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.


1 Answer 1


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: $$P=\left[\sum_{j=0}^{n-m}\left(\binom{n}{j}\left(p_3^j(p_1+p_2)^{n-j}-(p_2+p_3)^jp_1^{n-j}\right)\right)\right]^2.$$

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    $\begingroup$ The reasons I don't believe your answer are (1) it shouldn't be a perfect square and (2) in any case $m\ge n \ge 2$ it yields the wrong value. For instance, with $m=n=2$ there is only one term in the sum where $j=n-j=0,$ whence the sum equals $0,$ so you are trying to tell us $P=0.$ But enumeration of the four possible voting patterns shows $P$ must be $1/2.$ $\endgroup$
    – whuber
    Aug 18, 2021 at 12:15
  • $\begingroup$ @whuber No doubt you are right, but the special cases you mention aren't anywhere near my approximation assumptions. I have assumed that $n\gg m,$ and that the second-largest order statistic is distributed approximately the same as the maximum order statistic. $\endgroup$ Aug 18, 2021 at 13:54
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    $\begingroup$ Fair enough. But when you make an approximation, it's a good idea to be explicit about the conditions under which it might apply and, if possible, to provide error estimates or bounds. $\endgroup$
    – whuber
    Aug 18, 2021 at 13:58
  • $\begingroup$ My stats chops aren't up to providing error estimates, I don't think. Do you have any ideas for bounding the error on my estimate? $\endgroup$ Aug 18, 2021 at 14:08
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    $\begingroup$ I would start by recognizing the hypergeometric function as being both (a) an incomplete Beta integral and (b) a Binomial CDF. The latter can be approximated with the Normal CDF. $\endgroup$
    – whuber
    Aug 18, 2021 at 14:54

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