the chances of the first boy being born on a Wed-Mon & the second boy being born on Tuesday, which is 6/7 * 1/7, that is, 6/7 for boy 1 and 1/7 for boy 2.

But, a much simpler way I'd do it would be to see that there are 6/7 * 6/7 possible ways for two boys to NOT be born on Tuesdays at all, and just subtract that from 1.

So, does that make me weird for looking at it that way. Probability isn't my strong suit.

Re:Let's try it without reading TFA (1)

Zelos (1050172) | more than 4 years ago | (#32728130)

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)?

Re:Let's try it without reading TFA (1)

FeepingCreature (1132265) | more than 4 years ago | (#32728148)

That is beautiful - and yes; of the three (out of four) events that can make you eligible for inclusion in the 900 (boy, then girl; girl, then boy; boy, then boy), two include a girl. What makes this counterintuitive is that you treat children as unordered, but they're not - in fact, your (boy, girl) case actually mixes (boy, girl) and (girl, boy), which are both equally probable.

Re:Let's try it without reading TFA (1)

CProgrammer98 (240351) | more than 4 years ago | (#32728178)

yes.

Rubbish (0)

Dynamoo (527749) | more than 4 years ago | (#32727860)

Someone who comes up with convoluted maths like that is probably the same type of person who thinks that they have a system to beat the roulette wheel.

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..

Re:Rubbish (1)

Maddog Batty (112434) | more than 4 years ago | (#32727880)

Maths fail..

.. this is the same as saying

No it isn't. Go and read the article again.

Re:Rubbish (2, Interesting)

Securityemo (1407943) | more than 4 years ago | (#32727928)

Well, if it isn't the same as saying that, then it's a distortion of information expression or an error in method, in the sense that the mathematics don't fit reality when it's your intention that they do so?

Re:Rubbish (4, Informative)

guyminuslife (1349809) | more than 4 years ago | (#32727936)

"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.

Re:Rubbish (1)

guyminuslife (1349809) | more than 4 years ago | (#32727954)

Actually, I made a mistake up there...on the first line, I meant to take the Tuesday thing out. For reasons that are explained in the article.

Re:Rubbish (2, Informative)

ArsenneLupin (766289) | more than 4 years ago | (#32727948)

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

Nope, it is equivalent to "I have just tossed a 10 pence coin twice, and I tell you that it has come up heads at least once, what is the probability that it has come up heads twice".

The 2/3 vs 1/3 probability hinges on the fact that the ordering of the kids is not defined.

If the kid's father told you "my oldest child is a boy", then you would be right.

Unfortunately, any defined order can play that role ("the first of his kids that I met in person", "the first of his kids that he mentioned", ...), which makes this problem so hard to grasp. Depending on exactly in which context he mentioned that one of his kids was a boy may change the probability of the other being a boy too from 1/2 to 1/3 or any value in between.

Re:Rubbish (4, Informative)

Joce640k (829181) | more than 4 years ago | (#32727958)

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?"

Fail (0)

Anonymous Coward | more than 4 years ago | (#32728120)

The condition "only one of which is boy born on Tuesday" places constraints on the sex of BOTH children.

Re:Rubbish (2, Informative)

CProgrammer98 (240351) | more than 4 years ago | (#32728150)

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

Umm no, if you look at the original problem, "one of my 2 kinds is a boy, what's the probability both are boys" is like saying "Ive've tossed a coin twice, at least one of which was heads, what;s the probability that both were heads?" - so here the possibilities are H H / H T/ T H of which only one of the 3 is both heads so it's 1/3rd.

It's not saying, "the first guy was a boy - or in your analagy, the first coin toss was heads" - where you would be right, the chance of the second child being a boy, or the second coin toss being heads would be 50% - it's a different problem.

Re:Rubbish (2, Insightful)

eloisefreya (1844486) | more than 4 years ago | (#32728158)

I agree - by saying "one of whom" doesn't in English preclude the fact that both of them could be boys born on Tuesday!

The difference between a man and a woman (5, Funny)

dimethylxanthine (946092) | more than 4 years ago | (#32727862)

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.

When the OP Is a spoiler (1)

CuteSteveJobs (1343851) | more than 4 years ago | (#32727864)

> 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.

its a bunch of stupid (0)

Anonymous Coward | more than 4 years ago | (#32727872)

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...

What's counterintuitive about it? (1)

Securityemo (1407943) | more than 4 years ago | (#32727888)

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:What's counterintuitive about it? (2, Informative)

Maddog Batty (112434) | more than 4 years ago | (#32727940)

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.

Re:What's counterintuitive about it? (1)

Securityemo (1407943) | more than 4 years ago | (#32728028)

...Ah, now I see it. But it's still ambigious wether the speaker means that the other child couln't be born on a tuesday. "One and exactly one" would have solved the problem, right? Or expressing the problem in lojban (presumably; i don't know any lojban either. Anyone up for a shot at it?)

Re:What's counterintuitive about it? (3, Insightful)

X0563511 (793323) | more than 4 years ago | (#32728200)

This is a question written in purposely misleading English.

This, in other words, is a shit question.

Re:What's counterintuitive about it? (0)

Anonymous Coward | more than 4 years ago | (#32727952)

"One" in this context is being taken to mean "exactly one" not "one or more". The difficulty in the problem comes from this language trick and not the probability calculations.

Re:What's counterintuitive about it? (1)

iamweasel (1217570) | more than 4 years ago | (#32727988)

That's what I thought as well. While I'm not a native english speaker, I don't think "one of whom" rules out the possibility that the other is as well.

Re:What's counterintuitive about it? (1)

BoberFett (127537) | more than 4 years ago | (#32728026)

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 same choice, because birth order wasn't in question so the options are really:

Boy, girl
Boy, boy

Summary misstates the problem (1, Informative)

bjourne (1034822) | more than 4 years ago | (#32727946)

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.

