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Suppose you roll a dice 100 times, How many times would you expect the most common number to show up.

I.e. roll a dice 100 times and document the frequency of each value, then repeat this process infinitely many times and take the mean of the highest frequency from each trial.

Is there a way to derive a formula or approach to calculate such a value? thanks.

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  • $\begingroup$ Mathematica gives 21.997 from a recursion for the ways to get a $k-$fold mode of $max$ after $tot$ $n-$faced dice-rolls: $$\texttt{ f[tot_,n_,max_,k_]:=f[tot,n,max,k]=If[max<=tot/n,0,}$$ $$\texttt{Binomial[n,k]Product[Binomial[tot-j max,max],{j,0,k-1}]Sum[f[tot-k max,n-k,m,j],{m,1,max-1},{j,1,n-k}]];}$$ $$\texttt{ f[tot_,n_,max_,n_]:=If[max==tot/n,Product[Binomial[j max,max],{j,1,n}],0];}$$ $$\texttt{ f[tot_,1,max_,k_]:=If[And[max==tot,k==1],1,0]}$$ $$\texttt{Sum[max f[100, 6, max, j], {max, 0, 100}, {j, 1, 6}]/6^100 // N}$$ $\endgroup$
    – user44143
    Commented Jul 9, 2020 at 4:47
  • $\begingroup$ @MattF. interesting, thanks. 100 rolls is quite arbitrary but I wonder of there is any significance to this number. $\endgroup$
    – Jonah765
    Commented Jul 9, 2020 at 16:24
  • $\begingroup$ It is oddly close to 22. If the question is reopened and the current answer is unaccepted, I might play around more. $\endgroup$
    – user44143
    Commented Jul 9, 2020 at 17:13
  • $\begingroup$ @MattF. seems as though a moderator closed it, not much I can do. It was my error posting in the wrong place originally. If you decide to look further into it perhaps let me know $\endgroup$
    – Jonah765
    Commented Jul 9, 2020 at 17:52

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According to the multinomial probability mass function formula, the expected maximum frequency in $n$ rolls of a fair die is $$e_n=\frac1{6^n}\sum\frac{n!}{x_1!\cdots x_6!}\,\max(x_1,\dots,x_6),$$ where the sum is taken over all $n$-tuples $(x_1,\dots,x_6)$ of nonnegative integers such that $x_1+\dots+x_6=n$. There seems to be no reason for the existence of a simpler expression for $e_n$.

Mathematica computes $$e_{30}=\frac{3063261583291047469655}{383808888404050968576}$$ in about 7 sec, and $$e_{40}=\frac{936567872552422596737147305735}{92829823186414819915547541504}$$ in about 33 sec. It will likely take too long to compute $e_{100}$.

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  • $\begingroup$ thanks so much for the response, this was really informative. $\endgroup$
    – Jonah765
    Commented Jul 8, 2020 at 21:42

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