If I spin the roulette (numbers from 0 to 36) N consecutive times, how many
different numbers should I expect to get?
(If N=1, your formula should give 1 as an answer, but for N=2, it should be a little under 2, and for all other N, the formula should be less than the minimum of N and 37.)
Another question: how many times should you spin the roulette, so chances are better than one in a million, that all 37 numbers will have appeared?
Part 1 is the same situation as in Playing Lotto, and the solution is similar: plagiarizing my own post on that,
Each of the 37 numbers has 36/37 probability of not being hit on a particular roll, and (36/37)^N of not being hit on any of the n spins.
While the choices are not independent, that is, if one number is known not to have occurred, the others are all the more likely to have occurred, the expected value of the total of all these 1's (present) and 0's (absent) is still the sum of the expected values of each event (i.e., the event that the first number, second number, etc. is present).
So the expected number of nonhit numbers is 37*(36/37)^N, and the expected number of different numbers hit is then 37*(1(36/37)^N)

Posted by Charlie
on 20050323 15:05:14 