Follow-up: Mismatched Pennies
A correspondent writes:
Greetings,
I think that in the long run both strategies are equivalent. This game doesn’t favor any player.
Demonstration
Chaim Expected Gains = 3 * 1/4 + 1 * 1/4 = 1
Kyle Expected Gains = 1/4 * 2 + 1/4 * 2 = 1
This is so if both of us pick H half of the time, and pick T half of the time.
But!
If I know Kyle is going to pick H half of the time and T half of the time, I should adjust my strategy. I can do better by always picking H; the payout would then be
C: 3*1/2 = 3/2
K: 2*1/2 = 1
Net 1/2 in my favor!!
Conversely, if I am picking H half of the time and T half of the time, Kyle should adjust his strategy and choose T all of the time; this comes out to
C: 1*1/2 = 1/2
K: 2*1/2 = 1
Net 1/2 in Kyle’s favor– rats!
John von Neumann’s celebrated result is that both players have an optimal strategy, one that cannot be exploited by the other player. If we both play optimally, is the game balanced?
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Shawn said,
November 19, 2011 at 4:07 pm
That’s almost right. If the payoffs are like the following:
h
t
H
3, 0
0, 2
T
0, 2
1, 0
If we call the row player A and the column player B, then the mixed-strategy Nash equilibrium is for A to pick H 1/2 of the time and T 1/2 of the time and for B to pick h 3/4 of the time and t 1/4 of the time. Let’s see the expected payoffs. A is going to get an expected payoff of [(3) * (1/2) * (3/4)] + [(0) * (1/2) * (1/4)] + [(0) * (1/2) * (3/4)] + [(1) * (1/2) * 1/4)] = 1.25 B is going to get [(0) * (1/2) * (3/4)] + [(2) * (1/2) * (1/4)] + [(2) * (1/2) * (3/4)] + [(0) * (1/2) * 1/4)] = 1
Shawn said,
November 19, 2011 at 6:28 pm
Actually, that’s not quite right either. Sorry about that, I was in a bit of a rush when I typed out my previous comment and got myself confused. The payoffs are h t H 3,0 0, 2 T 0, 2 1, 0 The mixed-strategy equilibrium is for A to pick H 1/2 of the time and T 1/2 of the time, but it is for B to pick h 1/4 of the time and t 3/4 of the time. If A plays that strategy, player B is going to get an expected payoff of 2 half the time when B chooses h and an expected payoff of 0 the other half of the time, and player B is going to get an expected payoff of 0 half the time he chooses t and an expected payoff of 2 the other half of the time, for a total expected payoff of 1, regardless of what strategy he plays. 2*(1/2) + 0 *(1/2) = 1 + 0 = 1 0*(1/2) + 0 *(1/2) = 0 + 1 = 1 Any weighted average of 1 and 1 is going to result in 1. Therefore, player B has no profitable deviation. Now let’s check to see if player A has a profitable deviation. If B plays the 1/4 h and 3/4 t strategy, then when A plays H, he gets the expected payoff of H against 1/4 h and 3/4 t: 1/4 * 3 + 3/4 * 0 = 3/4 When A plays T, he gets the expected payoff of T against 1/4 h and 3/4 t: 1/4 * 0 + 3/4 * 1 = 3/4 Any weighted average of 3/4 and 3/4 is going to result in 3/4. Therefore, player A has no profitable deviation, and each player is playing a best response to each other. Therefore, it is indeed a Nash equilibrium.
Shawn said,
June 5, 2012 at 5:32 pm
Yale University has a nice game theory site on their Open Yale Courses. Video and audio of lectures can be downloaded from the page, and some of the lectures include explanations of how to find mixed-strategy Nash equilibria in the manner I mentioned above. http://oyc.yale.edu/economics/econ-159