Comments on: Follow-up: Mismatched Pennies http://mathfactor.uark.edu/2007/06/follow-up-mismatched-pennies/ The Math Factor Podcast Site Fri, 08 Aug 2014 12:52:06 +0000 hourly 1 https://wordpress.org/?v=4.9.25 By: Shawn http://mathfactor.uark.edu/2007/06/follow-up-mismatched-pennies/comment-page-1/#comment-1028 Tue, 05 Jun 2012 22:32:36 +0000 http://mathfactor.uark.edu/2007/06/08/follow-up-mismatched-pennies/#comment-1028 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

]]> By: Shawn http://mathfactor.uark.edu/2007/06/follow-up-mismatched-pennies/comment-page-1/#comment-894 Sat, 19 Nov 2011 23:28:39 +0000 http://mathfactor.uark.edu/2007/06/08/follow-up-mismatched-pennies/#comment-894 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.

]]> By: Shawn http://mathfactor.uark.edu/2007/06/follow-up-mismatched-pennies/comment-page-1/#comment-893 Sat, 19 Nov 2011 21:07:30 +0000 http://mathfactor.uark.edu/2007/06/08/follow-up-mismatched-pennies/#comment-893 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
 

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