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Let's say we have a tournament with 80000 people in it, where each rolls a 10 sided dice 9 times. The highest sum of the 9 dice rolls wins the tournament. How do we calculate the average sum of dice rolls that will win the tournament? Assume the dice are fair and that each roll is independent.

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Start by computing $c_{m,n}$ for all ($n\in[1,90] $ and $m\in [1,9]$) which is the number of combinations of $m$ numbers in $\{1,\ldots,10\}$ summing up to exactly $n$.

This can be done easily by recursion $$c_{m,n}=\sum_{i=1}^{10} c_{m-1,n-i}$$

With the boundary conditions:

$$\forall i\in \{1,\ldots,10\}: c_{1,i}=1$$ $$\forall i \in \{11,\ldots,90\}: c_{1,i}=0$$

The probability of some player summing at most $x$ is $$p_x = \frac{\sum_{i=1}^{x} c_{9,i}}{10^9}$$

Next, compute the actual expectation:

$$\mathbb E=\sum_{i=9}^{90} \sum_{p=1}^{80000} {80000 \choose p} i (\frac{c_{9,i}}{10^9})^p\cdot p_i^{80000-p}$$


EDIT:

I ran a short Python code to solve this. The code is here, and the result for the parameters you gave is $\approx 82.99$.

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