tear-your-hair-out probability puzzle

serendip · 16-06-2010 12:57PM #1

Problem 1 (with answer):

I have two children. One of my children is a boy. What is the probability that my other child is a boy?

Answer: 1/3

Many people find this counter intuitive, but if you work through the cases, you'll see that it is indeed 1/3.

Now, the real reason for this post ...

Problem 2 (solution to be provided later):

I have two children. One of my children is a boy born on a Tuesday. What is the probability that my other child is a boy?

Good luck!

Fremen · 16-06-2010 03:46PM

13/27.
*yawn* :pac:

Probability of at least one boy born on a tuesday is 27/196.
Probability of two boys AND at least one boy born on a tuesday is equal to probability of at least one boy born on a tuesday given that there are two boys tines the probability that there are two boys, which is 13/196. Now apply Bayes' theorem.

Problems like this are very sensitive to different interpretations of the wording, so I wouldn't be surprised if someone else gets a different but justifiable answer.

LeixlipRed · 16-06-2010 04:09PM

Exactly, question is pretty darn vague. Anyway, Spain are playing, no maths allowed :cool:

alan4cult · 16-06-2010 05:37PM

Is the second child allowed to be born on a Tuesday too?

When you say "One of my children is a boy born on a Tuesday." do you mean it in a sense that the particular child I picked out was born on a Tuesday or only one of my children was born on a Tuesday?

i.e at least one born on a Tuesday or only one born on a Tuesday?

16-06-2010 05:39PM

Its the same...still a third..

serendip · 16-06-2010 06:29PM

alan4cult wrote: »

Is the second child allowed to be born on a Tuesday too?

When you say "One of my children is a boy born on a Tuesday." do you mean it in a sense that the particular child I picked out was born on a Tuesday or only one of my children was born on a Tuesday?

My intention is exactly and only what is stated. I have two children. One of my children is a boy born on a Tuesday. I haven't stated anything about the other child, who may or may not have been born on a Tuesday.

serendip · 16-06-2010 06:33PM

Deleted User wrote: »

Its the same...still a third..

Nope. Very counter intuitive. But Nope, I'm afraid.

Fremen · 16-06-2010 06:37PM

serendip wrote: »

My intention is exactly and only what is stated. I have two children. One of my children is a boy born on a Tuesday. I haven't stated anything about the other child, who may or may not have been born on a Tuesday.

I guess that should read "At least one of my children is a boy born on a tuesday".
Told you this would happen

serendip · 16-06-2010 06:51PM

Fremen wrote: »

13/27.

Yup. I get the same answer but by a different method (just counting possibilities).

But most people's intuition is that the Tuesday seems to be independent and hence irrelevant. I find it very difficult to overcome this intuition.

Here's a suggestion.

Say you pose Problem 1 to me. If some factor truly is independent, then I should be able to ask you about it, and learn nothing about original problem (Problem 1, here).

For example, if you ask me Problem 1, I might ask you whether it rained in France yesterday. And you could answer accurately and I would learn nothing as to the original problem.

However, if you ask me Problem 1: "I have two children. One of my children is a boy" and I ask you "Was that boy born on a Tuesday?", then you cannot answer without revealing more information. If you have only one boy, you could answer "yes" or "no" -- at which point the probability of the other child being a boy goes to 0. But if you have two boys, then you cannot answer. Which boy? You didn't choose a boy in posing Problem 1, so the question cannot be answered. You'd have to say "which one?"

serendip · 16-06-2010 07:12PM

Fremen wrote: »

I guess that should read "At least one of my children is a boy born on a tuesday".
Told you this would happen

Fair enough.

RobBaxter · 06-07-2010 05:02PM

Yes it's definitely 13/27

did it out on excel to make sure!

pH · 12-07-2010 08:14PM

Here's a good writeup on it, but once again it's absolutely important to specify how the information is acquired.

http://www.sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong

Not so fast, says probabilist Yuval Peres of Microsoft Research. That naïve answer of 1/2? In real life, he says, that will usually be the most reasonable one.

Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy. If I specifically selected him because he was a boy born on Tuesday (and if I would have kept quiet had neither of my children qualified), then the 13/27 probability is correct. But if I randomly chose one of my two children to describe and then reported the child’s sex and birthday, and he just happened to be a boy born on Tuesday, then intuition prevails: The probability that the other child will be a boy will indeed be 1/2. The child’s sex and birthday are just information offered after the selection is made, which doesn’t affect the probability in the slightest.

