Game Theory Voting I am having some difficulty in solving the following problem. I was wondering whether someone would be kind enough to sketch a solution or even better to solve the whole game. Thanks

Suppose that 100 people live in a village of whom 51 support the
  conservative candidate and 49 support the liberal candidate. Villagers
  get a payoff of 10 if their candidate gets elected, -10 if the other
  one wins. Voting costs villagers 1 unit. Those who do not vote evade
  these costs but get the payoff related to the winning candidate ( +/-
  10).
  
  
*
  
*Find a Nash Equilibrium (in mixed strategies) in which all conservatives use the    same strategy and all liberals use the same
  strategy. In case of    equal number of votes toss of a coin decides. 
  (So for instance in the case of equal number of votes, if villager i
  is voting, then its payoff would be of 1/2*(10-1); while if villager
  i is NOT voting, then its payoff would be of 1/2*10)
  
*What is the expected number of villagers who will vote in this equilibrium?
  

 A: 
sketch a solution

Suppose that the assumption (made in a comment) that all liberals (with probability $q=1$) vote is correct.
Then a conservative will be indifferent between voting and non-voting if the cost and the gain from voting balance each other out. The cost of voting is $1$. The gain of voting is either $0$ or $10$. It is only $10$ if his vote is either the tying or winning vote. (Note that I am ignoring the question's statements regarding payoffs in a tie; I think they are wrong and I've taken a more logical route.)
Let $p$ be the (independent) probability that a conservative votes. The probability that his/her vote would be the tying one is $\binom{50}{48}p^{48}(1-p)^2$. The probability that his/her vote would be the winning one is $\binom{50}{49}p^{49}(1-p)$.
So, we look to solve
$$1=10\left[\binom{50}{48}p^{48}(1-p)^2+\binom{50}{49}p^{49}(1-p)\right].$$
Being lazy, we plug this into WolframAlpha, and lo and behold, we get two solutions: $p\approx0.898255$ and $p\approx0.997891$.
This is somewhat unexpected. However, I am almost completely sure that with $p\approx0.898255$, the liberals' strategy $q=1$ wouldn't be optimal anymore. So, we can ignore that case.
This leaves us with showing that, given $p\approx0.997891$, $q=1$ is indeed Nash. For this we need to show that one liberal will not be better off (in expectation) by non-voting.
And (more laziness) that's where this half of the answer ends... It requires more work, but my guess is that it would turn out just fine. (And the expected number of voters in this, yet to be established, Nash equilibrium would be $51p+49\approx99.892441$.)

I'm actually not that sure. Anyway, perhaps commenters may provide some further insight.
