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Consider the following calculation of Nash equilibria from Wikipedia:

                  Player B plays H  Player B plays T
Player A plays H    −1, +1          +1, −1
Player A plays T    +1, −1          −1, +1

To compute the mixed strategy Nash equilibrium, assign A the probability p
of playing H and (1−p) of playing T, and assign B the probability q of
playing H and (1−q) of playing T.

E[payoff for A playing H] = (−1)q + (+1)(1−q) = 1−2q
E[payoff for A playing T] = (+1)q + (−1)(1−q) = 2q−1
E[payoff for A playing H] = E[payoff for A playing T] ⇒ 1−2q = 2q−1 ⇒ q = 1/2

We have calculated the strategy for player 2, but we haven't used their payoffs, only those for player 1! How is this possible?

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That seems to be a badly phrased exercise which implicitly relies on being a zero-sum game. – Peter Taylor Apr 10 '12 at 10:32

Here comes a more or less complete citation of an earlier answer of mine:

This equilibrium can be found by noting that a strategy pair $(σ_1,σ_2)$ is a mixed strategy NE, iff every player is indifferent between the pure strategies played with positive probability in an equilibrium and each player weakly prefers the strategies played with positive probability to those played with zero probability.

Thus assuming player $j$ plays heads with probability $p$, player $i$ would obtain: $u(H,p)=p−(1−p)=2p−1$ playing $H$, and obtain $u(T,p)=−p+(1−p)=1−2p$ playing $T$. These two utilities are equal only if $p=1/2$. Assuming $i$ plays heads with probability $q$, by similar reasoning we obtain $q=1/2$.

Also as you noted above every finite game has a mixed strategy equilibrium.This can be shown by using correspondences (set valued functions) and kakutanis fixed point theorem. For an outline of the proof, you can have a look at the wikipedia page on Nash equilibrium. Although, to warn you, that site leaves out an tremendous amount of detail.

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Thanks for clarifying the Nash equilibrium issue. The issue I don't understand is it is possible to calculate a players optimal strategy without considering their payoffs. We found q=1/2, but we only used player 1's payoffs! – Casebash Apr 10 '12 at 9:41
Yes but this follows exactly from the fact that I stated in the second paragraph. In order for player 1 to be indifferent between her pure strategies, p has to be such that p=1/2. Otherwise she would not be indifferent and we would not have a mixed NE by the fact stated in my answer above. It is important to note that we can rule out any "degenerate" mixed strategy pair (i.e. a pure strategy), since for any such pair, one player always has an incentive to deviate. Hence any mixed strategy equilibrium must be completely mixed (i.e. every strategy is played with a strictly pos probability). – Henrik Apr 10 '12 at 13:38

The reasoning is as follows. Player 2 should choose $q$ such that player 1 will be indifferent between playing $H$ and playing $T$. We ex ante know that this reasoning will work out because we know that 1) there exists no pure equilibrium in the game therefore there will be a mixed equilibrium 2) According to the following theorem: The strategies in the support of Nash equilibrium (a strategy which is played with positive probability) always yield the same payoff.

[one can showthis as follows. assume that one strategy in the support of a NE yields more payoff than the other strategy in the support of a NE, then one would 'transfer' probability from the strategy which gives lower payoff than to the one which gives higher payoff.]

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