Finding Mixed Strategy Nash Equilibria Okay, so I was working through this problem:

Now, I understand the computations. What I don't understand is why the solution says that each player will play H with probability p=2/3. I would have thought each player would play T with probability p=2/3 and play H with probability p=1/3.
 A: $\newcommand{\tuple}[1]{{\left\langle{#1}\right\rangle}}$ Let mixed strategy be a pair $\tuple{p,q} \in [0,1]\times [0,1]$, where $p$ and $q$ are the probabilities that player 1 and player 2 would pick strategy $H$. Given your game
$$\begin{array}{c|cc}&H&T\\\hline H&1,1&0,0\\T&0,0&2,2\end{array}$$
we have that the payoffs are given as follows (the players are symmetrical)
\begin{align}
v(p,q) &= 1\cdot pq + 2\cdot(1-p)(1-q).
\end{align}
We will fix $q$ and try to maximize the payoff of the first player by adjusting $p$ (it cannot be an equilibrium if there is a strictly better strategy). The thing is, that if $q > \frac{2}{3}$ then $v(1,q) > v(p,q)$ for any $p \in [0,1)$, because $(3q-2) > 0$ and so the following function is strictly increasing with regard to $p$:
$$v(p,q) = (3q-2)p + (2-2q) < q = v(1,q). $$
Similarly, if $q < \frac{2}{3}$, then $v(\bullet,q)$ is strictly decreasing and attains the maximum at $0$. Finally, for $q = \frac{2}{3}$ the payoff does not depend on $p$. All this implies, that the best strategies are $p = 0$ or $p = 1$ unless $q = \frac{2}{3}$, and then it doesn't matter. In other words, if $q \neq \frac{2}{3}$ then any Nash equilibrium satisfies $p = 0$ or $p = 1$, in any other case the player would have an incentive to change his strategy.
As the players are symmetrical, the same reasoning implies that for any Nash equilibrium we have $q = 0$ or $q = 1$ unless $p = \frac{2}{3}$. The reason behind the number $\frac{2}{3}$ is that the other player would have no incentive to change his strategy. For $\frac{1}{3}$ of the first player the second would be inclined to move to $0$, and then the first player would also move to $0$. And indeed, for $p = \frac{2}{3} = q$ we obtain a stable state, i.e. a Nash equilibirum.
I hope this helps $\ddot\smile$
