1. A useful reduction (SAT -> game)
Theorem. SAT-solutions correspond to mixed-strategy equilibria of the following game (each agent randomizes uniformly on support)
SAT Formula:
(x1 or -x2) and (-x1 or x2 )
Solutions:
x1=true, x2=true
x1=false,x2=false
Game: x1 x2
+x1
-x1
+x2
-x2
(x1 or -x2)
(-x1 or x2)
default x1
-,-
-,-
0,-
0,-
2,-
2,-
-,-
-,-
- ,1 x2
-,-
-,-
2,-
2,-
0,-
0,-
-,-
-,-
- ,1
+x1
-,0
-,2
1,1
-,-
1,1
1,1
-,0
-,2
- ,1
-x1
-,0
-,2
-,-
1,1
1,1
1,1
-,2
-,0
- ,1
+x2
-,2
-,0
1,1
1,1
1,1
-,-
-,2
-,0
- ,1
-x2
-,2
-,0
1,1
1,1
-,-
1,1
-,0
-,2
- ,1
(x1 or -x2)
-,-
-,-
0,-
2,-
2,-
0,-
-,-
-,-
- ,1
(-x1 or x2)
-,-
-,-
2,-
0,-
0,-
2,-
-,-
-,-
- ,1
default
1,-
1,-
1,-
1,-
1,-
1,-
1,-
1,-
0,0

2. Proof sketch:
Playing opposite literals (with any probability) is unstable because the other player will prefer to play a variable move . For example, if the row player plays –x1 then the column player will play x2 which would prompt the first player to change his/her move to default. (Note as soon as one player plays default, the other player’s best response is to also play default, leading to the Nash equilibrium at (default, default).)
If you play literals (with probabilities), you should make sure that
for any clause, the probability of the literal being in that clause is high enough, and
for any variable, the probability that the literal corresponds to that variable is high enough
(otherwise the other player will play this clause/variable and hurt you)
A player won’t play any literal with a probability greater than 1/n because the other player will prefer to play a variable strategy which has a large negative payoff for the first player. For example if the row player plays –x1 with a probability of 2/3 and –x2 with a probability of 1/3 then the column player would be motivated to play x2 which is a big hit to the row player’s payoff.
So equilibria where both randomize over literals can only occur when both randomize over same SAT solution
A player would not play a variable strategy because the other player would play the default strategy which results in a negative payoff for the first player . The first player would then decide to play the default move with a zero payoff for both players.
Similarly for the clause stategies.
The default move is needed because then it is true that there is a satisfying assignment for the SAT formula iff there is more than one mixed Nash equilibrium in the game. To see why, note that if there is no satisfying assignment, then any mixed strategy that one of the players uses would violate one of the clauses and would motivate the other player to play a clause move which leads to the default Nash; the default Nash is the only Nash and starting from any state, a sequence of best replies will lead us there. If there is a satisfying assignment, then not only will (default,default) be a Nash, but the mixed strategy profile where each player puts probability 1/n on each literal of the satisfying assignment will also be a Nash.

3. Complexity of mixed-strategy Nash equilibria with certain properties
This reduction implies that there is an equilibrium where players get expected utility 1 each iff the SAT formula is satisfiable
Any reasonable objective would prefer such equilibria to 0-payoff equilibrium
Corollary. Deciding whether a “good” equilibrium exists is NP-hard:
1. equilibrium with high social welfare
2. Pareto-optimal equilibrium
3. equilibrium with high utility for a given player i
4. equilibrium with high minimal utility
Also NP-hard (from the same reduction):
5. Does more than one equilibrium exists?
6. Is a given strategy ever played in any equilibrium?
7. Is there an equilibrium where a given strategy is never played?
(5) & weaker versions of (4), (6), (7) were known [Gilboa, Zemel GEB-89]
All these hold even for symmetric, 2-player games

4. Counting the number of mixed-strategy Nash equilibria
Why count equilibria?
If we cannot even count the equilibria, there is little hope of getting a good overview of the overall strategic structure of the game
Unfortunately, our reduction implies:
Corollary. Counting Nash equilibria is #P-hard!
Proof. #SAT is #P-hard, and the number of equilibria is 1 + #SAT
Corollary. Counting connected sets of equilibria is just as hard
Proof. In our game, each equilibrium is alone in its connected set
These results hold even for symmetric, 2-player games