This is the problem with this question, for most reasonable interpretations of the English used the correct answer to both questions is 1/2

Now if you want to get really specific - and make the question counter-intuitive and bullet proof here it is:

Case 1

I have a room full of mothers of 2 children families, I ask all the mothers with at least one boy to stand, and those with 2 girls to sit, now I pick a woman still standing at random, what's the probability her other child is a boy?

Now the answer is the counter-intuitive 1/3.

However :

Case 2
In the same room, all mothers standing, I ask one to tell me the sex of one of her children - she answers "boy" (but she could just as easily have said "girl") - what is the probability the other child is a boy?

The answer now is the more intuitive 1/2.

In both cases we have the same information (2 children, at least one's a boy) however we get two different answers - as how (and indeed when) the information was received makes a difference. It's very hard to read into the wording of the OP that we're talking about case 1, most people would, unless the selection criteria was made explicit infer we're talking about something more similar to case 2, hence for most reasonable readings of the problem, 1/2 is probably the better and more correct answer.

Rubecula · 16-07-2010 12:03PM

I am intrigued how you get the 1/3 for the first part.

You have two children One is male the other could be either male or female. That is a possibility of even chance either way.

that is a 1:2 chance the second child is also a boy. By 1/3 are you also bringing in the possibility of a hermaphrodite?

The odds will always be the same. Either the child is male or it is female. Odds will only alter through some sort of genetic addition to the equation, which you haven't mentioned.

Now I am confusing myself.:D

gosplan · 16-07-2010 12:17PM

Rubecula wrote: »

I am intrigued how you get the 1/3 for the first part.

You have two children One is male the other could be either male or female. That is a possibility of even chance either way.

I'm going with the poor wording angle as well.

Read it as 'I have two children and at least one on them is a boy, what are the chances that the other is also a boy'.

Blackhorse Slim · 16-07-2010 12:22PM

serendip wrote: »

Problem 1 (with answer):

I have two children. One of my children is a boy. What is the probability that my other child is a boy?

Answer: 1/3

Wrong. The answer is zero. If your other child is a boy, the second part of your statement is wrong.

Rubecula · 16-07-2010 12:23PM

Thank heavens for that I thought I was just stupid for a moment LOL:D

Fremen · 17-07-2010 03:38PM

Rubecula wrote: »

I am intrigued how you get the 1/3 for the first part.

You have two children One is male the other could be either male or female. That is a possibility of even chance either way.

that is a 1:2 chance the second child is also a boy. By 1/3 are you also bringing in the possibility of a hermaphrodite?

The odds will always be the same. Either the child is male or it is female. Odds will only alter through some sort of genetic addition to the equation, which you haven't mentioned.

Now I am confusing myself.:D

Originally, there were for possibilities. We'll use a shorthand where the first letter stands for the sex of the first kid, and the second letter the sex of the second kid. Originally, the possibilities were:
BB, BG, GB and GG.

Once you're told one kid is a boy, you know that GG is impossible, and there are only three possibilities remaining. Hence one in three.

liamw · 18-07-2010 07:14PM

Fremen wrote: »

Originally, there were for possibilities. We'll use a shorthand where the first letter stands for the sex of the first kid, and the second letter the sex of the second kid. Originally, the possibilities were:
BB, BG, GB and GG.

Once you're told one kid is a boy, you know that GG is impossible, and there are only three possibilities remaining. Hence one in three.

Aren't the BG, GB possibilities really only one possibility IF we don't care about the order in which they were born?

Grudaire · 18-07-2010 10:18PM

In this case we have two slots _,_ which can be filled as follows:
M,M
M,F
F,M
F,F

However in this case we can measure either of two things;
P[Other Child Male | At least one Male Child]
Sample space: M,M - M,F - F,M

or

P[Other Child Male | First Child is Male]
Sample Space: M,M - M,F

I would have to say the 1/2 is the answer I would have given to the problem as it is worded in the OP...

Fremen · 19-07-2010 08:47AM

liamw wrote: »

Aren't the BG, GB possibilities really only one possibility IF we don't care about the order in which they were born?

