Christopher Makler
Stanford University Department of Economics
Econ 51: Lecture 11
1
2
Heads
Tails
Heads
Tails
1
1
,
-1
-1
,
1
1
,
-1
-1
,
Each player chooses Heads or Tails.
If they choose the same thing,
they both "win" (get a payoff of 1).
If they choose differently,
they both "lose" (get a payoff of -1).
Circle best responses.
What are the Nash equilibria of this game?
1
2
Heads
Tails
Heads
Tails
1
-1
,
-1
1
,
1
-1
,
-1
1
,
Each player chooses Heads or Tails.
If they choose the same thing,
player 1 "wins" (gets a payoff of 1)
and player 2 "loses" (gets a payoff of -1).
If they choose differently,
they player 1 "loses" (gets a payoff of -1)
and player 1 "wins" (gets a payoff of 1).
Circle best responses.
What are the Nash equilibria of this game?
Equilibria with mixed strategies are sometimes the only equilibrium!
A
B
X
Y
1
2
5
4
5
0
0
4
4
4
Mixed strategy for player 1:
probability distribution over {A, B}
Belief for player 1:
probability distribution over {X, Y}
2
If the other player is playing a mixed strategy,
your expected payoff from playing one of your strategies
is the weighted average of the payoffs
\({1 \over 6}\)
\({1 \over 3}\)
\({1 \over 2}\)
\(0\)
Player 2's strategy
\({1 \over 6} \times 6 + {1 \over 3} \times 3 + {1 \over 2} \times 2 + 0 \times 7\)
\(=3\)
\({1 \over 6} \times 12 + {1 \over 3} \times 6 + {1 \over 2} \times 0 + 0 \times 5\)
\(=4\)
\({1 \over 6} \times 6 + {1 \over 3} \times 0 + {1 \over 2} \times 6 + 0 \times 11\)
\(=4\)
Player 1's expected payoffs from each of their strategies
\(X\)
\(A\)
1
\(B\)
\(C\)
\(D\)
\(Y\)
\(Z\)
6
6
,
3
6
,
2
8
,
7
0
,
12
6
,
6
3
,
0
2
,
5
0
,
6
0
,
0
9
,
6
8
,
11
4
,
\(X\)
\(A\)
1
2
\(B\)
\(C\)
\(D\)
\(Y\)
\(Z\)
6
6
,
3
6
,
2
8
,
7
0
,
12
6
,
6
3
,
0
2
,
5
0
,
6
0
,
0
9
,
6
8
,
11
4
,
If you are playing a mixed strategy, and the other player is playing a pure strategy, your expected payoff is the weighted average given the way you are mixing.
\({1 \over 6}\)
\({1 \over 3}\)
\({1 \over 2}\)
\(0\)
Player 2's strategy
\({1 \over 6} \times 6 + {1 \over 3} \times 6 + {1 \over 2} \times 8 + 0 \times 0\)
\(=7\)
\({1 \over 6} \times 6 + {1 \over 3} \times 3 + {1 \over 2} \times 2 + 0 \times 0\)
\(=3\)
\({1 \over 6} \times 0 + {1 \over 3} \times 9 + {1 \over 2} \times 8 + 0 \times 4\)
\(=7\)
Player 2's expected payoffs given each of 1's strategies
1
2
0
0
,
1
1
,
1
1
,
4
4
,
Top
Middle
Left
Right
Bottom
4
4
,
0
0
,
Are any of player 1's strategies
dominated by a pure strategy?
Are any of player 1's strategies
dominated by a mixed strategy?
1
2
0
0
,
1
1
,
1
1
,
4
4
,
Top
Middle
Left
Right
Bottom
4
4
,
0
0
,
Are any of player 1's strategies
dominated by a mixed strategy?
Suppose player 1 plays Top with 50% probability,
Middle with 0% probability,
and Bottom with 50% probability.
Prob. \( {1 \over 2}\)
Prob. 0
Prob. \( {1 \over 2}\)
\(\mathbb{E}(u)\)
If player 2 plays Left, what is player 1's expected utility?
If player 2 plays Right, what is player 1's expected utility?
This is better than the utility from playing Middle no matter what player 2 does => Middle is dominated by \(({1 \over 2},0,{1 \over 2})\)
We write this mixed strategy \(({1 \over 2},0,{1 \over 2})\)
2
2
1
2
0
0
,
1
1
,
1
1
,
4
4
,
Top
Middle
Left
Right
Bottom
4
4
,
0
0
,
Prob. \( {1 \over 2}\)
Prob. 0
Prob. \( {1 \over 2}\)
\(\mathbb{E}(u)\)
This is better than the utility from playing Middle no matter what player 2 does => Middle is dominated by \(({1 \over 2},0,{1 \over 2})\)
2
2
Note that Middle is never a best response to Left or Right. There's a formal result in that in games like this, if a strategy is never a best response, it must be dominated;
so if it's not dominated by a pure strategy,
it must be dominated by some mixed strategy.
