Solution to Exercise 3.7 on Battleship from Part II of Fergusons Game Theory
3.7#15 Battleship
As noted in the problem description, Player I has two invariant strategies, [1, 2] and [2, 5] . To descri
Section 1.5
1.5#2 Player I holds a black Ace and a red 8. Player II holds a red 2 and a black 7. If the chosen cards
are of the same color Player I wins, if they dier Player II does. The amount won is
Solutions to Homework 2, Math 432 Spring 2013 Exercises from Parts I and II of Fergusons
Game Theory
Section 5.5
5.5#3 Triplets.
Player must turn over exactly three coins. If position x corresponds to
Solutions to Exercises from Part I of Fergusons Game Theory
Section 1.5
1.5#2 Take-Away Game where you can remove 1-6 chips.
(a) The empty pile is a P-position, and clearly piles of sizes 1 through 6
(The value of V can also be obtained by plugging in p = 7/10 into either column 1s or column
4s linear function). Is strategy is (7/10, 3/10) and IIs strategy is (1/2, 0, 0, 1/2).
(b) Reduce by domina
(d) The equation (16) only gives an optimal strategy for Player II if the assumption that Player I
has an optimal strategy giving positive weight to each of the rows. holds (rst paragraph of
section 3
4.5#1 (a) Paretto-optimal outcomes: t(4, 0) + (1 t)(3, 2), 0 t 1, and t(3, 2) + (1 t)(2, 3), 0 t 1.
(b) Paretto-optimal outcomes: t(3, 1) + (1 t)(1, 2), 0 t 1.
4.5#3 (a) We need to solve the game
1
2
B
does not have a saddle point, so the safety level of Player II is
vII = Val(B ) =
16
0344
=
,
04+34
5
and its MM-strategy is q = (1/5, 4/5).
(0, 0) (2, 4 )
and hence there
(2 , 4 ) (3 , 3)
is one PS
Solutions to Homework 3 Exercises from Part II and Part III of Fergusons Game Theory
Section 4.7
4.7#1 Consider the game with matrix
0
A = 1
9
7
4
3
2
8
1
4
2 .
6
(a) A Bayes strategy against (1/5, 1/
(a) If n = 4 and Player I begins with S in the rst square, Player II responds with S in the fourth
square, resulting in a reasonable string, both of whose empty squares are unsafe.
(b) If n = 7, we cl