Yes, you could say that. The reason I split the outcomes into four distinct events is that each of the four events is equally likely. If you split the possible outcomes into the following three distinct events

GG
BB
*boy and girl, ignoring order*,

then the events are no longer equally likely, which would make my explanation less intuitive.

liamw · 19-07-2010 10:59AM

Fremen wrote: »

Yes, you could say that. The reason I split the outcomes into four distinct events is that each of the four events is equally likely. If you split the possible outcomes into the following three distinct events

GG
BB
*boy and girl, ignoring order*,

then the events are no longer equally likely, which would make my explanation less intuitive.

Actually what I'm saying is that, given that the problem does not specify a birth order, then there are only two other possibilities (not three): either the other child is a girl or the other child is a boy.

BB or BG (same as GB) = 1/2

So, where am I going wrong?

Fremen · 19-07-2010 12:15PM

liamw wrote: »

So, where am I going wrong?

As you said, the problem doesn't specify an order, so it's not reasonable to talk about the "other" child as you're doing.

In order to demonstrate the reasoning behind the problem, I tacitly specified an order. Once we've specified an order, this allows us to talk about the "other" child. This is the sense in which GB and BG are different. It could be that the left letter represents the younger and the right the older, or the left the taller and the right the shorter of the siblings.

pH · 19-07-2010 06:27PM

liamw wrote: »

Actually what I'm saying is that, given that the problem does not specify a birth order, then there are only two other possibilities (not three): either the other child is a girl or the other child is a boy.

BB or BG (same as GB) = 1/2

So, where am I going wrong?

Just because there are 2 possibilities, it doesn't mean they're equally likely. As to last weeks lotto, I either won it or I didn't, but that doesn't mean that each one is equally likely.

So yes, there are 2 possibilities, it's either a girl or a boy, but you can't just make a jump from there to 50/50, you need to look at the frequency of each occurrence and assign a probability to it.

If you accept for this puzzle that boys and girls are equally likely to be born and survive then you'd expect overall:

25% of 2 child families to be 2 girls
25% of 2 child families to be 2 boys
50% of 2 child families to be a boy and a girl.

Now there's another way (incorrect) of looking at this:

A 2 child family has 2 possibilities, 2 boys, 2 girls or a boy and a girl, that's 3 options, hence each has a 33% chance of occurring, this as I said above is wrong as you can't jump for 3 possibilities to equal (33%) probabilities unless you know in advance they're equally likely.

Rubecula · 19-07-2010 10:22PM

This is one of those mathematical problems that has no definitive right answer then, as it depends on the wording and the understanding of that wording.

The answer is EITHER 1/2 or 1/3 depending on your translation of the question.

Is this how you see it?

Of course if we include the very rare chance that the child is a hermaphrodite then we can have?

BG
GB
HB
BH
GH
HG
BB
GG

Which then makes the odds 1/8 but if you have to include the rarety of the H possibility then the percentage for BG GB etc is not going to be affected by a great deal.

I am not a mathematician but at this point I still think that in the original question I would favour the answer 1/2. Although I will not disagree with anyone who finds a different set of odds to be correct.

c-note · 20-07-2010 11:18PM

i have looked at the graphical solution to the problem but i dont like the idea of it.

heres how i workd it out:

there are four gender combination possible
BB BG GB GG

each of these has 49 weekday combinations per gender combination
MM MT MW MTh MF MS MSu TM TT TW TTh etc etc etc

This give a total of 196 gender-weekday combination.
we are only interested in the ones where at least one is a Tuesday boy

For BB There are 13 eligable combinations
MT, All the tuesdays(7), WT, ThT, FT, ST, SuT

For BG there are 7 eligable combinations:
TM, TT, TW, TTh, ETC

For GB there are also 7 eligable combinations:
MT, TT, WT, Tht, etc.

For GG there are no eligable combnations.

this gives a total of 27 eligable combinations 13+7+7
of which 13 are a BB combination (what we want)

13/27
nice puzzle:)

Rubecula · 21-07-2010 12:31PM

Nice puzzle indeed and with so many possible answers.

I am intrigued now. How many answers are possibly out there?

c-note · 21-07-2010 08:05PM

^there is only one correct answer.

i was thinking though...
how would this change the problem:

given a child has a 10% chance of being left handed (gender independant)

I have two children. At least one of my children is a boy who is left handed. what is the probability that my other child is a boy.?

i get

19/39

does anyone agree with this?

also if i told you i that boy was right handed, what are the chances that my other child is a boy?

