---

title: 博弈论 斯坦福game theory stanford week 6-2
tags: note
notebook: 6- 英文课程-15-game theory
---

博弈论 斯坦福game theory stanford week 6-

1

In the following two-player Bayesian game, the payoffs to player 2 depend on whether 2 is a friendly player (with probability pp) or a foe (with probability 1-p1−p). See the following payoff matrices for details.

With probability pp, the payoff matrix is:

Friend Left Right
Left 3,1 0,0
Right 2,1 1,0
while with probability 1-p1−p, the payoff matrix is:

Foe Left Right
Left 3,0 0,1
Right 2,0 1,1
Player 2 knows if he/she is a friend or a foe, but player 1 doesn't know. If player 2 uses a strategy of Left when a friend and Right when a foe, what is true about player 1's expected utility?

a) It is 33 when 1 chooses Left;

b) It is 3p3p when 1 chooses Left;

正确 
**(b) is true.**

If 1 chooses Left, with probability pp player 2 is a friend and chooses Left and then 1 earns 33, and with probability (1-p)(1−p) player 2 is a foe and chooses Right and then 1 earns 0. Thus, the expected payoff is 3p + 0(1-p) = 3p3p+0(1−p)=3p.

c) It is 2p2p when 1 chooses Right;

d) It is 11 when 1 chooses Right;

第 2 个问题

Consider the conflict game:

With probability pp, the payoff matrix is:

Strong Fight Not
Fight 1,-2 2,-1
Not -1,2 0,0
and with probability 1-p1−p, the payoff matrix is:

Weak Fight Not
Fight -2,1 2,-1
Not -1,2 0,0
Assume that player 1 plays fight when strong and not when weak. Given this strategy of player 1, there is a certain p^p
∗
such that player 2 will prefer 'fight' when p < p^p<p
∗
, and 'not' when p>p^p>p
∗
. For instance, in the lecture p^p
∗
was 1/3.

What is p^p
∗
in this modified game? (Hint: Write down the payoff of 2 when choosing Fight and Not Fight. Equalize these two payoffs to get p^p
∗
):

c) 2/3

正确 
(c) is true.

Conditional on 1fighting when strong and not fighting when weak, the payoff of 2 when choosing Not is -1p+0(1-p)−1p+0(1−p) and the payoff of 2 when choosing Fight is (-2)p + 2(1-p)(−2)p+2(1−p).
Comparing these two payoffs, 2 is just indifferent when -1p+0(1-p) = (-2)p + 2(1-p)−1p+0(1−p)=(−2)p+2(1−p), thus p^* = 2/3p 
∗
=2/3, above which 2 prefers Not and below which 2 prefers to Fight.

d) 1/2

b) 1/3

a) 3/4