1. GEK1544 The Mathematics of Games
Suggested Solutions to Tutorial 9
1. Consider the payoff diagram of a zero sum game between A and B.
B1 B2 B3
A1 −4 6 −6
A2 −1 4 3
A3 −2 −3 −1
(a) Argue by eliminating dominated (or weaker) strategies that there is a pure pair of
strategies (each for each player) for this game. Explain what should happen with rational
play, and what payoff will result?
(b) Now change the A3 B3 payoff from −1 to 4 :
B1 B2 B3
A1 −4 6 −6
A2 −1 4 3
A3 −2 −3 4
Show that there are no dominated strategies. Then use the Maxi-mini method to find a
saddle-point solution. That is, the situation where
Maxi-mini (A) = −Maxi-mini (B) .
Answer. (a) Compare A2 with A3 :
B1 B2 B3
A2 −1 4 3
A3 −2 −3 −1
we see that −1 > −2 ; 4 > −3 ; 3 > −1. Hence A3 is a weaker strategy when compared to
A2 . We can eliminate A3 . Continue to eliminate weaker strategies, the diagram becomes
B1 B3
A2 −1 3
Thus the pure pair of strategies are A1 and B1 , where the payoff for A is −1, and for B,
+ 1.

2. (b) Maxi-mini (A) = −1 = −Maxi-mini (B) .
2. Consider the payoff diagram of a zero sum game between A and B.
B1 B2 B3
A1 3 −2 2
A2 −1 0 4
A3 −4 −3 1
(a) Argue by eliminating dominated (or weaker) strategies that the only meaningful
strategies are A1 , A2 , B1 and B2 .
(b) Show that the Maxi-mini method does not yield a saddle point.
(c) Using the probability method, show that A can avoid having to make an expected
payment of more than 1/3 by choosing A1 with probability 1/6 and A2 with probability
5/6, no matter what B chooses. B can ensure an expected gain of at least 1/3 by using
a randomized strategy of choosing B1 with probability 1/3 and B2 with probability 2/3,
no matter what A chooses.
Suggested solution. (a) As −1 > −4, 0 > −3 and 4 > 1, we can eliminate A3 .
Afterward, we find that 2 > −2 and 0 > −4, strategy B3 is also eliminated.
(b) Using the 2 × 2 table obtained in (a) :
B1 B2
A1 3 −2
A2 −1 0
the Maxi-mini for A equals to −1 , while the Maxi-mini for B equals to 0 . Hence we
don’t have a saddle point.
(c) Applying the mixed strategy method, let p be the probability that A plays according
to strategy A1 , [ therefore (1 − p) the probability that A plays according to strategy A2 ] .
Likewise, set q be the probability that B plays according to strategy B1 , [ therefore (1−q)
the probability that B plays according to strategy B2 ] . We have the following tables.
B1 B2 B1 B2
A1 3 −2 A1 p q p (1 − q)
A2 −1 0 A2 (1 − p)q (1 − p)(1 − q)
=⇒ X(A) = 3p q − 2 p (1 − q) − (1 − p) q
= 6p q − 2p − q
2 1
= 6 pq − p − q
6 6
1 1 1
= 6 p− q− − .
6 3 3

3. If A want to make X(A) > − 1 , A may try to increase p so that p > 6 . But B knows of
3
1
1
A’s intention, and B will take q so that q − 3 is negative, say q = 0 . It results a more
negative X(A) then − 1 . Thinking along this line, the optimal value for p is 61.
3
Similarly, we work to find X(B) , and argue along the same line for B .
3. An airline loses two suitcases belonging to two different travelers. Both suitcases
happen to be identical and contain identical antiques. An airline manager tasked to settle
the claims of both travelers explains that the airline is liable for a maximum of $100 per
suitcase, and in order to determine an honest appraised value of the antiques the manager
separates both travelers so they can’t confer, and asks them to write down the amount
of their value at no less than $2 and no larger than $100. He also tells them that if
both write down the same number, he will treat that number as the true dollar value of
both suitcases and reimburse both travelers that amount. However, if one writes down a
smaller number than the other, this smaller number will be taken as the true dollar value,
and both travelers will receive that amount along with a bonus/malus: $2 extra will be
paid to the traveler who wrote down the lower value and a $2 deduction will be taken
from the person who wrote down the higher amount.
What is (are) the Nash equilibrium ?
Suggested solution. Any equal numbers (x, x) with 2 < x ≤ 100 cannot be a Nash
equilibrium. This is because the first traveler can gain by changing the quote to, say x−1,
while the second traveler stays at x.
Any two different numbers (x, y) with 2 ≤ x ≤ 100 and 2 ≤ y ≤ 100 , cannot be a Nash
equilibrium. This is because, say x < y, then the situation (x, x) will give the second
traveler a higher payoff.
It follows that (2, 2) is the only Nash equilibrium . To check that it is a Nash equilibrium,
suppose it changes to, say (3, 2) (the first traveler cannot quote an amount smaller than
2), the payoff for the first traveler is
2 − 2 = 0.
That is, it drops from the equilibrium payoff of $ 2. Likewise, we check the other case.