Re:Summary misstates the problem (4, Insightful)

tomtomtom777 (1148633) | more than 4 years ago | (#32728152)

"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?

Slight of hand? (1)

mlush (620447) | more than 4 years ago | (#32727972)

Devlin started by listing the children’s sexes in the order of their birth: Boy, girl Boy, boy Girl, boy

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

Re:Slight of hand? (1)

Seth024 (1241160) | more than 4 years ago | (#32728038)

But your permutations aren't equally likely!

Prob[boy,girl] = Prob[boy,boy] = 1/3

Prob[boy,girl] = Prob[Named Boy, girl]
but Prob[boy,boy] != Prob[Named Boy, boy]

It's the "you're wrong or you're right => 1/2" all over again.

Re:Slight of hand? (1)

mlush (620447) | more than 4 years ago | (#32728118)

But your permutations aren't equally likely!

I'm not sure what your getting at here.

The other sibling can be born either before or after the Tuesday Boy. that's a 50:50 chance.

The other sibling can either be a boy or a girl that's ~50:50

this gives 4 outcomes
Other child is a girl born before the Tuesday Boy
Other child is a boy born before the Tuesday Boy
Other child is a girl born after the Tuesday Boy
Other child is a boy born after the Tuesday Boy

Two outcomes result in the other child being a boy two a girl

which of these is more likely?

Prob[boy,girl] = Prob[boy,boy] = 1/3 Prob[boy,girl] = Prob[Named Boy, girl] but Prob[boy,boy] != Prob[Named Boy, boy] It's the "you're wrong or you're right => 1/2" all over again.

0.5 (1)

greekBruin (998483) | more than 4 years ago | (#32727974)

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?"

Re:0.5 (1)

guyminuslife (1349809) | more than 4 years ago | (#32728190)

No, it's slightly less than 0.5. If you add more caveats, it gets closer to 0.5.

Re:0.5 (0)

Anonymous Coward | more than 4 years ago | (#32728206)

No, it's not 0.5 and yes, adding further restrictions would change the probability.

Why, because we have selected families with at least 1 boy, so we have 2/3rd that are boy/girl and 1/3rd boy/boy. Out of the families with 1 boy and 1 girl, the probability that the boy is born on a Tuesday is 1/7, but the boy/boy families have more probability that one of the boys is born on a Tuesday (13/49, since you basically get two chances to fulfill the condition). So it becomes actually very intuitive that the probability of the other child being a boy increases when you add this condition.

another way of looking at it? (1)

A Nun Must Cow Herd (963630) | more than 4 years ago | (#32727992)

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.

Other problems (2, Insightful)

Exitar (809068) | more than 4 years ago | (#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?

Re:Other problems (1)

A Nun Must Cow Herd (963630) | more than 4 years ago | (#32728046)

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.

Re:Other problems (0)

Anonymous Coward | more than 4 years ago | (#32728100)

That's 1/P not P ;-)

General case (0)

Anonymous Coward | more than 4 years ago | (#32728010)

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}.

Principles of Restricted Choice (3, Insightful)

tangent3 (449222) | more than 4 years ago | (#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.

It's easy (0)

RenHoek (101570) | more than 4 years ago | (#32728016)

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?"
 

Re:It's easy (1)

RenHoek (101570) | more than 4 years ago | (#32728040)

Ah right.. I get it.. while previous statement is 100% correct and true, the Tuesday statement does put down restrictions on the second child and thus reducing the 50% chance..

Re:It's easy (1)

MMC Monster (602931) | more than 4 years ago | (#32728154)

Exactly. The second child is not a boy born on a Tuesday.

Re:It's easy (1)

Looce (1062620) | more than 4 years ago | (#32728054)

I don't know about the chance of it landing heads up, but if you threw it up, it has a rather high chance of being damaged by your stomach acid...

Problems with the 1 boy 2 children question (0)

Anonymous Coward | more than 4 years ago | (#32728020)

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

The real problem (-1, Flamebait)

Meneth (872868) | more than 4 years ago | (#32728036)

The real problem is that so many people are be fooled by these logical hoops to think that the answer is anything than 50% (or 49%, if you take worldwide statistics into account). I had higher hopes for humanity.

Johnny's Mom Has 3 Kids... (5, Funny)

mim (535591) | more than 4 years ago | (#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.

OK I'm stupid (1)

abigsmurf (919188) | more than 4 years ago | (#32728084)

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...

He is wrong (1)

Endophage (1685212) | more than 4 years ago | (#32728102)

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:

  • Boy, Girl
  • Boy, Boy
  • Girl, Boy

If I annotate this it may become more clear why I think they are wrong.

  • Boy (known, older), Girl (unknown, younger)
  • Boy (known, older), Boy (unknown, younger)
  • Girl (unknown, older), Boy (known, younger)

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

Science and Intuition defeating Fun Math (1, Insightful)

Effugas (2378) | more than 4 years ago | (#32728106)

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%
Door 3: 50% ...when, really, we have:

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.)

I love mathematicians... (5, Funny)

itsdapead (734413) | more than 4 years ago | (#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!

one or only one? (0)

Anonymous Coward | more than 4 years ago | (#32728134)

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.

The other problem posed in TFA (1, Insightful)

Looce (1062620) | more than 4 years ago | (#32728144)

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.

The real question (0)

Anonymous Coward | more than 4 years ago | (#32728208)

I have two children, one of whom is a boy born on a Tuesday.

What is the probability I am actually their father?

Communicating badly... (0)

Anonymous Coward | more than 4 years ago | (#32728210)

is not cleverness.
http://xkcd.com/169/ [xkcd.com]

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