Puzzle 1: Kids’stuff

BBC Radio 4 recently aired an episode of their popular panel show, The Infinite Monkey Cage where they analysed the odds of various games from roulette to Monopoly. You can listen to it here. At the end of the program, they signed off with a puzzle.

Suppose you meet a man with two children. He tells you one of the children is a boy. What is the probability that the other child is also a boy?

Think about it a minute. I am sharing my answer below.

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Assuming the chances of a boy or a girl being born are 50-50, if a man has 2 children they could be:

  1. {older boy, younger boy}
  2. {older boy, younger girl}
  3. {older girl, younger boy}
  4. {older girl, younger girl}

All 4 of these cases are equally likely. So they each have a probability of ¼.

Now we are told that one of the children is a boy, this immediately eliminates the chances of option 4 {older girl, younger girl} while still telling us nothing about which of the first three options is true. So now we have 3 possible cases all of which are equally probably. Hence the probability of option 1 {older boy, younger boy} is 1/3.

Another alternative way of thinking about this is using Baye’s Theorem. Baye’s theorem says that suppose you have 2 events A and B, then the probability of A happening given that we know B has happened is given by

P(A given B) = P(A and B both happened) / P(B happened)

Where P() gives the probability.

So if we consider

A = Case that both children are boys, that is {older boy, younger boy},

Then, P(A happened) = P({older boy, younger boy}) = 1/4

B = Case that at least one of the children is a boy

Then, P(B happened) = P({older boy, younger girl}) + P({older girl, younger boy}) + P({older girl, younger boy}) = ¼ + ¼ + ¼ = ¾

P(A and B both happened) = P(A happened) = P({older boy, younger boy}) = ¼

So the final probability P(A given B) = (¼) /(3/4) = 1/3 again.

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