Jimmy Snyder said:
Here is a brute force proof that the probability is 13/27.
Actually, it isn't. It falls into the error described by http://fox-lab.org/papers/Fox&Levav(2004).pdf . Just read the abstract if you don't believe me.
A proper solution does not count cases, it sums the probability that each case would produce the observed result. If every case that is possible can't produce any other result, the two methods are the same. This is true for many other simple probability problems, which is why people mistakenly believe that counting is correct. It isn't.
To do it right, assign a number to each of the 196 (originally) equally probable outcomes in your grid, representing the probability that the parent of that family would say "One is a boy born on a Tuesday."
Obvioulsy, 196-27=169 get a zero this way. We don't really want to ignore them, we add a zero for them. It amounts to the same thing, I know, but it helps conceptually to recognize that you add their contribution.
One cell, "b3b3," gets a 1 this way. Assuming the parent is going to give us some information in this form, there is only one thing that parent can say.
The rest get a number somewhere between 0 and 1, inclusive. I called it P. There are 12 P's in the blue cells you outlined, and 14 in the red cells. That makes the answer (1+12P)/(1+12P+14P), as I said before.
There's one contingincy you haven't calculated. What is the probability if we don't know what he would have done. In other words, what is the answer to the question as it was put?
(1+12P)/(1+26P). If we don't know, we can only represent the answer as a function of what it is we don't know. The problem statement does not allow us to state categorically what P is, but assuming anything other than 1/2 represents a bias on the part of the parent, and we can't assume a bias. We can assume the lack of a bias, which means P=1/2, and the answer is 1/2.
Look up the
Monty Hall Problem on Wikipedia. Some mathematicians do the same thing for it. The answer depends on what bias Monty Hall has, toward opening whatever door you saw him open. The probability your door has the car is P/(1+P), and that the other door has it is 1/(1+P). Since P/(1+P)<=1/(1+P) for any P, you should always switch. But the laws in the USA require P=1/2, so the answers really are 1/3 and 2/3.
There is one other nasty side-effect if you assume P=1. When you ask the same question but with "...born on a Wednesday," you need to put a different number in each cell, and the numbers in each cell can't add up to more than 1. Same for any other day. Since at most two numbers can be non-zero, they must add up to 1. So if you put a 1 in cell b3b4 for "Tuesday," you must put a 0 in it for "Wednesday." The maximum probability you can get for this second question is 11/25. Then 9/23, and 7/21, etc., for other days. But if they are all to be the same, it can only be 1/2.
All you need to do to see I am right, is to look at what I said about the Law of Total Probability. That is a relationship that cannot be broken in probability. It essentially says that the average of the answers for all possible days
MUST be 1/2.
+++++
For peteb: you overlooked one part of the solution you linked to (with emphasis added):
Julie Rehmeyer said:
Everything depends, [Yuval Peres of Microsoft Research] points out, on why I decided to tell you about the Tuesday-birthday-boy. If I specifically selected him because he was a boy born on Tuesday (and if I would have kept quiet had neither of my children qualified), then the 13/27 probability is correct. But if I randomly chose one of my two children to describe and then reported the child’s sex and birthday, and he just happened to be a boy born on Tuesday, then intuition prevails: The probability that the other child will be a boy will indeed be 1/2. The child’s sex and birthday are just information offered after the selection is made, which doesn’t affect the probability in the slightest.
The first case corresponds to my P=1. The second, to P=1/2. What part of the statement you made do you think shoudl suggest to us that you "specifically choose him because he was a boy born on Tuesday?" Because the implication most people get, is that it is an observation.
In the simpler problem, where the dabated answers are 1/2 and 1/3, there are three approaches. The most naïve is that you are talking about a
specific child, so the "other" child has a 50% chance to be a boy. The intermediate approach is that there is no "specific" child, and that is partially true. People who take this approach compare it to saying the older chidl is a boy to emphasize what choosing a specific child does. But what they are doing, instead, is choosing a
specific gender, which is just as wrong. And in BG families, where there is only one boy, specifying "boy" is the same as specifying "older." The best answer is that however the statement was chosen, it was done randomly. Unless, of course, the problem statement is explicit about how one child, or one gender, was chosen.
