What is the probability that there is a re-election If there are $n$ voters and $k$ contestants ($n>k$) in an election. What is the probability that there is a re-election?
Re-election will happened only if two or more lead contestants have got equal number of votes. It is also assumed that any number of voters ($0$ to $n$) may appear on election day for voting or you may assume number of valid votes may vary between $0$ to $n$.
My approach: Any voters may vote any of the contestants or he may not appear for voting or he may cast an invalid vote. In this case, number of ways $n$ voters may cast their votes is $(k+1)^n$. This way I can calculate the total number of events in the sample space,
$$n(S) = (k+1)^n$$
But I don't know how can I calculate the number of favorable outcomes, $n(E).$
 A: Let $A_m$ denote the event that $m$ votes are valid and let $A$ denote the event that there is NO re-election. Then
$$  P(A) = \sum_{m=0}^n P(A|A_m)P(A_m) $$
where $P(A_m) = \binom{n}{m}p^m(1-p)^{n-m}$ assuming that votes are independent and their valid with probability $p$. 
Now let us look at $P(A|A_m)$ for fixed $m$. If $m=0$, then $P(A|A_m) = 1$ trivially, hence we can assume $m\ge 1$. Counting the number of favorable outcomes boils down to basically writing $m=a_1 + a_2 + \ldots + a_k$, where $a_j\ge0$ denotes the number of votes on candidate $j$, and the maximal value is unique. The number of ways of writing $m$ as the sum of $k$ nonnegative integers is actually a number theoretic question, there is no closed expression known even for the case when $k$ can vary.
Although for some concrete fixed values of $n$ and $k$ one can write a program to count all the cases...
A: This is just a partial answer, but I can't conveniently fit it in a comment box.  I will do the problem for the case $n=1000,\ k=3$ that the OP mentioned in a comment.
We are to assume that there are $4^{1000}$ equally likely outcomes, so we just have to count the number of outcomes that result in a tie.  
First, there is the possibility of a three-way tie, where each candidate gets $v$ votes, and $1000-3v$ members do not vote, for $0\leq v\leq333.$  The number of ways this can happen is $$\sum_{v=0}^{333}\binom{1000}{v,v,v,1000-3v}\tag1$$ 
For the two-way ties, there are three ways to determine which two candidates tie for the lead.  This time, each gets $v$ votes, where $1\leq v\leq500$.  The third candidate gets $w$ votes, where $0\leq w\leq \min\{v-1,1000-2v\}$ and $1000-v-w$ members don't vote.  This gives $$3\sum_{v=1}^{500}\sum_w\binom{1000}{v,v,w,1000-2v-w}\tag2$$
where the inner sum is over all $w$ with $0\leq w <=\min\{v-1,1000-2v\}$
Offhand, I don't know of any convenient way of computing the numbers in $(1)$ and $(2)$, although one can do it with a computer program of course. If you try to do this for general $k$, you'll get a $(k-1)$-fold sum at the end, and it will take someone smarter than me to deal with it.
EDIT
I forgot to bound $w$ by $1000-2v$ in $(2)$.  Corrected.
I wrote a python script to compute this, and amazingly there's a substantial probability, although the assumptions aren't very realistic.  I got a probability of $0.00904$.  Here's the script:
from math import factorial

def choose(n, *args):
    if len(args)==1:
        m =  args[0]
        if m>n: return 0
        args += (n-m,)
    assert sum(args) == n
    answer = factorial(n)
    for m in args:
        answer //= factorial(m)
    return answer

a = sum(choose(1000,v,v,v,1000-3*v) for v in range(334))
b = sum(choose(1000, v,v,w,1000-2*v-w) 
      for v in range(1,501) 
      for w in range(min(v,1001-2*v) ))
print((a+b)/4**1000)