99/298

. anyone agree?

hivizman · 22-07-2010 03:55PM

c-note wrote: »

^there is only one correct answer.

i was thinking though...
how would this change the problem:

given a child has a 10% chance of being left handed (gender independant)

I have two children. At least one of my children is a boy who is left handed. what is the probability that my other child is a boy.?

i get
19/39
does anyone agree with this?

Yes.

c-note wrote: »

also if i told you i that boy was right handed, what are the chances that my other child is a boy?
99/298
. anyone agree?

I make it

99/279

, which simplifies to

11/31

.

In general, if the members of a population can satisfy two independent conditions A and B with probabilities a and b respectively, and two members of the population are drawn at random, then, given that at least one of the two satisfies both condition A and condition B, the probability that the other satisfies condition A is:

(2a-ab)/(2-ab)

For example, in the case of two children, of whom at least one is a boy born on a Tuesday, a = 1/2 and b=1/7, so the probability that the other child is a boy is:

(2*1/2-1/2*1/7)/(2-1/2*1/7) = (1 - 1/14)/(2 - 1/14) = 13/27.

If there is just one condition, then we can take b as equal to 1, and the formula becomes a/(2-a). So, if at least one child is a boy, and the probability of being a boy is 1/2, the probability that the other child is a boy is 1/2 / (2 - 1/2) = 1/3.

hivizman · 24-07-2010 08:06PM

Following on from my previous post, as the second condition gets less likely, the probability b tends to zero, and the formula tends simply to a, the probability of the first condition.

For example, suppose the mother says "I have two children, and at least one of them is a boy born on 1 January". Then, ignoring leap years and assuming that births are distributed uniformly across days of the year, the probability b is 1/365, and the formula gives the probability of the other child's being a boy to be 0.499657, which is very close to 1/2. What is happening is that, as the probability of the second condition gets smaller, the mother comes closer to identifying specifically one of her children, so the probability that the other child is a boy (or a girl) comes close to being unconditional rather than conditional on the information about the first child.

Din Taylor · 29-07-2010 10:04PM

Rubecula wrote: »

This is one of those mathematical problems that has no definitive right answer then, as it depends on the wording and the understanding of that wording.

The answer is EITHER 1/2 or 1/3 depending on your translation of the question.

Afraid not - this is a classic subtle example of how conditional probability works. See http://en.wikipedia.org/wiki/Conditional_probability. This example is used in a lot in Statistics "101" courses at third level. With reards to the wording: strictly speaking the OP should have said at least one of the children is a boy. Otherwise could be interpreted as one and only one child is a boy giving the probability of the other child being a girl as one.

If it were 1/2 then the fact that one of the children is a boy wouldn't give us any other information but here it does as the GG combination of two children is out. I actually like the OPs explanation of statistical independence.

OP - I really like the second question as well. I used Bayes' Theorem myself. Took me ages to "count" all the possibilities.

pH · 30-07-2010 12:19AM

Din Taylor wrote: »

Afraid not - this is a classic subtle example of how conditional probability works. See http://en.wikipedia.org/wiki/Conditional_probability. This example is used in a lot in Statistics "101" courses at third level. With reards to the wording: strictly speaking the OP should have said at least one of the children is a boy. Otherwise could be interpreted as one and only one child is a boy giving the probability of the other child being a girl as one.

No. Even the "at least" wording is at best ambiguous, because as the question in the OP is worded it appears the information is being offered to you, you haven't been made aware of any explicit rule by which the "offerer" decided to give you the information.

For example, I toss two coins, keep one hidden, see the other one is a head and tell you in the form "I have tossed at least on head."

Everyone (I hope) can agree that the probability of the other coin being a head or tail is indeed 50/50.

This seems identical in wording to the boy girl question 1 - you've told me the sex of one of your children without specifying why you told me (and under what conditions you would have said nothing)

Note - I'm not disagreeing that you can get the question into the form where the answer is 1/3 - however to do so you need an explicit questioning step - for most normal readings of the question as asked by the OP (even including the words "at least") the correct answer is 1/2.

tear-your-hair-out probability puzzle

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