9
$\begingroup$

I was studying games when one of the players seems to be indifferent between two or more pure strategies because he gets the same payoff with each strategy. We say that there are infinite Nash equillibria in the game but I am unable to calculate it and get an intuitive idea of how is it happening. For example in the following game $$\begin{array}{c|a|c} & L_2 & R_2& \\ \hline L_1 & (3,1) & (0,1) \\ \hline R_1 & (0,1) & (4,1) \\ \hline \end{array}$$

Seemingly, P2 should not even bother about anything even if he is playing pure strategy or randomizing. But here if we assume that P2 plays L with probability q and P1 plays L with probability p then making them indifferent we see that we cannot calculate p but q comes out to be 4/7. How is it possible or what is the interpretation ? If I am doing it wrong then what is the correct way to determine equilibria in this case ?

Another example is

$$\begin{array}{c|a|c} & L_2 & R_2& \\ \hline L_1 & (1,1) & (0,0) \\ \hline R_1 & (0,0) & (0,0) \\ \hline \end{array}$$

How many Nash's equilibria are possible in this ?

Improve this question
$\endgroup$

3 Answers 3

Reset to default
4
$\begingroup$

Well the intuition is quite straightforward: as you have mentioned yourself, player 2 is indifferent between playing any action in pure strategies, or randomizing in any possible way. So, for a mixed strategy equilibrium to exist, player 2 needs to play L w/ probability 4/7. If this is the case, it is irrelevant what Player 1 player---player 2's outcome is the same. As a result, in pure strategies the Equilibria are L,L and R,R and, in Mixed strategies, q=4/7 and p can take any value between 0 and 1. Hence, there exist infinite possible Nash Equilibria (p just has to obey the fundamental laws of probability).

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ I did not follow you when you say that P1 wants P2 to play L with prob. 4/7. Won't P1 benefit more if P2 plays R more so that P1 can get a payoff of 4. $\endgroup$
    – Sub-Optimal
    Commented Sep 27, 2015 at 17:27
  • 1
    $\begingroup$ Player P2 who is completely indifferent between L and R ends up randomizing while player P1 can choose actually any mix. Isn't it counter-intuitive ? $\endgroup$
    – Sub-Optimal
    Commented Sep 27, 2015 at 17:47
  • 1
    $\begingroup$ A Nash Equlibrium in pure strategies is a degenerate version of a mixed strategies when the players play an action with probability 1. The idea of a Mixed Strategy is that for P1 to be indifferent between her actions, she needs player 2 to play her actions with a probability such that P1 is indifferent between playing her actions with any probability. So, they simultaneously provide each other with a random strategy such that the other is indifferent between the two actions on expectation. In short, they are both indifferent between the two actions and therefore any random combination. $\endgroup$
    – ChinG
    Commented Sep 27, 2015 at 18:10
  • 1
    $\begingroup$ The counter-intuitive part comes from the fact how mixed strategies are found. When we say "P1 wants P2 to play so and so..." we don't mean thats the best thing for P1. Each player "wants/must" in mxed strategies make the other indifferent. Thats why P1 wants P2 to play L with prob. 4/7 to make him indifferent. Its just the way these equilibria are defined. A comment doesn't allow space to explore this further. Also keep in mind that there is an infinite amount of numbers between 0 and 1 and here the strategies are arbitrary probabilites - numbers- between 0 and 1. $\endgroup$
    – BB King
    Commented Sep 27, 2015 at 20:55
4
$\begingroup$

The second game has two equilibria, $(L_1, L_2)$ and $(R_1,R_2)$. It is straightforward to note that these two are Nash equilibria. On the other hand, if one player plays $R_i$ with probability strictly greater than 0, then it is in the other's best interest to play $L_j$ with probability 1. This last observation shows that no other equilibrium exits.

Finally, there is nothing wrong with having infinitely many equilibrium points. Your simplest example is $$ \begin{array}{|c|c|c|} \hline & L_2 & R_2 \\ \hline L_1 & 0,0 & 0,0 \\ \hline R_1 & 0,0 & 0,0 \\ \hline \end{array} $$ for which every strategy profile is a Nash equilibrium.

Improve this answer
$\endgroup$
2
$\begingroup$

Here is one way using the theorem below from the book Game Theory And Mechanism Design by Y. Narahari. Refer to chapter 7 of the book for notation used in this answer.

Theorem 7.1: The mixed strategy profile $(\sigma_1^*, \sigma_2^*,\ldots,\sigma_n^*)$ is a mixed strategy Nash equilibrium if and only if $\forall i \in N$,

  1. $u_i(s_i, \sigma_{-i}^*)$ is the same $\forall s_i \in \delta(\sigma_i^*)$, and
  2. $u_i(s_i, \sigma_{-i}^*) \geq u_i(s_i', \sigma_{-1}^*)~\forall s_i \in \delta(\sigma_i^*); \forall s_i' \notin \delta(\sigma_i^*)$.

Let $(x^*, y^*)$ be the mixed strategy Nash equilibrium, then the following equations hold $$ \begin{align} u_1(L, y^*) &= u_1(R, y^*) \\ u_2(x^*, L) &= u_2(x^*, R) \end{align} $$ Thus $3y^* = 4(1-y^*) \implies y^* = \frac{4}{7}$ and $x^* \times 1 + (1-x^*)\times 1 = x^* \times 1 + (1 - x^*) \times 1$. The second equation is satisfied for any $x^* \in [0,1]$. Thus the game has infinite Nash equilibria of the form $\left(x^*, \frac{4}{7}\right)$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.