Trouble with boys

The first thing to remember about probability questions is that everyone finds them mind-bending, even mathematicians. The next step is to try to answer a similar but simpler question so that we can isolate what the question is really asking.

So, consider this preliminary question: "I have two children. One of them is a boy. What is the probability I have two boys?"

This is a much easier question, though a controversial one as I later discovered. After the gathering ended, Foshee's Tuesday boy problem became a hotly discussed topic on blogs around the world. The main bone of contention was how to properly interpret the question. The way Foshee meant it is, of all the families with one boy and exactly one other child, what proportion of those families have two boys?

To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3.


The given solution included sloppy logic. Here are a few ways to use proper logic to analyze this puzzle.

Method 1: If BB, GG, BG, and GB refer to genders of children without regard to birth order, then BG and GB are the same, and so are duplicated. The correct possibilities are BB (boy, boy), and BG (boy girl), given that we already know there is one boy. Unfortunately, we can't just jump to the conclusion that we have a 1/2 chance of there being two boys. We need to make sure we are being precise. Namely, we need to make sure the sample populations are the same type of population.

Consider our problem, we have a household with 2 children, one of which is a boy, and one of which is currently unknown. Therefore, our sample population must consist of 2 children, where one is a boy, and one is unknown. Naturally, when you sample all households with 2 boys, you'll get all those households. So the next probability of getting a household with 2 boys is:

25% * 100% = 25%

Where we have taken the format of:

(fraction of all households that have two boys) * (fraction of these households which meet the one-boy-one-unknown critera) = (resulting fraction of all households that have two boys and meet the one-boy-one-unknown assumption via random sampling)

Now, for households with 2 girls, they won't make it into this sample, as they have no boys:

25% * 0% = 0%

Finally, households with a girl and a boy. These make up half of all households. However, to get the one-boy-one-unknown required sample set, we have to sample these households. Half of the time, we'll end up picking the girl as our "known" child, and so have to remove that household from the overall population. There other half will be considered as they will be one-boy-one-unknown households:

50% * 50% = 25%

Observer, only 25%+25% of the original households made the cut, so our "total" is 50%.

BB: 25%/50% = 50% chance of having two boys when you have one boy and one unknown child.

BG: 25%/50% = 50% chance of having one boy one girl, when you have one boy and one unknown child.

Method 2:Let BB, BG, GB, and GG denote children genders AND the specified birth order. Each has a 25% chance of appearing in a house in the overall population. However, we are given that we have a boy in one household, we are not told however whether this boy is the younger or the older of the children. Therefore, let's look at our real possibilities, where we let "b" denote the child we know about in the birth order:

bB

Bb

bG

Gb

If we knew all these probabilities were equal, then we could say that there's a 50% chance (again) of there being two boys, we know that one child is a boy, and nothing else. However, I don't feel like working through the logic to show all those have the same probability (probably take the route of looking at sub populations where all eldest children are boys, etc). I won't take that route as there is a much simpler one to take.

Method 3:Each child has a 50% chance of being a boy, and a 50% chance of being a girl (actually, not strictly true, but we are assuming it for this problem). One child is revealed as a boy. Therefore there is a 100% chance that he is a boy. A second child has a 50% chance of being a girl, and a 50% chance of being a boy. So:

Boy boy: 100% * 50% = 50%

Boy girl: 100% * 50% = 50%

There we go, 50% chance of there being two boys. If two events are independent (and baring research showing that having one child of a given gender affects the gender of future children, these are independent) then the net probability is simply the product of the individual events' probabilities. This was the answer I got immediately upon reading this question, and seeing a result of 1/3 was a major WTF moment.

The 2/4 is the score I would give to someone that turned this in as homework to me. A failing grade, but one that demonstrated at least a basic knowledge of how probabilities work.