# [FOM] 2 problems in the foundations of statistics

Shipman, Joe jshipman at surveyusa.com
Sat Aug 6 16:43:13 EDT 2005

```My original post did not spark a discussion as concrete as I would have like, so let me try again, focusing entirely on a specific example.

Two basketball teams have split their first two games. What is the chance they will split their next two games?

I am being purposely vague about what "what is the chance" means, but I am expecting people who answer this question to make reasonable assumptions about this and defend them.

I am aware of at least two methods to solve the problem, which involve formulas in the parameters a,b,c,d and address the more general problem "when an event has occurred a times in a+b trials, what is the chance it will occur c times in the next c+d trials?". The particular problem above is the simplest nontrivial case, a=b=c=d=1.

Method 1 (Naïve estimation):
The probability of success in a single trial is p=1/2, so the probability of 1 success in 2 trials is 2p(1-p) = 1/2. (By symmetry, the chance of 0 successes and the chance of 2 successes each equal 1/4).

Method 2 (Bayesian uniform prior):
Assume all probabilities of success p between 0 and 1 are equally likely a priori. The chance we will see 1 success in the first 2 trials is the integral of 2p(1-p)dp from p=0 to 1, or 1/3. Therefore, the a posteriori density for p, rather than being uniform, is 2p(1-p)dp/(1/3) = 6p(1-p)dp. We integrate this times 2p(1-p) to estimate the chance of 1 success in the next 2 trials. This is the integral of 12(p^2)((1-p)^2) dp, which evaluates to 12/5 - 24/4 + 12/3, or 2/5. (By symmetry, the chance of 0 successes and the chance of 2 successes each equal 3/10).

Can anyone suggest any sense in which some SPECIFIC real number other than 0.5 and 0.4 can be justified as an answer to the question posed? If not, can anyone explain to "the man in the street" why 0.4 (or, alternatively, 0.5) is the "right answer"?

I don't want an argument that "there is no right answer", I want you to make the most reasonable assumptions you can that make the question numerically answerable, and defend them.

Joseph Shipman

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