Game Theory
Your Task
 Split
into pairs and label one person A and
the other B.
 You are theoretically about to play a
game involving money. Player A has two
strategies of playing this game and player
B has three strategies. The matrix on the
next slide shows the possible winnings for
A depending on the strategies chosen by
each player. This is called a pay-off
matrix.
Your Task
B
X
Y
Z
P
1
-2
2
Q
3
2
4
A
Work in pairs to decide:
1) Which strategies give A the most amount of money?
2) Which strategies give B the most amount of money?
3) Which strategy allows for one person to lose the same
amount of money that the other person wins?
Play-Safe Strategy
 To
find a play-safe strategy, we look for the worst
possible outcome for each option, and then
choose the option in which the worst possible
outcome is least bad.
Why?
X
Y
Z
Minimum
Outcome
P
1
-2
2
-2
Q
3
2
4
2
3
2
4
B
A
Why?
Maximum
Outcome
Stable Solution
 If
neither player can improve their strategy if the
other plays safe, the game has a stable
solution.
 Does our game have a stable solution?
B
Minimum
Outcome
X
Y
Z
P
1
-2
2
-2
Q
3
2
4
2
3
2
4
A
Maximum
Outcome
This is hence called the value of the game.
The position of this point is called the saddle point.
Optimal Mixed Strategies
 In
many two-person zero-sum games, there is no
stable solution. The optimal strategy is therefore
found by using two or more options with a fixed
probability of choosing each option. This is
called a mixed strategy.
 We will look at games with a 2x2 payoff matrix.
Your Task


Look at the example on your hand-out of how to work
out the value of the game and optimal mixed strategy
of both players for a 2x2 pay-off matrix.
Read through and understand the example, then
apply your knowledge to the following question, in the
same way, to find a solution!
Your Question: A two-person zero-sum game has the following 2x2
pay-off matrix:
B
X
Y
P
-3
4
Q
2
1
A
Find the value of the game and the optimal mixed strategy of both
players.
The Solution
Let A choose option P with probability p and
option Q with probability 1-p.
If B chooses option X, expected pay-off for
A = -3p + 2(1 - p) = 2 - 5p
If B chooses option Y, expected pay-off for
A = 4p + 1(1 – p) = 3p + 1
Equating the expected pay-offs we have
2 – 5p = 3p + 1
1 = 8p
p = 1/8
The Solution
A should choose option P with probability 1/8 and
option Q with probability 7/8.
Value of game = 2 – 5p = 2 – (5 x 1/8) = 11/8
Let B choose option X with probability q and
option Y with probability 1 – q
If A chooses option P, the expected pay-off for
B = -3q + 4(1 – q) = 4 – 7q
Equating this to the value of the game, we get:
4 – 7q = 11/8
q = 3/8
B should therefore choose option X with
probability 3/8 and option Y with probability 5/8.
Game Theory in Real Life
“Mediation
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and the
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Can you think
of any more
examples?
Quotes from http://blogs.cornell.edu/info2040/2011/09/21/solving-reallife-problems-with-game-theory/
Evaluation
1)
2)
3)
4)
What do you feel are the advantages of
Game Theory?
What is difficult about the idea of Game
Theory?
How do you think the work we have
done today could be made more
difficult?
What have you learnt from today’s
lesson?
A Well-Known Example
Please click the following link to see the
YouTube video:
http://www.youtube.com/watch?v=iZKErrv
VMaY&feature=player_detailpage