4
$\begingroup$

You know that a couple has two children. You go to the couple's house and one of their children, a young boy, opens the door. What is the probability that the couple's other child is a girl?

If you list all possibilities for the sexes of two children, $BB, BG, GB, GG$, you see that $2$ of the $3$ pairs that have B (for boy) in them also have a girl, so the answer one could argue is $2/3$.

On the other hand, one could argue that the answer is $1/2$, since the probability that any one child is a girl is $1/2$, and intuitively (?) should be independent of the gender of its siblings.

Some background to possibly justify posting it here: the question was asked at an interview for an actuarial/insurance type position, and the interviewer said the answer was $2/3$, whereas my friend who was being interviewed (and has a masters in math) thought the answer was $1/2$, even after the interviewer explained his logic. My friend felt that the interviewer wasn't taking into account the fact that it is not equally likely that a boy will open the door in the $BB$ versus the $BG$ combination, and one has to take into account that fact. I have no idea which is the correct answer, both sound somewhat convincing to me (I have a Ph.D. in math, but I won't mention from where in an effort to avoid embarrassing my degree granting institution!). Anyways, any help would be appreciated and I apologize if this is too simple a question for this forum.

$\endgroup$
4

6 Answers 6

Reset to default
7
$\begingroup$

This was also posted on MathOverflow, where the answer of 1/2 was confirmed several ways. My favorite is this: There are really three independent "choices" are being made: the sex of the first (say, older) child, the sex of the younger child, and whether the older or younger child opens the door. So there are eight equally likely outcomes, and by considering them it's easy to get 1/2.

I want to mention a related game, which I believe I first saw in Martin Gardner's Aha! Gotcha.

I have three cards; one is black on both sides, one is red on both sides, and one is black on one side and red on the other. I will let you select a card at random and look at one side, and I'll bet you even money that the other side is the same color.

This sounds fair. Suppose you see black; now the card is either the black-red (which would win the game for you) or black-black (which wins for me), and at the outset, you were equally likely to have picked either one. But in fact I win the game 2/3 of the time. As with the puzzle, considering all possible outcomes, there are 6 total sides of cards. You are equally likely to be looking at any of the 3 black sides, but only one of them (the black side of the black-red card) wins the game for you. Another argument is to note that if you pick the black-black card, you are twice as likely to see a black side as if you pick the black-red card, so I am twice as likely to win.

I recommend you find this interviewer and offer to play this game with him. You could even sweeten the odds a little to make it sound more attractive. With high probability, you can clean him out. Don't forget to kick back a "finder's fee" to your friend, or at least a nice dinner (as consolation for the job).

$\endgroup$
3
  • $\begingroup$ You say the answer is 1/2 and seem to argue that the answer is 2/3. $\endgroup$
    – Aryabhata
    Aug 18, 2010 at 20:10
  • 3
    $\begingroup$ He says the answer is 2/3 for the cards example, which is correct, not for the boys/girls paradox, which should be 1/2 and he so states in the opening paragraph. $\endgroup$
    – Abel
    Aug 18, 2010 at 20:24
  • $\begingroup$ Ok, I clarified. $\endgroup$
    – Nate Eldredge
    Aug 18, 2010 at 20:32
4
$\begingroup$

Let A be the event "Family has at least one girl". Let B be the event "Boy opens door"

Pr(A and B)=1/4 *1/2 + 1/4*1/2 = 1/4

AND

Pr(B)=1*1/4 + 1/4*1/2 +1/4*1/2 = 1/2

So Pr(A|B) = 1/4 / ( 1/2) = 1/2.

$\endgroup$
3
$\begingroup$

(Pasting my mathoverflow answer...)

We will assume all the obvious implicit assumptions (eg. random child being boy of girl is 50/50, boys and girls open the door uniformly, etc.).

If you had a slightly different question, i.e. if you asked the couple if they have at least one boy, and the answer is yes, then the chance of the other one being a girl is 2/3. Intuitively, the probability is not 1/2 because in this case the answer depends on both the children, i.e. it is a function of both of them considered together.

However, if you asked the couple to pick a child at random, then she/he bears no information about the other child, and consequently his/her gender does not give you any information about the sibling.

Your case is the second case, where the child opening the door is selected at random, and she happens to be female. This does not bear any information regarding the other child.

So answer is 1/2 and your friend is correct.

Mathematically,

$P(Other\ is\ B|G\ opens\ door) = P(BG|G\ opens\ door) =$ $\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2*1/2}{1/4+1/2*1/2} = 1/2$

This can be graphically understood as below -

As a digression, a twist in the question can be brought about - if you take probabilities for a boy and girl to be different for opening the door.

Eg. suppose boy opens with probability $p$, girl with $q=1-p$, in a family with BG.

Then $P(Other\ is\ B|G\ opened\ door) = P(BG)/P(G\ opened\ door) = $

$\frac{P(BG\ and\ G\ opens)}{P(GG\ and\ G\ opens\ door) + P(BG\ and\ G\ opens\ door)} = \frac{1/2*q }{1/4+1/2*q} = \frac{2q}{2q+1}$.

(This means $q = 0$ implies $P(BG|G\ opens)=0$. That makes sense, since girls don't open the door if there is a boy, so definitely the other one is girl too.)

EDIT: Table added.

$\endgroup$
2
$\begingroup$

1/2. Imagine there are four houses in a row, each with two children. One has two boys; one has two girls; one has an older boy and a younger girl; one has a younger boy and an older girl.

Now, you go up to one of these houses at random and a child (chosen at random, say by coin flip) comes to the door. In four of the possible scenarios, this child is a boy. In two of those scenarios, the other child is also a boy. The answer is then two divided by four, or one-half.

$\endgroup$
2
  • $\begingroup$ +1, this would be a good way of explaining the situation to someone who isn't mathematically inclined (like the interviewer). $\endgroup$
    – Cam
    Aug 22, 2010 at 1:50
  • $\begingroup$ Thanks! Thanks again, because comments have to be fifteen characters or more! $\endgroup$
    – Michael Lugo
    Aug 22, 2010 at 2:42
1
$\begingroup$

The answer is 1/2 since whether first or second child opens the door is independent of if he is a boy or a girl.

$\endgroup$
0
$\begingroup$

Here's a simple way to think about it...

We know one of the children is a boy (he opened the door). Therefore there are 3 possible sets of children (numbered for later reference), which you can trivially derrive from the Boy-Girl probability tree:

  • B1B2
  • B3G4
  • G5B6

So there are 4 possibilities for the second child:

  • B1 (B2 opened the door)
  • B2 (B1 opened the door)
  • G4 (B3 opened the door)
  • G5 (B4 opened the door)

Then the chance is obviously 1/2 that the other child is a girl.

$\endgroup$

You must log in to answer this question.