Answers
Note that in all the games, I refer to p as the probability that Player 1 goes U, and q as the
probability that Player 2 goes L.
Game 1
Player 2
Player 1
U
D
L
2,2
1,1
R
0,1
1,3
There are 2 Pure equilibria (U,L) and (D,R) There is one nondegenerate mixed NE with p =
and q = 12 .
2
3
Game 2
Player 2
Player 1
U
D
L
1,1
2,-1
R
-1,2
0,0
(D,R) is the unique NE. (Note this is also a dominant strategy equilibrium as both players are
playing their strictly dominant strategies)
Game 3
Player 2
Player 1
U
D
L
5,0
0,5
R
0,3
3,0
There are no Pure NE. But as this is a finite game, we know from the Nash Existence Theorem
that there must exist a NE. If it is not in pure strategies, then it must be in nondegenerate
mixed strategies. Indeed, this NE is where p = 58 and q = 83 .
1
Game 4 - A little tricky
Player 2
Player 1
U
D
L
1,0
1,-1
R
1,0
2,2
There are 2 Pure NE (U,L) and (D,R). This game is tricky because people intuitively think that
(U,L) should not be a reasonable prediction of human behaviour as Player 1 has a weakly
dominant strategy to play D, and Player 2 has a weakly dominant strategy to play R. However,
your task is to find NE, and the definition of NE means that (U,L) is NE, regardless of any
intuitive argument. The reasoning goes like this (you should see that this is the reasoning you
used when you did the underlining!):
Player 1 thinks - Given that Player 2 will play L for sure, my best responses are both U and D.
Player 2 thinks - Given that Player 1 will play U for sure, my best responses are both L and R.
There are no nondegenerate mixed here. A rule of thumb is that there are an odd number of NE
in these grids, but I am showing you here that this is not always the case.
Game 5 - Hard
Player 2
Player 1
U
D
L
2,0
1,-1
R
0,0
1,2
There are 2 Pure NE (U,L) and (D,R). Notice that Player 2 has a weakly dominant strategy to
play R. You probably found that there is a strange mixed NE with p = 1 and q = 12 , meaning
that Player 1 plays U for sure, while Player 2 mixes over L and R with equal probability. This is
indeed a mixed NE. Understand what this means, it means that to make Player 2 indifferent,
Player 1 must play U for sure. This should be obvious as Player 2 has a weakly dominant
strategy to play R, so the only way you can make Player 2 have L as a best response is if Player
1 plays U for sure. To keep Player 1 indifferent, Player 2 must play q = 21 , which he is only
willing to do if he is indifferent, which he is when p = 1. This is the standard nondegenerate
mixed NE reasoning.
However, the full answer
is that there is a continuum of non-degenerate mixed NE characterised
by p = 1 and q ∈ 12 , 1 . To see why, consider the mixed equilibrium where both players are
indifferent, with p = 1 and q = 12 . Now consider any q > 12 . Notice that this makes Player 1
strictly prefer to play U, so in any NE Player 1 will do that. But if Player 1 plays U, Player 2 is
indifferent between L and R and is willing to mix with any q, but we need q ≥ 12 to get Player 1
to go U. Note that for any q < 12 , this ceases to be NE because Player 1 will have an incentive to
play D rather than U. This strange solution has emerged because of the strange formulation of
payoffs, involving one player with a weakly dominant strategy.
So the answer to this question is that there are 2 Pure NE and a continuum of nondegenerate
mixed NE.
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