One of the groups I belong to on LinkedIn posted an interesting problem that was published in a New Scientist article on the Gathering for Gardner meeting. The problem goes like this:
I have two children.
One is a boy born on a Tuesday.
What is the probability I have two boys?
Can the fact that you were told that one of the boys was born on a Tuesday tell you anything about whether both children are boys?
Indeed it can!
You can go to the end of the New Scientist article to see how to solve the problem by enumerating all the possible cases. However, if you want to see how to solve it using Bayes’ Theorem, stay right here.
We want to compute the probability that both children are boys given that we know that one child is a boy born on Tuesday and the other child’s sex and weekday of birth are unknown:
P(B1 ^ B2 | (B1Tu ^ S2W2) v (B2Tu ^ S1W1))
where B1 ^ B2 represents that they are both boys.
B1Tu ^ S2W2 represents that child #1 is a boy born on Tuesday and the sex and weekday of child 2 is unknown, and similarly for the last term.
Use Bayes theorem to flip the context around…
This gives us
P((B1Tu ^ S2W2) v (B2Tu ^ S1W1) | B1 ^ B2) * p(B1 ^ B2) /
P((B1Tu ^ S2W2) v (B2Tu ^ S1W1))
This is in the form AB/C, we will handle C first using the sum rule.
P((B1Tu ^ S2W2) v (B2Tu ^ S1W1)) = P((B1Tu ^ S2W2)) + P((B2Tu ^ S1W1)) – P((B1Tu ^ B2Tu))
Its the fact that the OR is not disjoint that makes this tricky.
Now continue to decompose each of the three terms using the product rule…
P((B1Tu ^ S2W2)) = P(B1Tu) P(S2W2) = (1/2*1/7)(1) = 1/14
P(S2W2) = 1 since the child has a gender and was born on some day of the week. Technically one can just marginalize over S2 and W2.
Similarly
P((B2Tu ^ S1W1)) = P(B2Tu) P(S1W1) = (1/2*1/7)(1) = 1/14
and
P((B1Tu ^ B2Tu)) = P(B1Tu)P(B2Tu) = (1/14)(1/14) = 1/(14)^2
We now have for the denominator
C = 1/14 + 1/14 – 1/(14)^2 = 27/(14)^2
Now on to the numerator A
P((B1Tu ^ S2W2) v (B2Tu ^ S1W1) | B1 ^ B2)
It decomposes similarly
P((B1Tu ^ S2W2) | B1 ^ B2) + P((B2Tu ^ S1W1) | B1 ^ B2) – P((B1Tu ^ B2Tu) | B1 ^ B2)
For the first term:
P((B1Tu ^ S2W2) | B1 ^ B2) = P(B1Tu | B1) P(S2W2 | B2)
Lets break this down more carefully
P(B1Tu | B1) = 1/7
P(S2W2 | B2) = 1, since the boy had to be born on one of the seven days (marginalization again).
So
P((B1Tu ^ S2W2) | B1 ^ B2) = 1/7
The second term is by symmetry
P((B2Tu ^ S1W1) | B1 ^ B2) = 1/7
The last term is
P((B1Tu ^ B2Tu) | B1 ^ B2) = P(B1Tu | B1) P(B2Tu | B2) = 1/7 * 1/7 = 1/(7)^2
So we have for the term A
A = 1/7 + 1/7 – 1/(7)^2 = 13/(7)^2
Last we have B = p(B1 ^ B2) = P(B1) P(B2) = 1/2* 1/2 = 1/4
The numerator is then
AB = 13/(7)^2 * 1/4 = 13/(14)^2
All together
P(B1 ^ B2 | (B1Tu ^ S2W2) v (B2Tu ^ S1W1)) = 13/(14^2) / 27/(14^2) = 13/27
Not quite as easy as enumerating cases, but far easier that designing a Bayes net.
As I mentioned above the same result can be obtained by counting all the possible cases.
This is far from the 1/3 probability that we would have obtained had we been told that one child was a boy. If 13/27 was intuitively obvious to you, you are far better than me. We are pretty poor at working with probabilities, which is why we are so easily fooled by mis-represented statistics. The fact that we are told that one child is a Boy born on Tuesday does provide extra information, and surprisingly this information can assist you in inferring the probability the other child is a boy as well. Here we see that it is almost a 50% chance that the other child is a Boy! Being told that the Boy was born on Tuesday implies that the person may very well have had a choice of which Boy to talk about. This little piece of information changes the probability that the other child is a Boy by about 16%.
Exercise:
Show that had you been told that one child was a Boy, then the probability that the other child is a Boy is 1/3.
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Enjoy,
Kevin
