Oops, it has been kindly pointed out to me that Bernie's questions is about sampling *without* replacement, which I somehow managed to miss. Sorry about that. In this case I don't think there's any neat closed formula for the expected number of trials before first choosing one of the M items out of N. It's just E(M;N) = E(# trials to the first success) = 1*[1]*(M/N) + 2*[(1-(M/N))]*(M/(N-1)) + 3*[(1-(M/N)*(1-M/(N-1))]*(M/(N-2)]+. . . + (N-M+1)*[(1-(M/N))*(1-M/(N-1))*...*(1-M/(N-M+1))]*(M/(N-(N-M)) where the last fraction is of course = 1. ----------------------------------------- There might be an interesting asymptotic formula for E(k*M;k*N) as k -> oo. (My guess, however, is that this just approaches the expectation for sampling *with* replacement, i.e., N/M.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele