What's the intuition behind mixed strategies?

Question

Ok, so this issue has been bothering me ever since we covered mixed strategies in my college course on Game Theory.

I understand that they're a way for a player to create a belief about what the other player does in the face of uncertainty (that is, when we know that there's no clear dominant strategy that they're going to choose).

However, I can't really wrap my mind on what these probabilities are truly saying. To make my point clear, I'll set up a generic simultaneous game as an example:

$$ \begin{pmatrix} a,b & c,d \\ e,f & g,h \\ \end{pmatrix} $$ where the first row represents strategy 1 for player 1, and the second row strategy 2. Analogously, the first column is strategy 3 for player 2 and the second column is strategy 4.

Now, if I want to find out what player 1 thinks player 2 could play, then I have to calculate what would be the probability of player 3 choosing strategies 3 and 4 that would leave player 1 indifferent between choosing his/her strategies.

That is, assuming $P($$s_2$$=3) = p$ for player 3:

$$U^e(1)=U^e(2)$$

iff

$$ap + c(1-p) = ep + g(1-p)$$

And assuming I didn't make a dumb mistake manipulating this algebraic expression, this gets you the following value of p:

$$p = \frac{g-c}{(a-e)+(g-c)}$$

As you can see, this means that the probability player 1 imposes on player 2 playing strategy 3 depends on the difference in payoff that player 1 has from playing strategy 2 instead of 1 (given that player 2 is playing strategy 4) and on the difference in payoff from playing strategy 1 instead of 2, when player 2 is playing strategy 3.

How should I interpret this, though? Why should these differences in payoffs induce player 1 to deduce a different probability of player 2 playing strategy 3? If $d(|g-c|)$>0, that is, if player 1 now receives a higher payoff from playing $s_1=2$ rather than $s_1=1$ (given $s_2=4$), why should that make player 1 more likely to think player 2 should now play strategy 3 with a higher probability?

I feel like I'm close to figuring it out, but I can't quite grasp it yet. Any help would be really appreciated. Thanks!

Answer 1

You are on the right track, but it may be helpful to draw some distinctions that are usually omitted on a first pass. For simplicity (but without loss of generality), assume two opponents.

A Nash equilibrium in mixed strategies requires: a) each player $i$ has a belief $\theta_{-i}$ about the strategies chosen by the opponent; b) the belief is correct and thus $\theta_{-i} = \sigma_j$; c) each player plays a best reply to his correct beliefs.

It is incorrect to assume that "these differences in payoffs induce player 1 to deduce a different probability of player 2 playing strategy 3". Instead, one should argue that $\sigma_i$ can be a best reply (and hence be played) only if $i$ is indifferent over the strategies in the support of $\sigma_i$, and therefore (if $i$'s conjecture has to match $j$'s mixed strategy) it must be the case that in equilibrium $j$ uses a different mixed strategy.

In short, for $i$ to have correct beliefs about $j$, it is necessary that $j$ plays different strategies. Nothing compels $j$ to do so, but if she does not randomise in the proper way, players will not be at an equilibrium, in the sense that at least one of a)-b)-c) above is false.

Answer 2

Think of it in these terms: if my opponent knows my mixed strategy, can she exploit it? For example, in Rock-Paper-Scissors, an unexploitable mixed strategy is to play each of R, P, and S with probability 1/3. With that, you are guaranteed to get a return of 0 in the long run. But suppose an opponent has watched your matches and observed that you play Rock with probability 0.4, Paper and Scissors with probability 0.3 each. Your opponent can then open with Paper, and your average return will be less than 0.

Or think about it from the other perspective: ask "Is my opponent playing an unexploitable mixed strategy?" and if the answer is no, then ask "How can I exploit my opponent's strategy."