From: thornley@deca.cs.umn.edu (David H. Thornley)
Subject: Re: stats help
Organization: University of Minnesota, Minneapolis, CSci dept.
Date: Tue, 29 Dec 1992 18:44:01 GMT

In article <BzzvrB.KCD@mentor.cc.purdue.edu> poutsmaj@mace.cc.purdue.edu (unknown) writes:
>I have a question regarding basic probability theory:  (I'll use baseball
>terms because that's what I'm using)  Suppose I have two teams EXACTLY
>equal to each other such that, when they play, the probability of either
>team winning is .500000 etc.  If they were to play, say, 1620 games (10
>seasons) against each other, what is the probability that they will each
>win exactly half?

Roger answered this as a formula.  If I had a copy of Stirling's formula
for estimating factorials handy I'd give you an order-of-magnitude estimate.
Without that, consider it very low.

>What would be the distribution of possible outcomes?

This is *very* close to the normal distribution with mean 810 and standard
deviation sqrt(0.5*0.5*1620) = 0.5*sqrt(1620), which is about equal to
20, a little over actually.  If you are familiar with the normal distribution,
fine.  If not, it's a bell-shaped distribution, and it means that team A
will win between 1600 and 1640 about two-thirds of the time (actually, this
gives you a good order-of-magnitude estimate of equal wins, given that you
have about 40 possibilities for about two-thirds of the chances - I'd call
it between two and three percent), and between 1580 and 1660 about 95% of
the time, and between 1560 and 1680 almost all the time.

>If I have two teams that play and each win half, what is the probability
>that the two are exactly equal ?

Now it gets tricky.  What do you mean "the probability is exactly equal"?
Probabilities are expressed as real numbers between zero and one (inclusive),
so what's the probability that one real number is equal to another one, if
they weren't derived the same way?

Further, if you believe in classical probability, the probability that the
probabilities are equal is either 1 or 0.  If, on the other hand, you're
Bayesian, your question does make sense.  If, like Roger Maynard, you think the
Jays have either a 0% or 100% chance of repeating next year, it's just that
you don't know which, you're classical.  If you think they have maybe a 15-20%
chance (or other number to taste), you're Bayesian.

Note also that this is a statistical problem, not a probability problem,
since you have gone from deducing probabilities to inferring them.

Using classical statistics, you derive confidence intervals for the desired
probability, which is a sort of word game you play; using Bayesian statistics,
you just call it another probability distribution.  In either case, if I
remember aright the confidence interval is calculated the same way as the
distribution above, so you have a two-thirds chance that the first team is
good enough to win 810 +/- 20 games, or that its winning percentage is
50% +/- 1.2% (if my arithmetic is OK) - except that you say "confidence"
not "chance" if you're classical.

>Any insight you can give me to this 
>problem would be helpful (I realize that this is probably elementary 
>probability theory, but that was one math class I never had).
>
Elementary probability and statistics, that is.

David Thornley
The world looks a bit different with dilated pupils.


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