A Probability Brain Teaser

A little probability brain teaser for all of you… I was given this last week and I’m still having problems working out what the answer is!

A family has 2 children. When you knock on the door one of the children, a boy, opens the door. What is the probability that the other child is a boy?

I’ve got an answer in my head but I can’t quite figure out which one is correct so it would be great to hear your thoughts on what you think the answer is!

7 thoughts on “A Probability Brain Teaser

  1. 50%. That the first child is a boy doesn’t affect the odds. Before the door was opened it was a 1/3 possibility that both of the children were boys. But when it turns out that the first child was a boy the odds goes up to 50% that both of the children are boys.

  2. I think it is only “Monty Hall” if there is additional selection material, the decision to open the empty box – in this case something like “boys must answer the door before girls”. Therefore, as Robin says the odds of the sex of the second child don’t change

  3. Hi, I’m actually studying for sales and trading right now and I think I figured it out. It’s conditional on the fact that the first door is boy. So its probability of a boy GIVEN the first is a boy. That equals to P(B and B)/P(B). If we write out all the options, we have BB, BG, and GG. That’s 3 options, so the P of Boy in 1st and Boy in 2nd is 1/3. The overall probability of boy is 1/2. So based on our conditional formula it’s 1/3 divided by 1/2, which yields us 2/3.

  4. hi guys the answer s 1/2. This is a very simple question.
    let B1 be the child that opens the door and B2 be the other boy who does not open the door. Their probabilities are independant. B1 had 50% chance of been a boy prior to opening the door, P(B1)=1/2. P(B2) = 1/2.
    We are after,
    P(B2|B1) the probability that B2 is a boy given that B1 is a boy.
    P(B2|B1) = P(B2) = 1/2

    or using conditional probability to complicate things:
    P(B2|B1) = [P(B2 n B1) / P(B1)
    = [P(B2)*P(B1|B2)] / P(B1)
    P(B2|B1) = [P(B2)*P(B1)] / P(B1) due to independance of B1 from B2
    P(B2|B1) = P(B2) = 1/2

    I provide this next solution just cos christina tried to apply it.
    rephrase the question as, what is the probability that both children are boys. Outcome space, let B=Boy, G=Girl then Space = BG, BB, GB, GG.
    assume equal likely,
    P(GG) = 1/4, P(GB) = 1/4, P(BB) = 1/4, P(GB) = 1/4

    We aftre the probability that both childrens are boys given that we saw a boy P(BB|B),

    P(BB|B) = [P(B|BB)*P(BB)] / [ P(B|BB)*P(BB) + P(B|BG)*P(BG) + P(B|GB)*P(GB)
    + P(B|GG)*P(GG)].

    Where P(B|BB) is the probability of a boy given that both childrens are boys, which has to be 100% so P(B|BB) = 1.

    Where P(B|BG) is the probability of a boy given that one is a boy and the other a girl, which has to be 50% so P(B|BG) = 1/2.

    Similar reasoning for P(B|GB) = 1/2, P(B|GG) = 0.

    therefore,
    P(BB|B) = [1*1/4] / [ 1*1/4 + 1/2*1/4 + 1/2*1/4 + 0*1/4].
    P(BB|B) = [1/4] / [ 1/2] = 1/2

    Christina, am trying to get into quantitative finance, any chance you can tell me how I can go out it cos I have been applying without any interviews. cheers

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