Chapter 12 : Subtleties of conditional probability

Conditional probabilities are quite subtle. Apart from the common mistake of confusing $\mathbf{P}(A\ \pmb{\big|} \ B)$ for $\mathbf{P}(B\ \pmb{\big|} \ A)$, there are other points one sometimes overlooks. In fact, most of the paradoxical sounding puzzles in probability are based on confusing aspects of probability. Let us see one.

 

There are four possibilities $BB, BG,GB,GG$, of which $GG$ has been eliminated. Of the remaining three, two are favourable, hence the chance is $2/3$ that the other child is a girl. This is a possible solution. If you accept this as reasonable, here is another question.

 

Does the addition of the information about the boy change the probability? One opinion is that it should not. The other is to follow the same solution pattern as before. Write down all the $2\times2\times7\times7$ possibilities: $BBss$ (boy, boy, sunday, sunday), $BBsm$, etc. The given information that one is a boy who was born on Sunday eliminates many possibilities and what remain are $27$ possibilities $BGt*$, $GB*t$, $BBt*$, $BB*t$ where $*$ is any day of the week. Take care to not double count $BBtt$ to see that there are $27$ possibilities. Of these, $14$ are favourable (i.e., the other child is a girl), hence we conclude that the probability is $14/27$.

Is the correct answer $14/27$ as calculated here or is it $2/3$ (since the information of the day of birth of the boy is irrelevant, why should we change our earlier answer of $2/3$?)?

We leave it as food for thought. If you want a hint, the point is that to compute conditional probabilities, it is not enough to know what the person said, but also what else he could have said. Not realizing this point is the main source of confusion in many popular puzzles in probability.