Probability of a draw between at least two options in an election with n voters and m options I hope that this question has not been asked before, but i couldn't find an answer to it online or by myself. I came by that question during a project of mine:
If i have an election with n voters and m possible choices, and the choices are equally likely selected by each voter, what would be the probability of at least 2 options receiving the same amount of votes (which is higher than the amount of votes for the other options, hence a shared majority between two or more choices)?
I assumed this would be just combinatorics, but couldn't solve it by myself.
Any insights are greatly appreciated. Thanks in advance, and again sorry if this has been asked before.

Each candidate has probability $1/m$ of being voted for by any voter, a constant. However, each candidate's total vote count is not independent of the other candidates' vote counts, because each voter can only vote once. If a voter votes for candidate $A$, the vote cannot also be cast for candidate $B.$ This is a multinomial distribution, and the joint probability that each candidate $i$ will have $y_i$ votes is given by
\begin{align*}
p(y_1,y_2,\dots,y_m)
&=\frac{n!}{y_1!\,y_2!\cdots y_m!}\,(1/m)^{y_1}(1/m)^{y_2}\cdots(1/m)^{y_m}\\
&=\frac{n!}{y_1!\,y_2!\cdots y_m!}\,\frac{1}{m^n},
\end{align*}
since $\sum y_i=n.$ But if we assume that $n\gg m,$ we can examine each candidate's distribution as approximately binomial (and independent of the other candidates' distributions). This would enable us to use order statistics to answer the question. Under this approximation, a single candidate's distribution would be
$$f(y_i)=\binom{n}{y_i}(1/m)^{y_i}(1-1/m)^{n-y_i}.$$
Order statistics with discrete distributions are somewhat annoying - the analysis is quite different from continuous distributions. The wiki page has a good section on how to discover the distribution of the maximum order statistic. Following that procedure, we define:
\begin{align*}
p_1&=P(Y<y)\\
&=F(y)-f(y)\\
p_2&=P(Y=y)\\
&=f(y)\\
p_3&=P(Y>y)\\
&=1-F(y).
\end{align*}
Note that
$$F(y):=\sum_{x=0}^{y}\binom{n}{x}(1/m)^x(1-1/m)^{n-x}.$$
This has rather messy closed-form expression involving the regularized hypergeometric function:
$$F(y)=\left(\frac{m-1}{m}\right)^{\!\!n}\left(\!\left(\frac{m}{m-1}\right)^{\!\!n}-(m-1)^{-y-1}(y+1)!\binom{n}{y+1}\,_2\tilde{F}_1\left(1,-n+y+1;y+2;\frac{1}{1-m}\right)\!\right).$$
In any case, we compute the next three quantities regarding
$Y_{(m)}:=\max(Y_1,Y_2,\dots,Y_m)$ as
\begin{align*}
P(Y_{(m)}\le y)&=\sum_{j=0}^{n-m}\binom{n}{j}p_3^j(p_1+p_2)^{n-j}\\
P(Y_{(m)}<y)&=\sum_{j=0}^{n-m}\binom{n}{j}(p_2+p_3)^jp_1^{n-j}\\
P(Y_{(m)}=y)&=P(Y_{(n)}\le y)-P(Y_{(n)}<y)\\
&=\sum_{j=0}^{n-m}\binom{n}{j}p_3^j(p_1+p_2)^{n-j}-\sum_{j=0}^{n-m}\binom{n}{j}(p_2+p_3)^jp_1^{n-j}.
\end{align*}
To finish up, we note that the $(m-1)$th order statistic is going to be very similar - we can approximate it as the same probability as what we just calculated. We wind up with:
$$P=\left[\sum_{j=0}^{n-m}\left(\binom{n}{j}\left(p_3^j(p_1+p_2)^{n-j}-(p_2+p_3)^jp_1^{n-j}\right)\right)\right]^2.$$
