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Unformatted text preview: Stat 155 Fall 2009: Solutions to Homework 3
(was due October 1, 2009) 1. The payoﬀ matrix is shown below, with player I’s choices along the rows,
and player II’s choices along the columns. Black
8 Using equalizing strategies, as discussed in class, we will solve for the
optimal p to ﬁnd the value of the game where p is the probability that
Player I chooses the Black Ace, and 1 − p is the probability that Player
I chooses the Red 8. Equating Player I’s expected payoﬀs under the two
cases of Player II playing Black or Red, we get:
1 · p − 7(1 − p) = −2p + 8(1 − p)
We solve this to get p = 5/6. This gives the value of the game to be −1/3.
If Player II’s optimal strategy is (q, 1 − q), where q is the probability that
Player II chooses Black, then we can use the same method to solve for q,
getting q = 5/9. Thus,
Player I’s optimal strategy: (5/6, 1/6),
Player II’s optimal strategy: (5/9, 4/9), and
Value of the game = −1/3.
2. Payoﬀ matrix:
1 Scissors 1
−1 0 Let player I’s strategy be (p1 , p2 , p3 ), where p1 + p2 + p3 = 1. Then we
can set up the equations to get:
p2 − p3 = p3 − p1 = p1 − p2 , and using the constraint on the pi , this is
easily solved to give us the optimal strategy (1/3, 1/3, 1/3) for player I
(and player II, as a matter of fact).
3. First, let us set up the payoﬀ matrix for the original game (without Olaf’s
suggested side payment). Alex’s choices are along the rows and Olaf’s
along the columns:
2 −10 110
Solving for the optimal strategies, we get that they should both use the
same strategies of (24/27, 13/37), and the value of the game works out to
be about +32.16 cents, thus favoring Alex. If he pays Olaf 42 cents before
each game, the game’s value changes to −9.84 cents, and thus will favor
Olaf. It will still be an unfair game. ...
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This note was uploaded on 10/02/2009 for the course STAT 87531 taught by Professor Muralistoyanov during the Spring '09 term at University of California, Berkeley.
- Spring '09