robdickerson.net

I don’t usually read Jeff Atwood, but Paul pointed out to me this post because he knows I like math. The post poses this 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?

Commenters argued contentiously between 1:2 and 2:3 (and there were some arguments over semantics as well, but I’m focusing on the math right now). Atwood made this follow-up post claiming the correct answer was 2:3.

I’m not so sure this is necessarily correct.

Atwood’s reasoning is that the possible child combinations are: BB, GB, BG, and GG. We know one child is a girl, so that eliminates BB, and 2:3 remaining options have a boy. The sticking point many people have is: why are GB and BG two separate cases?

Atwood’s answer is to draw analogy with The Unfinished Game, a problem he formulates thusly:

Two players, Harry and Ted, place equal bets on who will win the best of 5 coin tosses. In each round, Harry always chooses heads (H), and Ted always chooses tails (T). Suppose they are forced to abandon the game after 3 coin tosses, with Harry ahead 2 to 1. What is the fairest way to divide the pot?

The “intuitive” answer is that the only possible continuations of the game are: H, TH, or TT (since Harry wins on the next H) so 2/3 of the pot should go to Harry since he wins 2:3 of the expected games. The real answer is that the HT and HH continuations affect the expected games equally “strongly” as TH and TT, so Harry actually has a 3:4 chance of winning. Jeff demonstrates this with a code simulation.

The implied argument is that, just like HT and HH are “equally strong” in the Unfinished Game, so are the GBs and BGs in the child gender problem.

Again, I don’t think this is necessarily the case.

In Atwood’s original problem, there is the complication of the mother deciding to tell you one of her children is female. I think the answer hinges on the behavior of the mother.

Here is my answer to Atwood’s original child gender problem. It’s a two-parter, depending on how the mother acts:

• Scenario 1: When deciding which child’s gender to report, the mother always picks a female if she can.

This yields the 2:3 answer.

If we do a large number of simulations where we randomly generate two genders sequentially, we’ll get BB, GB, BG, or GG equal numbers of times.

We know the mother told us one child is a girl. So 100% of the time we generate a BB, it never makes it to the mother reporting a female gender. So, that gender generation doesn’t count and we throw it out.

We know the mother will always pick G if she can, so 100% of the BGs, BGs, and GGs meet the assumptions of our scenario — they make it to the “mother reported a girl” phase. So, Atwood’s original logic applies and 2:3 times the second child will be a boy.

To demonstrate, here’s a simulation in Python 3.0:

import random
 
genders = ["B","G"]
 
def spawn(n):
    return [random.choice(genders) for i in range(n)]
 
def run(times):
 
    boy = 0
    girl = 0
 
    valid = 0
 
    for i in range(times):
 
        children = spawn(2)
 
        if not "G" in children:
            # This scenario is was already thrown out
            # by our initial assumption -- doesn't count
            continue
        else:
            valid += 1
 
        if "B" in children:
            boy += 1
        else:
            girl += 1
 
    boyRatio = float(boy) / valid
    girlRatio = float(girl) / valid
 
    print("Boy: %f" % boyRatio)
    print("Girl: %f" % girlRatio)

>>> run(10000)
Boy: 0.669774
Girl: 0.330226

• Scenario 2: The mother randomly picks a child whose gender to report.

This yields the 1:2 answer.

Imagine running the simulation as before. Just as in the previous simulation, we have to throw out scenarios where the mother reports B, since we were assuming we had arrived at the point where she reported G. Previously, since she always reported G in the BG and GB cases, we only had to throw out the BBs.

This time, however, when we get a BG, the mother will choose B or G to report at random. 50% of the time, the mother will report B, which means we will have to throw that simulation out — just as we do the BBs. Similarly, 50% of the BGs will not count. The BGs and GBs are both at “half strength” since half of them don’t make it to the “mother reported a girl” stage. So, we’ll get (BG or GB) the same amount of time we get GG, giving us a 50-50 shot at a boy.

Here’s the Python simulation for this case:

import random
 
genders = ["B","G"]
 
def spawn(n):
    return [random.choice(genders) for i in range(n)]
 
def run(times):
 
    boy = 0
    girl = 0
 
    valid = 0
 
    for i in range(times):
 
        children = spawn(2)
 
        random.shuffle(children)   
 
        childX = children[0]
        childY = children[1]
 
        if childX != "G":
            # This scenario is was already thrown out
            # by our initial assumption
            continue
        else:
            valid += 1
 
        if childY != "B":
            boy += 1
        else:
            girl += 1
 
    boyRatio = float(boy) / valid
    girlRatio = float(girl) / valid
 
    print("Boy: %f" % boyRatio)
    print("Girl: %f" % girlRatio)

>>> run(10000)
Boy: 0.495971
Girl: 0.504029

The crucial difference in the simulations is:

if not "G" in children:
            # This scenario is was already thrown out
            # by our initial assumption -- doesn't count
            continue

versus:

if childX != "G":
            # This scenario is was already thrown out
            # by our initial assumption -- doesn't count
            continue

So, as far as I can tell, the real answer is that we don’t know enough about the mother’s behavior to give a definitive answer of how this scenario plays out on average.

Of course, this kind of reasoning is pretty tricky to pull off correctly, so I wouldn’t be surprised if a good counter arises. But I’ve thought about this problem for a while, and, for now at least, I’m pretty thoroughly convinced that this solution is correct.

This entry was posted on Friday, January 2nd, 2009 at 4:05 am and is filed under math. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.