Section 10.1, Problem 1

Part a

\[\begin{array}{crccc} \ & \ & \ & \textbf{Colin} & \ \\ \ & \ & \textbf{C1} & \ & \textbf{C2} \\ \ & \textbf{R1} & 10 & \Longleftrightarrow & 10 \\ \textbf{Rose} & \ & \Uparrow & \ & \Uparrow \\ \ & \textbf{R2} & 10 & \Longrightarrow & 10 \end{array}\]

The game has a pure Nash equilibrium with a value of 10. Strategy R1 maximizes Rose’s value regardless of Colin’s strategy. While the value can be achieved at more than one combination of strategies, it is still a Nash equilibrium since neither party can benefit by departing from that strategy (i.e. either (R1, C1) or (R2, C2)).

Part c

\[\begin{array}{crccc} \ & \ & \ & \textbf{Pitcher} & \ \\ \ & \ & \textbf{Fastball} & \ & \textbf{Knuckleball} \\ \ & \textbf{Guesses fastball} & 0.400 & \Longrightarrow & 0.100 \\ \textbf{Batter} & \ & \Uparrow & \ & \Downarrow \\ \ & \textbf{Guesses knuckleball} & 0.300 & \Longrightarrow & 0.250 \end{array}\]

The game has a pure Nash equilibrium with a value of 0.250. Pitcher strategy Knuckleball minimizes the score regardless of batter strategy; given this pitcher strategy, batter strategy Guess knuckleball maximizes the score.

Section 10.2, Problem 2a

Referring to \(x\) as the portion of times that Rose plays strategy R1 and \(1-x\) the portion of the time that Rose plays strategy R2, her goal is to maximize the payoff \(P\). If Colin plays purely strategy C1, the expected value of \(P\) is \(10x + 5(1-x)\); if he plays purely strategy C3, the expected value of \(P\) is \(10x\). Thus, since \(x\) is a probability, the linear program for Rose is

\[\textrm{Maximize }P\] Subject to \[\begin{array}{c} P \leq 10x + 5(1-x) \\ P \leq 10x \\ x \geq 0 \\ x \leq 1 \end{array}\]

If \(y\) represents the portion of the time that Colin plays strategy C1, then the expected value of \(P\) is \(10y + 10(1-y) = 10\) if Rose plays purely strategy R1 and \(5y\) if Rose plays purely strategy R2. This means that, for Colin, the linear program is

\[\textrm{Minimize }P\] Subject to \[\begin{array}{c} P \geq 10 \\ P \geq 5y \\ x \geq 0 \\ x \leq 1 \end{array}\]

Geometric Solution

From this graph, any solution along or below the line \(P = 10x\) is a feasible solution in the range \(0 \leq x \leq 1\). The maximized value of \(P\) is \(P = 10\) at \(x = 1\) – if Rose plays strategy R1 100% of the time, she is guaranteed a maximum payoff of 10.

This graph shows that there is no optimal strategy for Colin – regardless of his mix of strategies, the furthest he can minimize the payoff is to \(P=10\). Due to this, Colin should likely play a strategy of pure C2, as this places him in the best position to take advantage of suboptimal play by Rose.

Algebraic Solution

For Rose, the intersection points of the above-stated constraints are shown below:

x P Feasible
0 0 Y
1 10 Y
0 5 N
1 10 Y

For Colin, the values are below:

x P Feasible
0 10 Y
1 10 Y
0 0 N
1 5 N

As in the geometric solution, the best strategy for Rose is \(x=1\) i.e. always playing strategy R1, and Colin’s strategy does not matter.

Section 10.3, Problem 3

Investor’s Game

For the investor, the variables of interest are

  • \(P\) = Payoff
  • \(x_A\) = Portion of the time to play alternative A
  • \(x_B\) = Portion of the time to play alternative B
  • \(x_C\) = Portion of the time to play alternative C

The linear program is then \[\textrm{Maximize }P\] Subject to \[\begin{array}{rl} 3000x_A + 1000x_B + 4500x_C - P \geq 0 & Economy\ 1\\ 4500x_A + 9000x_B + 2000x_C - P \geq 0 & Economy\ 2\\ 6000x_A + 2000x_B + 3500x_C - P \geq 0 & Economy\ 3\\ x_A, x_B, x_C \geq 0 \\ x_A, x_B, x_C \leq 1 \\ x_A + x_B + x_C = 1 \\ P \geq 0 \end{array}\]

This linear program can be solved using the lpSolve R package:

Success: the objective function is 4125

The optimal strategy for the investor is \(x_A = 0.25\), \(x_B = 0\), \(x_C = 0.75\), which yields an optimal payoff of \(P = 4125\).

