From a deck of n cards, we cut—randomly select with replacement—10 different cards before cutting a card we had seen before. We are trying to determine the probability of achieving this result for n from 10, the lowest n for which result is possible, up to n < 30.
{10\over10}\cdot{9\over10}\cdot{8\over10}\cdot{7\over10}\cdot{6\over10}\cdot{5\over10}\cdot{4\over10}\cdot{3\over10}\cdot{2\over10}\cdot{1\over10} \cdot{10\over10}\ =\ {9!\,10\over10^{10}}\\[16pt] \operatorname P n \propto {(n-1)!\,10\over (n-10)!\,n^{10}}
For n = 10, the calculation is simple and shown above. For n = 11, is the calculation more complex? No: 10!9/1110 works! However… this does not work for n = 12. 11 * 10 … 2 does not match the equation used above to derive the formula. It should be 11 * 10 * … 3, of course. With that error avoided, the formula above is produced, and this can easily produce a probability distribution. The expected value of n is then easily calculated to be ≈23.31.