# What's the intuition behind mixed strategies?

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!

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.