Probability for an equal number of voters for each candidate We just had city elections and, in one city, each of the two candidates got the same number of votes. 
Could you tell me what is the probability of such an event as a function of the number of voters ?
Thanks for helping.
 A: You will need to find some facts, or make some assumptions, about the electorate in order to get any kind of answer to this question.  Here is an attempt which may not be too ridiculously far from reality.
Suppose that there are $n$ electors, where $n$ is even, and every one of them casts a valid vote.  Assume that there are only two candidates, A and B, and that


*

*a proportion $p$ of voters will vote for A;

*a proportion $q$ will vote for B;

*a proportion $r=1-p-q$ is undecided and will choose between the candidates by tossing a fair coin.


We shall assume that $p,q\le\frac{1}{2}$, otherwise one of the candidates is sure to win and there cannot be a tie.  In what follows we shall assume that any numbers which need to be integers are in fact integers.  For a tie, candidate A, who definitely has $pn$ votes, must receive exactly $\frac{1}{2}n-pn$ of the $rn$ "random" votes.  Assume that the "random" votes are independent.  Then all we have to do is use a binomial distribution to get the tie probability
$$\binom{rn}{\frac{1}{2}n-pn}\Bigl(\frac{1}{2}\Bigr)^{rn}
  =\frac{(rn)!}{\bigl(\frac{1}{2}n(r+q-p)\bigr)!\bigl(\frac{1}{2}n(r+p-q)\bigr)!}
   \frac{1}{2^{rn}}\ .$$
If you would care to estimate values for $p,q$ and $r$ in your city, this gives a result in terms of $n$, which could be put into a simpler approximate form by using Stirling's formula.
Of course we have made a large number of assumptions in these calculations, and the result is highly speculative - but surely there is no real alternative unless we can get some hard data about the voters.
A: You'd have to have an even number of voters obviously, say $2n$. If you just assume random voting, there are $^{2n}C_n$ combinations of voting that give n votes to each candidate out of a total of $2^n$ and that's the probability: $^{2n}C_n / 2^n$
