Situation Expected payoffs
When Player 1 chose Rock Ep R =0· (0.4) +10· (0.3) -10· (0.3) 0
When Player 1 chose Scissors Ep S =-10(0.6) +0· (0.3) +10· (0.3) -1
When Player 1 chose Paper Ep P =10· (0.4)-10(0.3) +0· (0.3) 1
Therefore, Player 1 is going to put Paper as it draws the highest payoff of 1. But it is quite unrealistic assumption. In practical situation, Player 2’s announcement cannot be believed by rival because they are in zero-sum game so both players want to better off by deceiving the opponent. As a result, Player 1 has to be rational and has to keep the opponent indifferent. So, if we calculated second situation, Player 1’s payoff would be like below
(1) Rock: 0· p1 +10· p 2-10· (1- p 1- p 2) =20 p 2+10 p 1-10
(2) Scissors: -10· p 1 +0· p 2+10· (1- p 1- p 2) =-10 p 2-20 p 1+10
(3) Paper: 10· p 1 -10· p 2+0· (1- p 1- p 2) =10 p 2-10 p 1
3) Find P which maximize expected payoff
(1) 20 p 2+10 p 1-10 = (2)-10 p 2-20 p 1+10
(2) -10 p 2-20 p 1+10 = (3) 10 p 2-10 p 1
(1) 20 p 2+10 p 1-10 = (3) 10 p 2-10 p 1
So, each P-value should be as follows
Pr R (1) = 1/3
Pr S (1) = 1/3
Pr P (1) = 1/3
In conclusion, Player 1 has to use his option with probability of 1/3 each according to keep the opponent indifferent rule.
(b) Find the mixed-strategy equilibrium of this Rock-Scissors-Paper game.
I already draw mixed-strategy equilibrium of this game in question (a). So, I will skip the explanation. Player 1 and 2 have to use their option with probability of 1/3 each to keep the opponent indifferent rule.
6. In baseba...
... middle of paper ...
...(1-p) = -3p+1+4p -3 =p-2= -1.43
In the same way, if we calculate Batter’s expected payoff, it is 10/7
So, it shows that slowing the fastball cannot improve the pitcher’s expected payoff.
1. Find Nash equilibrium in mixed strategies for the following game by using the method of best-response analysis. Draw a best-response diagram and show the equilibrium mixture on the diagram. Also indicate each player’s expected payoff in equilibrium.
1) Give probability to each option
ROW Up(p) 4, 0 -1, 2
Down(1-p) 1, 1 2, -1
2) Derive player’s expected pay-off
Row’s Expected Payoffs Column’s expected payoffs
EpU(R) = 4(q) - 1(1-q) = 5q-1 EpL(C) = 0(p) + 1(1-p) = 1-p
EpD(R) = 1(q) + 2(1-q) = -q+2 EpR(C) = 2(p) – 1(1-p) = 3p-1
3) Find p and q which maximize expected payoff
(1) q value 5q-1=-q+2 q = 0.5
(2) p value 1-p=3p-1 P = 0.5
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