Minor correction:


The probability of being right if you predict it will hit on the nth play is the probability that it will not hit on the first n-1 plays and then hit on the nth play. This is

(1/40,000)(39999/40000)^n


Make that (1/40,000)(39999/40000)^(n-1) with the same conclusion. The idea is that the first play has the highest probability (1/40000) because all the other plays require n-1 consecutive failures immediately prior to the first success.