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Chapter 10: Mixed strategies Nash equilibria,
reaction curves and the equality of payoffs theorem
Nash equilibrium: The concept of Nash equilibrium can be extended in a natural manner to the mixed strategies introduced in Lecture 5. First we generalize the idea of a best response to a mixed strategy
Definition 1. A mixed strategy ̂σR is a best response for R to some mixed strategy σC of C if we have 〈σ̂ R, PRσC 〉 ≥ 〈σR, PRσC 〉 for all σR.
A mixed strategy σ̂ C is a best response for C to some strategy σR of R if we have
〈σR, PC ̂σC 〉 ≥ 〈σR, PC σC 〉 for all σC
We can then extend the definition to Nash equilbrium
Definition 2. The mixed strategies ̂σR, ̂σR are a Nash equilibrium for a two-player game with payoff matrices PR and PC if
̂ σR is a best response to ̂σC and ̂σC is a best response to ̂σR
or in other words 〈σ̂ R, PR ̂σC 〉 ≥ 〈σR, PR ̂σC 〉 for all σR. 〈̂σR, PC σ̂ C 〉 ≥ 〈σ̂ R, PC σC 〉 for all σR
Reaction curves: For games with two strategies one can compute the best responses and the Nash equilbria in terms of the reactions curves. To explain the idea let us start with a example
Example: Matching Pennies The payoff matrices are given by
PR =
, PR =
and let us write the mixed strategies as
σR = (p, 1 − p) σC = (q, 1 − q)
To find the best response to σC we compute
〈σR, PRσC 〉 =
p 1 − p
q 1 − q
p 1 − p
2 q − 1 1 − 2 q
= p(2q − 1) + (1 − p)(1 − 2 q) = (2q − 1) + p(4q − 2)
Since we are computing the best response for R to the strategy of C we consider the payoff (2q − 1) + p(4q − 2)
for fixed q and variable p with 0 ≤ p ≤ 1. This is a linear function of p and the maximum will depend on the slope of this function (here 4q − 2), whether it is positive, negative, or 0.
Best response for R:
- 4 q − 2 > 0 (or q < 1 /2) The slope is positive so the maximum is at p = 1.
- 4 q − 2 < 0 (or q > 1 /2) The slope is negative so the maximum is at p = 0
- 4 q − 2 = 0 (or q = 1/2) The slope is 0 so the maximum is at any p between 0 and 1.
To find the best response for C we compute
〈σR, PRσC 〉 =
p 1 − p
q 1 − q
p 1 − p
1 − 2 q 2 q − 1
= p(1 − 2 q) + (1 − p)(2q − 1) = (2p − 1) + q(2 − 4 p)
which we now consider has a function of the variable q and for fixed p. By maximizing over 0 ≤ q ≤ 1 we find
Best response for C:
- 2 − 4 p > 0 (or p > 1 /2) The slope is positive so the maximum is at q = 1.
- 2 − 4 p < 0 (or p < 1 /2) The slope is negative so the maximum is at q = 0
- 2 − 4 p = 0 (or p = 1/2) The slope is 0 so the maximum is at any q between 0 and 1.
To find the Nash equilibria we argue as follows.
- If q > 1 /2 then the best response is p = 0 but the best response to p = 0 is q = 0 which contradicts q > 1 /2 and this does not lead to a Nash equilbrium.
- If q < 1 /2 then the best response is p = 1 but the best response to p = 1 is q = 1 and again this does not lead to a Nash equilbrium.
- If q = 1/2 then the best response is any p and so if we choose p = 1/2 then the best response to p = 1/2 is any q, in particular q = 1/2.
Figure 2: Reaction curves for the battle of the sexes game
- 3 − 4 p < 0 (or p > 3 /4) The slope is negative so the maximum is at q = 0
- 3 − 4 p = 0 (or p = 3/4) The slope is 0 so the maximum is at any q between 0 and 1.
The best response curves are given in Figure ??
The equality of payoffs theorem The method used above to compute the Nash equilib- ria works well if there are 2 strategies but is not very useful if three or more strategies are used since the optimization problems become much more complicated. We present here a method which, in principle, allows to compute, all Nash equilibria of a game. But the reader should be warned that computations become quickly quite lengthy and involved. Some games do not have a Nash equilibrium in pure strategies (like rock-paper-scissors) or matching pennies but there is always one (and often many) if we consider mixed strategies.
