[net.math] correction to posted puzzle answer

lew@ihuxr.UUCP (08/29/83)

Argh! I misstated the formula for P(M,N) in my posting. To repeat:

The probability of correctly "sticking" on the highest of N numbers
by picking M and then stopping on the next number which is the highest
so far is:

	P(M,N) = M/N * sum i = M, N-1 of 1/i

In my posting of this answer I had 1/N instead of M/N. My description of
the derivation was correct.

As long as I'm posting, note that P(N/2,N) -> log 2 / 2 (.346574) as
N -> infinity. P(5,10) = .3728, not much worse than P(3,10) = .3987

	Lew Mammel, Jr. ihuxr!lew