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My understanding of mixed equilibrium is that it's basically an optimization problem: you try to tweak the probabilities of your actions so that your opponents gets the least payoff.

This involves solving two linear equations for a 2 x 2 matrix.

Is it a special case of mixed nash equilibrium? I was expecting that to be the case, but it seems easy to disprove it with a counterexample:

Say we have this payoff matrix:

  A         B

A' (1, 2) |  (3, 100)

B' (1, 2) | (4  100)

It's easy to see that the column player should always play B since that gives them the best payoff.

However, if you treat it is as a mixed equilibrium problem, with probability A = p and probability for B = 1-p,

you get:


p + 3 - 3p =  p + 4 - 4p

p = 1, which suggests that they should always play A and never play B. This seems to contradict the logical best move strategy (highlighted in bold)

Am I misunderstanding something?

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2 Answers 2

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Your definitions are not completely correct.

A Nash Equilibrium (NE) is a pair of strategies, not a strategy for a single player. When playing a NE, each one chooses a distribution so that the other player will not have a profitable deviation (and not the least payoff).

In your example, the column player can play A and indeed rows will not have a profitable deviation, but it's not an equilibrium strategy since you cannot pair it to any strategy of the row player. Regardless of the strategy of the row player, column will have a profitable deviation.

Having said that, indeed a pure NE is a special case of a mixed one, but when some of the players play pure, the condition is no longer indifference between his actions (only between the ones he plays with non-zero probability; which again emphasizes the fact that a NE is a pair of strategies).

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I am just a beginner to the field, but here is how I understand it.

There is one Nash equilibrium (NE) in pure strategies, which is easy to see. As you say, the strategy $B$ for player two dominates $A$, and so we know that a rational player $B$ will never play $A$. Similarly, $B'$ for player one dominates $A'$. Thus, in pure strategies the only solution is $(B',B)$, as you probably have figured out on your own.

Is there a NE in mixed strategies? I do not think so, and here is how I reason. I give an extensive answer in the hope that it might clarify any confusions, and also to make the answer more transparent in case I make a mistake.

Let the mixed strategy for player one, $\sigma_1$ be "play $A'$ with probability $p$, and $B'$ with probability $1-p$ (the complement of $\text{Pr}(A'))$. Let also $\sigma_2$ be the mixed strategy for player two, with $\text{Pr}(A) = q, \text{Pr}(B) = 1-q$.

First ask what is the expected utility of player one from playing $A'$ and $B'$. This will depend on what player two does:

$$ v_1(A',\sigma_2) = 1q + 3(1-q) = 3-2q,\\ v_1(B',\sigma_2) = 1q + 4(1-q) = 4-3q. $$

For player one to be mixing between his strategies, he must be indifferent to these outcomes, i.e., $v_1(A',\sigma_2) = v_1(B',\sigma_2)$. Using the above expressions for the respective utilities we get

$$ q=1 $$

I think this is what you calculated above! What does this mean? It means that player one will be willing to mix between his two pure strategies $A'$ and $B'$ iff player two plays $A$ with a probability of 1. If you think about what we said at the top of this post it kinda makes sense -- if player two plays $A$ then it doesn't matter what player one does.

We can make the same calculation for player two. Under what conditions will player two mix between his strategies? We have:

$$ v_2(A,\sigma_1) = 2 p + 2 (1-p) = 2,\\ v_2(B,\sigma_2) = 100 p + 100 (1-p) = 100.\\ $$

Player two mixes when $v_2(A,\sigma_1) = v_2(B,\sigma_1)$, i.e., when $2=100$. This is, of course, impossible. Player two will never be indifferent between his two strategies in this game, which makes sense which you also write in your post!

Remarks.

Is it a special case of mixed nash equilibrium?

Mixed equilibrium and mixed NE are the same thing as far as I know!

which suggests that they should always play A and never play B. This seems to contradict the logical best move strategy (highlighted in bold)

You need to be careful in distinguishing the strategies of both players. I think this confusion is what made it difficult. In NE, player one plays $B'$ and player two plays $B$. Unfortunately, it does not quite make sense to talk about both player playing $A$ or $B$.

Sorry for the elaborate answer, but hope that makes sense :)

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  • $\begingroup$ As YJT remarked in his answer, a pure NE is a special case of a mixed NE, so if you have found a pure NE you have automatically found a mixed NE as well. However, there are games with only mixed NE, so you need to be a bit careful with your definitions. $\endgroup$
    – AntimatterHedgehog
    Commented Aug 31, 2020 at 11:48
  • $\begingroup$ your answer suggests that player 2's optimal probability distribution is to always play A and never play B. Doesn't that seem incorrect? The optimal strategy is to always play B and never play A $\endgroup$
    – nz_21
    Commented Aug 31, 2020 at 13:47
  • $\begingroup$ Hmm, that's not really what I meant. I apologize if my answer could be reat that way! What I meant whas that the optimal strategies are pure strategies, $B'$ for player one, and $B$ for player two. However, when thinking about whether one of the players would be willing to use a mixed strategy, we find that player one would be willing to mix between $A'$ and $B'$ if player two plays $A$ with probability 1 (i.e., a pure strategy). But player two will never play $A$ since it's a dominated strategy, so that situation will never arise. $\endgroup$
    – AntimatterHedgehog
    Commented Aug 31, 2020 at 14:48

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