Game theory mixed strategy on a 3×3 Matrix. Finding which actions a player will randomize over.

economicsgame theory

Problem at Hand

I have spent a huge amount of time trying to find the mixed strategy equilibria of this game, and unfortunately, I keep getting to no answer.

My answers worked through

For (a), it is obvious that the pure strategy of this game is (3,3). It is a given when using the simple method of starring the best strategy given the other player's strategy,

(b) & (c) is where I start to get stuck. I am guessing I need to find the mixed strategy in order to know if the row player is randomizing over 2 or 3 of her actions.

For the expected payoffs of the row player, I wrote them accordingly:
The payoff of the strategy given probability $q$,
$$E(U) = (3)q_1 + (0)q_2 + 0(1-q_1 -q_2)$$
$$E(M) = (0)q_1 + (5)q_2 + (1)(1-q_1 – q_2)$$
$$E(D) = (0)q_1 + (1)q_2 + (5)(1-q_1 – q_2)$$

These simplify to
$$E(U) = (3)q_1$$
$$E(M) = (4)q_2 – q_1 +1$$
$$E(D) = -(4)q_2 -(5)q_1 + 5$$

At this point I solve for either q_1 or q_2:

$$E(M) = E(D)$$
$$(4)q_2 – q_1 +1 = -(4)q_2 -(5)q_1 + 5$$
$$8q_2 + 4q_1 = 4$$
$$2q_2 + q_1 = 1$$
$$q_1 = (1 – 2q_2)$$

This is the exact part I begin to get a little mixed up. I am not sure where to plug this in or, more specifically, where it is most appropriate to plug this in.

I decided to plug it into $E(U) = E(D)$:

$$(3)(1-2q_2) = -4q_2 – 5(1-2q_2) + 5 $$
$$3-6q_2 = -4q_2 -5 + 10q_2 + 5$$
$$3 = 12q_2$$
$$q_2 = 1/4$$

Now I plug in $q_2$ into the $q_1$ term:

$$q_1 = (1- 2q_2) = 1/2$$.

This leaves the probability of the row going player down to $(1- 1/2 – 1/4) = 1/4$. I have a few questions because I feel iffy about this game theory material.

  1. Is my work done correctly?
  2. With this information, would I be able to conclude the answer for (b) is she will randomize over M and D since they are both $1/4$?
  3. If that is not correct. How do I go about this? How can the row player randomize over 3 actions? It does not seem possible in this case. I have been searching and clawing for hours trying to grasp this. Any information would help tremendously.

Best Answer

There are three Nash equilibrium:

  1. $(U, L)$ (by the way $(3,3)$ in your answer is not a Nash equilibrium...)

  2. mixing $(M, P)$ and mixing $(C,R)$ with probability $(1/2, 1/2)$ each, and giving a probability of $0$ to rest. This is the case where the row player mixes exactly two of his strategy.

  3. mixing $(U, M, P)$ and mixing $(L, C,R)$ with probability $(1/2, 1/4, 1/4)$ each. This is the case where the row player mixes exactly three of his strategy.

The last Nash equilibrium may seem to be a bit surprising, because it seems that once the row player start to play $M$ or $P$, there is no chance of ever playing $U$ again. But in fact this strategy does just as well. Because if the row player for some reason switch to $U$, then the column player is forced to play $L$, which does just as good in expectation as compared to randomizing over $M, P$

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