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“I Have Two Children, And At Least One of Them Is A Boy”

13 May

20130513-130322.jpg

I’m reading Yudkowsky’s post My Bayesian Enlightenment and I thought I’d write out why (I think) his reasoning in this post is correct.

First, I’ll write the two versions of this question that he writes:

I remember (dimly, as human memories go) the first time I self-identified as a “Bayesian”. Someone had just asked a malformed version of an old probability puzzle, saying:

If I meet a mathematician on the street, and she says, “I have two children, and at least one of them is a boy,” what is the probability that they are both boys?

In the correct version of this story, the mathematician says “I have two children”, and you ask, “Is at least one a boy?”, and she answers “Yes”.

So why do they yield two different answers even though they are essentially, as in, plain-English-end-result-you-knew-what-I-meant the same? Because of the assumptions that go into each question. And you have to assume something, contrary to the oft-repeated dig “when you assume you make an ass out of u and me!!111!!one”

The first question — the mangled version — if we put it into prior probability format:

If I meet a mathematician on the street, and she says, “I have two children, and at least one of them is a boy,” what is the probability that they are both boys?

The prior probability is the statement “I have two children and at least one of them is a boy”. In this formulation, because of the “and”, the prior probability has to include at least one boy. This means there are only two options for the prior: Boy-Boy (BB) and Boy-Girl (BG). P(BB) + P(BG) = 1.00, meaning that only those two options exhausts all of our possibilities, so P(BB) = .5. The format of the question already answers itself; the answer is .5.

The second question — the actual version — if we put it into prior probability format:

In the correct version of this story, the mathematician says “I have two children”, and you ask, “Is at least one a boy?”, and she answers “Yes”.

The prior probability for this one is the statement “I have two children”. Now there are three possibilities for the prior probability: P(BB) + P(BG) + P(GG) = 1.00.

Then you come in and ask a question, adding more information to the prior probability: “Is at least one a boy?” and the mathematician says “Yes”.

Since we are adding information to the mathematician’s statement (i.e. the prior probability) we use BT:

P(BB) = .33
P(B | BB) = .???
P(B | BG) = .???
P(B | GG) = 0.0

Now we have to solve for the two remaining conditionals. Well, think about what the conditional probabilities are saying, and then realize that the conditional probabilities don’t necessarily have to add up to 1.00. So what is the probability that she would have at least one boy given that she has two boys? What is the probability that she would have at least one boy given that she has a boy and a girl?

You don’t even have to know what those probabilities are. They’re the same. And what happens when they are the same? Bayes Factor is 1. Which means that the prior probability doesn’t move. Meaning that the prior probability stays the same; P(BB | B) = .33.

The difference between the two formulations of the problem is that one has the information (the one boy) included in the prior probability and the other one does not.

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1 Comment

Posted by on May 13, 2013 in Bayes

 

One response to ““I Have Two Children, And At Least One of Them Is A Boy”

  1. PalmerEldritch

    July 12, 2013 at 7:16 am

    Well you got both answers correct but for totally the wrong reasons 🙂

    The prior probabilities (and they’re the same for both problems) are P(BB)=P(GG)=P(BG)=P(GB)= 1/4
    In other words, of 100 2 children families 25 will have 2 boys, 25 will have 2 girls and 50 will be of mixed gender.

    In problem 1 the condition is: “what is the probability someone who tells you the gender of one of their children will mention a boy?” Let’s call this Event B. Then
    P(Event B) = 1*1/4 = 1/4 if family is (BB)
    P(Event B) = 1/2*1/4 = 1/8 if family is (BG)
    P(Event B) = 1/2*1/4 = 1/8 if family is (GB)
    P(Event B) = 0*1/4 = 0 if family is (GG)
    Total P(Event B) = 1/2. Using Bayes Theory we get
    P{(BB)!Event B} =P(BB)*P{Event B!P(BB)}/P(Event B)=(1/4*1)/(1/2)=1/2

    In problem 2 the condition is: “what is the probability someone answers Yes to the question “Is at least one a boy?”. Let’s call this Event B. Then
    P(Event B) = 1*1/4 = 1/4 if family is (BB)
    P(Event B) = 1*1/4 = 1/4 if family is (BG)
    P(Event B) = 1*1/4 = 1/4 if family is (GB)
    P(Event B) = 0*1/4 = 0 if family is (GG)
    Total P(Event B) = 3/4. Using Bayes Theory we get
    P{(BB)!Event B} =P(BB)*P{Event B!P(BB)}/P(Event B)=(1/4*1)/(3/4)=1/3

    When information is volunteered (as in problem1 ) you have to consider what other information could have been given , but wasn’t, under the same circumstances and assign probabilities accordingly.

     
 
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