Theorem 3. (Nash Theorem) A game (with a finite number of strategies) always has at least one Nash equilibrium (̂σR, σ̂ C ) in mixed strategies.
Many interesting examples of games are symmetric.
Theorem 4. (Nash Theorem for symmetric games) For a symmetric game we have
(̂ σR, σ̂ C ) is a NE ⇐⇒ (̂σC , ̂σR) is a NE
Moroever there always exists at least one symmetric NE
(̂ σR, σ̂ C ) = (̂σ, ̂σ)
The computation of Nash equilibria is based on the simple observation.
Theorem 5. (Equality of payoff theorems) Suppose ̂σR is a best response to σC. Then we have
- If the strategy i and j are played with positive probability for R (that is we have σ̂ R(i) > 0 and ̂σR(j) > 0 ) then the payoff to play i and j against σC are identical.
̂ σR(i) > 0 and ̂σR(j) > 0 =⇒ PRσC (i) = PRσC (j)
- If the strategy i is played with positive probability (that is we have ̂σR(i) > 0 ) but the strategy j is played with probability 0 (that is ̂σR(j) = 0) then the payoff for R to play j against σC is less than or equal to the payoff to play i against σC.
̂ σR(i) > 0 and ̂σR(j) = 0 =⇒ PRσC (i) ≥ PRσC (j)
Proof. (i) It is best to argue by contradiction. Suppose that the strategies i and j are played with positive probability for R (that is we have ̂σR(i) > 0 and ̂σR(j) > 0) but the payoffs PRσC (i) and PRσC (j) are not equal. Let us say for example that we have PRσC (j) > PRσC (i). Then we argue that σ̂ R cannot be a best response to σC : since it it more favorable to play j than to play i if you choose a new mixed strategy where you play j with greater probability than ̂σR(i) and j with a smaller probability that ̂σR(j) and all the other strategies with the unchanged probabilities then your payoff in this new strategy against σC will strictly increase. So ̂σR was not a best response. (ii) Argue again by contradiction. Suppose that ̂σR(i) > 0 and ̂σR(j) = 0 but that PRσC (j) > PRσC (i). Then σ̂ R cannot be a best response to σC. To see this change your strategy ̂σR into a new strategy where you play now j with positive probability and i with probability 0. In this new strategy the payoff against σC is greater than for σ̂ R and so ̂σR was not a best response.
It is useful to give a name for the strategies which are played with positive probabilities:
Definition 6. The support of a mixed strategies σ is the set of pure strategies which are played with positive probability. We denote the support of σ by S(σ):
S(σ) = {i : σ(i) > 0)}
For example if σ = (1/ 7 , 2 / 7 , 0 , 0 , 4 /7) then S(σ) = { 1 , 2 , 5 } that is the mixed strategy σ the strategies played with positive probability are 1, 2, and 5.
while the payoff for C are given by
P (^) CT σR =
q 1 − q
2 − q 1 + 2q
Setting the payoff for R to be equal we find
3 − 5 p = 8p − 5 ⇒ p = 8/ 13
while setting the payoff for C to be equal gives
2 − q = 1 + 2q ⇒ q = 1/ 3
So we have a NE (̂σR, σ̂ C ) = ((1/ 3 , 2 /3) , (8/ 13 , 5 /13)).
Example: Symmetric 2-strategies game Consider the symmetric game with payoff matrices
PR =
a b c d
, PC =
a c b d
We already know that the game as 1 pure strategy NE is a > c, b > d or a < c, b < d and 2 pure strategy NE if a > c, b < d or a < c, b > d. To find when this game as a mixed Nash equilibrium we use the equality of payoff theorem. Denote σC = (p, 1 − p), the strategy for C, then the payoff for R is ( a b c d
p 1 − p
b + p(a − b) d + p(c − d)
Setting the payoff to be equal we find
b + p(a − b) = d + p(c − d)
or p[(a − c) + (d − b)] = d − b.
So we find
p =
(d − b) (a − c) + (d − b)
Since the game is symmetric if we denote that strategy for R by σR = (q, 1 − q) the payoff for C are (^) ( b + q(a − b) d + q(c − d)
and equating them gives the same solution q = (^) (a−(cd)+(−bd)−b). Note that p and q should be between 0 and 1 and this occurs only if a > c, b < d or a < c, b > d.
