How can this be an optimal strategy? I'm following a course on Game Theory and I'm trying to understand the following exercise:
Consider the game with game matrix
$$
G = \begin{pmatrix}-2 & 2 & -1\\1&1&1\\3&0&1\end{pmatrix}
$$
Show that player II has an optimal strategy giving positive weight to each of each of his columns.
There are two players in this game, player I and player II. Each element $G(i,j)$ in the game matrix resembles the payoff to player I if player I chooses strategy $i$ and player II chooses strategy $j$.
The correct solution is apparently that player II has the optimal mixed strategy $(1/4, 1/2, 1/4)$. When I asked my teacher why this is the case and how you can find such a solution, he told me that since $(1,0,0), (1/2, 1/2, 0)$, and $(0,0,1)$ are optimal strategies for player II, and since the convex combination of optimal strategies is also optimal, we can take 
$$
1/3*(1, 0,0) + 1/3 * (1/2, 1/2, 0) + 1/3*(0,0,1) = (1/4, 1/2, 1/4)
$$
such that $(1/4, 1/2, 1/4)$ is again an optimal strategy for player II.
My question is: Why are $(1,0,0)$ and $ (1/2,1/2,0)$ optimal strategies for player II? 
What I don't get is how for example $(1,0,0)$ could be an optimal strategy for player II. If I understand the concept of optimal strategies correctly, an optimal strategy should guarantee a player at least the value of the game, no matter what the other player does (I believe this is called the principle of indifference, player X looks for a strategy that makes player Y indifferent). 
If player II plays $(1,0,0)$, and player I plays $(0,1/2,1/2)$, the expected payoff for player II is $1/2 * 1 + 1/2 * 3 = 2$ which is higher than the value of the game. So how can this be an optimal strategy?
 A: You're quite right that $(1, 0, 0)$ and $(\tfrac12, \tfrac12, 0)$ are not optimal strategies for player II. The value of the game is $1$ to player I, so if the probabilities of player II's mixed strategy are $p,q,(1-p-q)$ then $$-2p + 2q - (1-p-q) \le 1 \\
p + q + (1-p-q) \le 1 \\
3p + (1-p-q) \le 1$$
The second inequality is, of course, trivial.
The other two rearrange to $$-p + 3q \le 2 \\
2p - q \le 0$$
The algorithmic way to find a solution with three non-zero weights would be to add the constraints $p \ge \varepsilon$, $q \ge \varepsilon$, $p + q \le 1 - \varepsilon$, and try small values of $\varepsilon$ (e.g. $\varepsilon = 10^{-n}$) until you find one which gives a consistent set of inequalities.
The quick and dirty approach is to observe that the inequality $-p + 3q \le 2$ certainly holds if $q \le \tfrac23$, the inequality $2p - q \le 0$ is just $p \le \tfrac q2$, and any value of $q$ which is strictly less than $\tfrac23$ therefore implies that $p + q < 1$. The "correct" mixed strategy which you were given is a member of this family of solutions $q < \tfrac23$, $p \le \tfrac q2$, and arguably the simplest member.
