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Probably not much of a hard question, but still annoys me that I can't find an easy answer.

Let's imagine we have an audience of n people and each can give a score from 0 to m (equally probable) and that score is an integer number. What is the probability that the average score for these people will be also integer?

I did try some basic calculations but did not progress too much here so I got nothing to "show". And if anything, this question is more of an .. academic interest to me. Perhaps there is a solution known already?

Alma Do
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  • There are 'easy' cases like $m=n-1$ or more generally $m=kn-1$. Can you try to work things out in those cases? – Ingix Aug 14 '18 at 09:49
  • Well.. there is no correlation between m and n (they are independent). But yes, you make a good point, these corner cases would be interesting to explore – Alma Do Aug 14 '18 at 09:51
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    I don't think it's that easy. Have a look at this question which basically asks a very similar thing as you do: https://math.stackexchange.com/questions/1746956/ – Andreas Aug 14 '18 at 10:08
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    @Andreas: The easy cases are when the number of possible choices ($m+1$) is a multiple of $n$, because then each reminder class is taken with the same probability, simplifying any calculation. Note that in the linked artice this is not the case, as the number of choices is one less than the modulus. Generally I agree that a simple formula is not to be expected for the general case. – Ingix Aug 14 '18 at 10:12
  • But may be there's at least a way to estimate this probability? – Alma Do Aug 14 '18 at 10:46
  • For example if $\frac{m}n \to \infty$, then your probability goes to $\frac1n$. – Ingix Aug 14 '18 at 12:23

1 Answers1

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I don't know if an easy closed formula exists generally.

If you want to calculate the probabilites numerically, note that it doesn't matter what score each person gives, it only matters in which reminder class$\pmod n$ the score lies. So each player chooses a reminder class$\pmod n$ according to a probability distribution $(p_0, p_1,\ldots,p_{n-1})$ (where $p_0$ is the probability that a score is from reminder class $0$ is chosen, a.s.o).

Since the scores where equally probable, with $m+1=qn+r, 0\le r < n$ we get for $i=0,1,\ldots,n-1$: $p_i=\frac{q+1}{m+1}$ if $i<r$ and $p_i=\frac{q}{m+1}$ otherwise.

Note that for $r=0$ this comes out as a nice symmetrical $p_i=\frac1n$ for all $i$.

So given 2 distributions $s_1,s_2$ of reminder classes, what is the distribution of the sum of 2 random variables $x_1, x_2$ distributed according to $s_1$ and $s_2$, resp?

That's easy to calculate:

$$P(x_1+x_2 \equiv i\pmod n) = \sum_{k=0}^{n-1}s_1[k]s_2[i-k],$$

where the index $i-k$ is taken$\pmod n$.

Given this formula, one can calculate the distribution$\pmod n$ of the sum of the scores from 2 persons, then use the formula again to add a third person, a.s.o. After all $n$ persons have been added, the probability that the sum is in reminder class $0$ is the answer to the question.

In the cases where the distribution is uniformly over all reminder classes, the formula above shows that the sum is also uniformly distributed over the reminder classes. That means adding more and more persons will not change that and the final answer is $p=\frac1n$.

It also shows that for fixed $n$ that $\lim_{m \to \infty} p = \frac1n$, because we have a fixed set of continuous calculations to do to find the answer, and the initial probability distribution goes to $(\frac1n,\ldots,\frac1n)$. In other words, if you have $n=10$ but allow scores up to $m=1000$, it doesn't matter much that the probability to choose a number of reminder class $0$ is $\frac{101}{1001}$ and not $\frac1{10}$.

Ingix
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