Expected value of the number of different numbers drawn in 37 rounds of roulette?

Question

I need help with this problem. What is the expected value of the number of different numbers drawn in 37 rounds of roulette? Is this possible to interpret as the number of records? So the expected value would be a partial sum of the harmonic series to the 37th term?

Answer 1

An American Roulette wheel has $38$ equal probability slots: $1$-$36$, $0$ and $00$. A European Roulette wheel has $37$ equal probability slots: $1$-$36$ and $0$. We will compute for the American wheel and for the European wheel, we just change $38$ to $37$. We will also compute the expectation for $n$ spins.

Using the Generalized Inclusion-Exclusion Principle:

Let $S(i)$ be the set of $n$ spins that miss slot $i$.

Let $N(j)$ be the the sum of the sizes of all intersections of $j$ of the $S(i)$. $$ N(j)=\overbrace{\,\,\binom{38}{j}\,\,}^{\substack{\text{choices for}\\\text{the $j$ slots}\\\text{missed}}}\overbrace{\vphantom{\binom{1}{1}}(38-j)^{\large n}}^{\substack{\text{number of maps}\\\text{from $n$ spins}\\\text{to $38-j$ slots}}} $$ The number of maps from $n$ spins to $38$ slots is $38^n$.

Thus, the probability of missing exactly $k$ of the slots is $$ \sum_{j=0}^{38}(-1)^{j-k}\binom{38}{j}\left(\frac{38-j}{38}\right)^{\large\!n}\binom{j}{k} $$ which means hitting exactly $38-k$ slots. Therefore the expected number of slots hit is $$ \begin{align} &\sum_{k=0}^{38}\sum_{j=0}^{38}(-1)^{j-k}\binom{38}{j}\left(\frac{38-j}{38}\right)^{\large\!n}\binom{j}{k}(38-k)\\ &=\sum_{k=0}^{38}\sum_{j=0}^{38}(-1)^{j-k}\binom{38}{j}\left(\frac{38-j}{38}\right)^{\large\!n}\left[38\binom{j}{k}-j\binom{j-1}{k-1}\right]\\ &=\sum_{j=0}^{38}\binom{38}{j}\left(\frac{38-j}{38}\right)^{\large\!n}\Big[38[j=0]-[j=1]\Big]\\ &=38-38\left(1-\frac1{38}\right)^{\large\!n} \end{align} $$ For a European Roulette wheel, replace $38$ by $37$