Probabilities, huh!

sanket asked a very interesting question in the comments to my previous post on Monty Hall Problem:

Assume that boys and girls are equally likely to be born. Let us say that a family has two children. Given that one of them is a boy, what is the probability that the other one is a boy too? (Source: One of Scott Aaronson’s (http://scottaaronson.com/blog/) lecture notes.)

Update: It turns out that after stating this problem this way here, I solved a different problem altogether. Thanks, Nikhil, for pointing it out in the comments. To keep my life simple, I’ll state a modified problem below — the one that I did solve.

Assume that boys and girls are equally likely to be born. Let us say that a family has two children. Given that one of them is a boy, what is the probability that the other one is a girl?

Most people would jump out with 1/2 as the answer. Of course, if the answer was that obvious the question wouldn’t exist. The answer is 2/3. Here I will describe two different ways of arriving at this, as well as the common mistake that leads people to 1/2.

I will start by defining the problem in more formal terms. We are doing an experiment of giving birth to two children. Let GG, GB, BG and GG denote the outcomes that the children born are girl-girl, girl-boy, boy-girl and girl-girl respectively.

So the sample space S is, S = { GG, GB, BG, BB }

Since each outcome is equally likely, the probabilities of the outcomes, P(GG) = P(GB) = P(BG) = P(BB) = 1/4

We are given that one of the child is a boy.

We are interested in the event E that the other child is a girl, E = { GB, BG }

But before seeing the correct solutions, let us see one of the the flawed solutions that generally leads people to the conclusion that the the probability is 1/2.

S = { GG, GB, BG, BB }

P(GG) = P(GB) = P(BG) = P(BB) = 1/4

E = { GB, BG }

P(E) = P(GB) + P(BG) = 1/4 + 1/4 = 1/2

Voila! Of course this is wrong, the first of correct solutions below should make it clear why it is so.

Now the correct solutions:

1. Let’s just count.

S = { GG, GB, BG, BB }

Since we are given that one of the children is a boy, our sample space reduces to

S’ = { GB, BG, BB}

See how GG is conspicous by its absence? That is because the outcome GG cannot occur (wow, a boy cannot be girl!)

In this reduced sample space, our probabilities become

P(GB) = P(BG) = P(BB) = 1/3

E = { GB, BG }

P(E) = P(GB) + P(BG) = 1/3 + 1/3 = 2/3

Huh!

2. Using Bayes’ Rule.

S = { GG, GB, BG, BB }

P(GG) = P(GB) = P(BG) = P(BB) = 1/4

E = { GB, BG }

Let F be the event that at least one of the child is a boy, F = { GB, BG, BB }

We are looking for probability of E given F, P(E | F)

Using Bayes’ rule,

P(E | F) = (P(E) * P(F | E)) / P(F)

Now, P(E) = P(GB) + P(BG) = 1/4 + 1/4 = 1/2

P(F) = P(GB) + P(BG) + P(BB) = 1/4 + 1/4 + 1/4 = 3/4

Because E is the even that there is one boy and one girl and F is the even that there is at least one boy, P(F | E) = 1

So, P(E | F) = (1/2 * 1) / (3/4) = 2/3

The observant might notice that we did not reduce the sample space in second solution. This is no trick: we could reduce the sample space and would still end up with the same answer; try it yourself. But there is no need to do any such thing when using the Bayes’ rule because the fact that the sample space is actually reduced is inherently captured. In fact, many a time it is not possible to simply “reduce the sample space and proceed” — Bayes’ rule really shines then.