A symmetric two strategies game with matrix PR =
a b c d
and PC = P (^) RT has a mixed strategy NE if a > c, b < d or a < c, b > d. The NE is symmetric and given by ( σ̂ R, ̂σC ) = (̂σ, ̂σ) with
̂ σ =
(d − b) (a − c) + (d − b)
(a − c) (a − c) + (d − b)
Example: Nash equilibria for Rock-scissors-papers The payoff matrices are
PR =
, PC =
Note that the game is a symmetric one so we should find a symmetric Nash equilibrium. The computation of Nash equilibria goes in several steps.
- Assume that one of the player use all his three pure strategies, for example take σC = (p 1 , p 2 , 1 − p 1 − p 2 ). Then the payoffs for R against this mixed strategy are given by
PRσC =
p 1 p 2 1 − p 1 − p 2
2 p 2 + p 1 − 1 1 − p 2 − 2 p 1 p 1 − p 2
We set the payoffs to be equal and find two equation
2 p 2 + p 1 − 1 = p 1 − p 2 → p 2 = 1/ 3 1 − p 2 − 2 p 1 − 1 = p 1 − p 2 → p 1 = 1/ 3
so we must have σC = (1/ 3 , 1 / 3 , 1 /3). Since the game is symmetric by reversing the roles of R and C we find then σR = (1/ 3 , 1 / 3 , 1 /3) and we have found a (symmetric) NE.
- We try next to find NE where one player plays only 2 of his strategies, say let us pick σC = (p, 1 − p, 0). Then the payoff for R is
PRσC =
p 1 − p 0
1 − p −p 0
We set the payoffs to be equal and find
2 p 1 − 1 = 3 − 6 p 1 − 6 p 2 4 p 2 − 2 = 3 − 6 p 1 − 6 p 2
or 8 p 1 + 6p 2 = 4 6 p 1 + 10p 2 = 5 and after some algebra we find σC = (5/ 22 , 8 / 22 , 9 /22). Since the game is symmet- ric, reversing the role of R and C we also find σR = (5/ 22 , 8 / 22 , 9 /22).
- Let us assume that C uses only his first two strategies (one finger and two fingers) σC = (p, 1 − p, 0). Then we have
PRσC =
p 1 − p 0
2 p − 1 2 − 4 p 0
If we set the payoffs to be equal we find 2p−1 = 2− 4 p or p = 1/2. For that choice the payoffs are then (0, 0 , 0). So we can assume that R is playing is first two strategies with positive probability. His third strategy yields a payoff which is not bigger. But now if R use his first two strategies using the symmetry of the game we find the same result for C. So we obtain a Nash equilibrium σR = (1/ 2 , 1 / 2 , 0), σC = (1/ 2 , 1 / 2 , 0).
- Let us assume that C uses only his first and third strategies, σC = (p, 1 − p, 0). Then we find PRσC = (2p − 1 , 0 , 3 − 6 p) and arguing as in the previous case we find a Nash equilibrium σR = (1/ 2 , 0 , 1 /2), σC = (1/ 2 , 0 , 1 /2).
- Let us assume that C uses only his second and third strategies, then we obtain in a similar way σR = (0, 1 / 2 , 1 /2), σC = (0, 1 / 2 , 1 /2).
- Finally let us assume that C plays a pure strategy, say 1. Then we have
PRσC =
and so the best response for R is to play 1. By symmetry we find the Nash equilib- rium σR = (1, 0 , 0), σC = (1, 0 , 0). One argues similarly with playing pure strategies 2 and 3 and one finds two more Nash equilbrium in pure strategies.
To summarize we have found 7 Nash equilibria, all of them symmetric, i.e., we have (̂σR = ̂σC ) = (̂σ, ̂σ) with ̂ σ = (5/ 22 , 8 / 22 , 9 /22), ̂ σ = (1/ 2 , 1 / 2 , 0), ̂σ = (1/ 2 , 0 , 1 /2), ̂σ = (0, 1 / 2 , 1 /2) σ̂ = (1, 0 , 0), σ̂ = (0, 1 , 0), ̂σ = (0, 0 , 1).
