Question

In a version of the game Russian roulette, made infamous in the movie The Deer Hunter, two players are faced with a six-shot pistol loaded with one bullet. The players ante $1000, and player I goes first. At each play of the game, a player has the option of putting an additional $1000 into the pot and passing, or spinning the chamber and firing (at his own head). If player I chooses the option of spinning and survives, then she passes the gun to player II, who has the same two options. If player I does not survive the shot, the game is over and Il gets the pot. If player I has chosen to fire and survives, she passes the gun to player I; if player II chooses to fire and survives, the game is over and both players split the pot. If I fires and survives and then Il passes, both will split the pot. The effect of this is that II will pay $500 to I. On the other hand, if i chooses to pass and II chooses to fire, then if Il survives, he takes the pot.

Above is the Russian roulette ex. shown in class

Answer 1

It Pau safe chose 2 0 lvs.ian >oouth

Player II also has one decision to make but it depends on what player I has done and

the outcome of that decision. That’s why it’s called a sequential game. The pure strategies for

player II are summarized in the following table.

II1 :If I spins, then pass; If I passes, then spin.

II2 :If I spins, then pass; If I passes, then pass.

II3 :If I spins, then spin; If I passes, then spin.

II4 : If I spins, then spin; If I passes, then pass.

In each of these strategies, it is assumed that if player I spins and survives the shot, then

player II makes a choice. Of course if player I does not survive, then player II walks away

with the pot.

The payoffs are now random variables and we need to calculate the expected payoff to player 1.

In player1 1 against player 2 1 ,the payoff to I is 1/2 with probability 5/6 and -1 with probability 1/6 .The expected payoff to I is then

I1 against II1 : 5/6 (1/2) +1/6(-1) =1/4

and

I12 against II1 :5/6(-2)+ 1/6(1) = -3/2

Strategy II3 says the following: If I spins and survives, then spin, but if I passes, then spin
and fire. The expected payoff to I is

I1 against II3 : 5/6(5/6(0)+1/6(1)) +1/6(-1) = -1/36 and

I2 against II3 : 5/6(-2)+1/6(1) = -3/2

Continuing in this way, we play each pure strategy for player I against each pure strategy
for player II. The result is the following game matrix:

Now that we have the game matrix we may determine optimal strategies. This game is
actually easy to analyze because we see that player II will never play II1, II2, or II4 because
there is always a strategy for player II in which II can do better. This is strategy II3 that gives
player II 1/36 if I plays I1, or 3/2 , if I plays I2. But player I would never play I2 if player II plays
II3 because ?3/2 < ? 1/36 . The optimal strategies then for each player are I1 for player I, and
II3 for player II. Player I should always spin and fire. if I should always spin and fire if I has arrived his shot.

The expected payoff to player 1 is -1/36.