# Pure vs mixed strategy Nash Equilibria

Just learning about Nash Equilibria. The pure strategy one is explained as an outcome where both/all players feel like they couldn't have done better given what the others were doing. Mixed strategy is one where even after announcing your strategy openly, your opponents can make any choice without affecting their expected gains. Is there a relationship between the two ? Is the pure strategy one some sort of special case for the mixed strategy one ? (I doubt it, since there can be multiple pure ones, while the linear equations of the mixed one can only give one or infinite number of results). I'd like to intuitively understand the relationship. Thanks !

If you like, you can think of a pure strategy as a mixed strategy in which a player has a 100% chance of picking a certain strategy.

The equilibrium definition is the same for both pure and mixed strategy equilibria ("even after announcing your strategy openly, your opponents can make any choice without affecting their expected gains"). The difference is that in a mixed equilibrium, you are announcing your probability distribution, not the strategy that it randomly produces.

Example: Rock-Paper-Scissors. There are no pure strategy equilibria: If I announce "I'm going to play definitely Rock!" then clearly my opponent will choose Paper; if I know they're going to play paper then I don't want to play Rock anymore, so this is not stable. However, if I announce "I'm going to secretly roll a die, play Rock if it shows 1-2, Scissors for 3-4, and Paper for 5-6!" then my opponent is equally happy with any choice he makes. If he therefore chooses the same strategy as me, then I am equally happy with any choice I make, so this is a mixed equilibrium.

Also, your statement "the linear equations of the mixed one can only give one or infinite number of results" isn't true - there are many games with an in-between number of equilibria (Chicken, for example, has 3 in a two-player version of the game). As an aside, a randomly-generated game will have a finite, odd number of equilibria with probability 1, but that's about all you can say about the number of equilibria.

• GMB, from what I understand, being in pure equilibrium doesn't mean that I could make any choice; it means that I couldn't have done better, but could have done worse. Take the intersection game, where either both cars cross at the same time and crash, or both stop, or one stops and the other goes. The pure strategy NE is the last, because if A knows that B will stop, then why should he waste time stopping, and if B knows that A will go then B wants to avoid crash. However A announcing that he will go doesn't mean that B can do anything. See where I'm confused ? B can have a worse outcome. – Balazs Rau Apr 6 '14 at 17:12
• I agree with your NE analysis. The "equally happy with anything" bit of my post referred specifically to mixed equilibria, sorry if that was unclear. It turns out that if $A$ plays a mixed-strategy equilibrium where he has nonzero chance of playing $k$ different strategies, then there is some set of $k$ strategies for $B$ that are all equally good (and therefore $B$ can randomize among them however he likes). In the special case of $k=1$ ("pure strategy equilibrium"), that means there is (at least) $1$ strategy for $B$ that stands up to scrutiny, which is not so informative. – GMB Apr 6 '14 at 17:20
• This is off on a bit of a tangent from your original post, though =) the answer to your original question is "a pure strategy is just a mixed strategy over exactly 1 strategy." – GMB Apr 6 '14 at 17:22
• Oh man, I feel a little thick... So how does "equally happy with anything" in the mixed case become "I could not have done better" in the general case ? – Balazs Rau Apr 6 '14 at 18:37
• @BalazsRau : Let's say you're playing a game where the players can pick between $n$ different strategies. In general, there might be mixed equilibria that only mix over $k$ of those $n$ strategies. In this case, there will be some linear combination of my $k$ strategies that will make you like some $k$ of your strategies equally well (and the other $n-k$ strategies less than that). In the pure case ($k=1$), this statement is uninteresting: it just says that you like $1$ of your strategies at least as well as all the others, which is obvious. – GMB Apr 6 '14 at 21:25

As far as I know, a mixed strategy vector is a randomized vector where each player can play some action with some probability. i.e., if you play action $a$ with probability $p_a$, I can play (for example) an action $b$ with probability $p_b$. So the set of actions is randomized. There is a random play! That's why it is called $\mathbf{mixed}$.

In pure strategy, if you play $a$ (with probability $1$), I can play for example the same action $a$ but with probability $1$. There is no random play! That's why it is called $\mathbf{pure}$.

In fact, mixed strategy is used to guarantee that every game has at least one Nash equilibrium. Randomize the set of strategy to make them convex sets and apply Brouwer fixed point Theorem to prove this result. A well known theorem due to Nash.

• You might be correct, but it's a little over my head. Someone else please tell me if I should accept. – Balazs Rau Apr 7 '14 at 5:12