Probability of number of unique numbers in $37$ Roulette Wheel spins. I was wondering if someone could help me answer the following question:
Calculate the probability that in $37 $ consecutive spins of a roulette wheel (using an European Roulette wheel) you will have $24$ different numbers. 
 A: We interpret the problem as asking for the probability that there are exactly $24$ distinct numbers in the $37$ spins. 
There are $37^{37}$ equally likely sequences of length $37$ over the alphabet $\{0,1,\dots,36\}$. It remains to count the good sequences, the ones that have exactly $24$ distinct numbers.
Which $24$ numbers? They can be chosen in $\binom{37}{24}$ ways. Now we count the number of strings of length $37$ formed out of these $24$ letters, and use all of them. The tricky thing is that we must make sure we don't count strings that use fewer than $24$ of these letters. 
For dealing with that, we can use Stirling numbers of the second kind or Inclusion/Exclusion. We describe the Inclusion/Exclusion approach. 
For any of the chosen sets of $24$, there are $24^{37}$ words of length $37$ that use our chosen alphabet.   
We must subtract the words that use $23$ or fewer of the letters. So a first step is to subtract $\binom{24}{23}23^{37}$. However, we have subtracted too much, we have for example subtracted the words that are missing two of the digits. There are $\binom{24}{2}22^{37}$ of these. So add back  $\binom{24}{2}22^{37}$. 
We have added back too much, for we have added back more than once the $\binom{24}{3}21^{37}$ words that only use $21$ of the digits. And so on.
