Basic Combinatorial Mathematics

Posted by Steven Suwatanapongched on January 2, 2009

I absolutely love reading CodingHorror.com. It’s easily one of my favorite blogs to read about problem solving from the viewpoint of a programming/developer. Recently, Jeff Atwood posted a problem:

Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?

I was surprised by the number of readers who responded who got it wrong, very wrong. Quite a few who commented answered one-half, 50%. Reasoning ranged from it’s always half, or because the ordering doesn’t matter– and that there are only 3 sets to choose from. The latter of which I want to discuss.

Note: B = Boy, G = Girl

Some argued that the combinations to work off of is: BB, GG, BG (3 sets) The full combination is: BB, GG, BG, GB (4 sets, where BG and GB are treated as separate sets)

The 3 set argument is that the BG, GB should be considered as one set, since order does not matter. But I beg to differ. Although order (older/younger) was not implied in the problem, it must be considered.

Here’s why: Having only 3 sets is basically saying that it’s equally likely that a couple with two children will have either a Boy-Boy, Girl-Girl, or Boy-Girl set. This is not true. It is a 50/50 chance they will have children of the same gender (Boy-Boy or Girl-Girl versus a Boy/Girl combo). So a full 4 set combo must be used to deduce the answer to the problem.

The answer then, since we know that at least one is a girl, the BB set needs to get tossed out, leaving GG, BG, GB. So then 2 out of the 3 sets can have boys, so there is a 2/3 chance that the couple’s second child is a boy.

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