[net.math] Bayes' theorem & coin problem

lew@ihuxr.UUCP (08/19/83)

The "Principle of Restricted Choice" would appear to be a restricted
form of Bayes' Theorem, which is the following:

If a sample space is divide into n subsets (called events), Hi,
and E is an event, then the conditional probability P( Hi | E ) is given by:

	P(Hi | E) = P(Hi) * P(E | Hi) / sum j=1,n of P(Hj) * P(E | Hj)

Note that an "event" is some set of sample points, not a single
sample point. There may be several ways for an event to occur.

In the coin problem, E is "picked gold first" and the Hi are,
"picked box 1", "picked box 2", and "picked box 3".
So P(H1) = P(H2) = P(H3) = 1/3 and

	P( gold first | box 1 ) = 0
	P( gold first | box 2 ) = 1/2
	P( gold first | box 3 ) = 1

so	P( box1 | gold first ) = 0
	P( box2 | gold first ) = 1/3
	P( box3 | gold first ) = 2/3

Of course, this is all just a rehashing of the analysis that rabbit!ark
gave, except that it allows you to retain the idea of picking a box, then
a drawer.

	Lew Mammel, Jr. ihuxr!lew