If two or more pure strategies are best responses given what the other player is doing, then any mixed strategy which puts probability on those strategies (and no others) is also a best response.
2
\({1 \over 6}\)
\({1 \over 3}\)
\({1 \over 2}\)
\(0\)
Player 2's strategy
\({1 \over 6} \times 6 + {1 \over 3} \times 3 + {1 \over 2} \times 2 + 0 \times 7\)
\(=3\)
\({1 \over 6} \times 12 + {1 \over 3} \times 6 + {1 \over 2} \times 0 + 0 \times 5\)
\(=4\)
\({1 \over 6} \times 6 + {1 \over 3} \times 0 + {1 \over 2} \times 6 + 0 \times 11\)
\(=4\)
Player 1's expected payoffs from each of their strategies
\(X\)
\(A\)
1
\(B\)
\(C\)
\(D\)
\(Y\)
\(Z\)
6
6
,
3
6
,
2
8
,
7
0
,
12
6
,
6
3
,
0
2
,
5
0
,
6
0
,
0
9
,
6
8
,
11
4
,
If player 2 is choosing this strategy, player 1's best response is to play either Y or Z.
Therefore, player 1 could also choose to play any mixed strategy \((0, p, 1-p)\).
1
2
Heads
Tails
Heads
Tails
1
-1
,
-1
1
,
1
-1
,
-1
1
,
Let's return to our zero-sum game.
\((p)\)
\((1-p)\)
What is player 1's expected payoff from Heads?
Suppose player 2 is playing a mixed strategy: Heads with probability \(p\),
and tails with probability \(1-p\).
What is player 1's expected payoff from Tails?
For what value of \(p\) would player 1 be willing to mix?
1
2
Heads
Tails
Heads
Tails
1
-1
,
-1
1
,
1
-1
,
-1
1
,
\((p)\)
\((1-p)\)
For what value of \(p\) would player 1 be willing to mix?
Now suppose player 1 does mix, and plays Heads with probability \(q\) and Tails with probability \(1 - q\).
\((q)\)
\((1-q)\)
For what value of \(q\) would player 2 be willing to mix?
A mixed strategy profile is a Nash equilibrium if,
given all players' strategies, each player is mixing among strategies which are their best responses
(i.e. between which they are indifferent)
Important: nobody is trying to make the other player(s) indifferent; it's just that in equilibrium they are indifferent.
1
2
Hawk
Dove
Hawk
Dove
-2
-2
,
0
4
,
2
2
,
4
0
,
Nodes:
Branches:
Initial node: where the game begins
Decision nodes: where a player makes a choice; specifies player
Terminal nodes: where the game ends; specifies outcome
Individual actions taken by players; try to use unique names for the same action (e.g. "left") taken at different times in the game
Information sets:
Sets of decision nodes at which the decider and branches are the same, and the decider doesn't know for sure where they are.
A "tree" representation of a game.
She chooses to give one of three gifts:
X, Y, or Z.
1
X
Y
Z
Player 1 makes the first move.
Initial node
Player 1's actions at her decision node
(and decision node)
Twist: Gift X is unwrapped,
but Gifts Y and Z are wrapped.
(Player 1 knows what they are,
but player 2 does not.)
After each of player 1's moves,
player 2 has the move: she can either accept the gift or reject it.
2
Accept X
Reject X
2
1
X
Y
Z
We represent this by having an information set connecting
player 2's decision nodes
after player 1 chooses Y or Z.
2
2
Player 2's actions
Player 2's decision nodes
Information set
Accept Y
Reject Y
Accept Z
Reject Z
Also: player 2 cannot make her action contingent on Y or Z; her actions must be "accept wrapped" or "reject wrapped"
Accept Wrapped
Reject Wrapped
Accept Wrapped
Reject Wrapped
After player 2 accepts or rejects the gift, the game ends (terminal nodes) and payoffs are realized.
1
0
1
0
2
0
2
0
3
0
–1
0
2
2
1
X
Y
Z
,
,
,
,
,
,
Accept X
Reject X
Accept Wrapped
Reject Wrapped
Accept Wrapped
Reject Wrapped
Terminal Nodes
Player 1's payoffs
Player 2's payoffs
In this game, both players get a payoff of
0 if any gift is rejected,
1 if gift X is accepted, and
2 if gift Y is accepted.
If gift Z is accepted, player 1 gets a payoff of 3, but player 2 gets a payoff of –1.