Economy’s Game

For the economy, the variables of interest are

  • \(P\) = Payoff
  • \(y_1\) = Portion of the time to play condition 1
  • \(y_2\) = Portion of the time to play condition 2
  • \(y_3\) = Portion of the time to play condition 3

The linear program is \[\textrm{Minimize }P\] Subject to \[\begin{array}{rl} 3000y_1 + 4500y_2 + 6000y_3 - P \leq 0 & Alternative\ A\\ 1000y_1 + 9000y_2 + 2000y_3 - P \leq 0 & Alternative\ B\\ 4500y_1 + 4000y_2 + 3500y_3 - P \leq 0 & Alternative\ C\\ y_1, y_2, y_3 \geq 0 \\ y_1, y_2, y_3 \leq 1 \\ y_1 + y_2 + y_3 = 1 \\ P \geq 0 \end{array}\]

The economy’s program is solved in the same way as the investor’s:

Success: the objective function is 4125

The optimal strategy for the economy is \(y_1 = 0.625\), \(y_2 = 0\), \(y_3 = 0.375\), which yields an optimal payoff of \(P = 4125\).

Section 10.5, Problem 3

As shown by the movement diagram below, a pure strategy exists – Rose playing R2 and Colin playing either C1 or C2 \[\begin{array}{crccc} \ & \ & \ & \textbf{Colin} & \ \\ \ & \ & \textbf{C1} & \ & \textbf{C2} \\ \ & \textbf{R1} & 0.5 & \Longrightarrow & 0.3 \\ \textbf{Rose} & \ & \Downarrow & \ & \Downarrow \\ \ & \textbf{R2} & 0.6 & \Longleftarrow & 1 \end{array}\]

Due to this, neither the equating expected values or method of oddments will return useful solutions; however, they are still conducted for demonstration purposes.

Equating Expected Values

For Rose, the expected value under each of Colin’s strategies are \[\begin{aligned} E(C1) &= 0.5x + 0.6(1-x) \\ E(C2) &= 0.3x + 1(1-x) \end{aligned}\]

where \(x\) is the portion of the time Rose uses strategy R1. Setting these equal to one another and solving,

\[0.5x + 0.6(1-x) = 0.3x + 1(1-x) \longrightarrow x = \frac{2}{3}; 1-x = \frac{1}{3}\] The value of the game is \[E(C1) = 0.5x + 0.6(1-x) = \frac{8}{15} \approx 0.5333\]

For Colin, the expected value under each of Rose’s strategies are \[\begin{aligned} E(R1) &= 0.5y + 0.3(1-y) \\ E(R2) &= 0.6y + 1(1-y) \end{aligned}\]

where \(y\) is the portion of the time Colin uses strategy C1. Setting these equal to one another and solving,

\[0.5y + 0.3(1-y) = 0.6y + 1(1-y) \longrightarrow y = \frac{7}{6}; 1-y = -\frac{1}{6}\]

Clearly this is a violation of the condition \(0 \leq y \leq 0\) that applies since \(y\) is a probability. Nonetheless, the value of the game is calculated: \[E(R1) = 0.5y + 0.3(1-y) = \frac{7}{12} - \frac{3}{60} = \frac{8}{15} \approx 0.5333\]

Method of Oddments

As above, the solution does not produce useful results:

  C1 C2 \([\Delta]\)
R1 0.5 0.3 0.2
R2 0.6 1 0.4
\([\Delta]\) 0.1 0.7 \(0.6 \neq 0.8\)

Section 10.6, Problem 2

\[\begin{array}{crccc} \ & \ & \ & \textbf{Colin} & \ \\ \ & \ & \textbf{C1} & \ & \textbf{C2} \\ \ & \textbf{R1} & (1,2) & \Longrightarrow & (3,1) \\ \textbf{Rose} & \ & \Downarrow & \ & \Downarrow \\ \ & \textbf{R2} & (2,4) & \Longleftarrow & (4,3) \end{array}\]

There is a stable Nash equilibrium at \((2,4)\) – neither player can utilaterally improve from this position.

Rose would rather Colin play C2, as it increases her potential payoff. To do this, she can issue a threat:

If Colin plays C1, Rose will play R1 This meets the critera for a threat:

The game then becomes \[\begin{array}{crccc} \ & \ & \ & \textbf{Colin} & \ \\ \ & \ & \textbf{C1} & \ & \textbf{C2} \\ \ & \textbf{R1} & (1,2) & \Longrightarrow & (3,1) \\ \textbf{Rose} & \ & & \ & \Downarrow \\ \ & \textbf{R2} & Eliminated & & (4,3) \end{array}\]

Thus Colin will choose strategy C2, and Rose will choose strategy R2, maximizing her payoff at \((4, 3)\).

Section 10.7, Problem 3

To get the table of payoffs, the probabilities must be matched in a 3-by-3 grid and multiplied by the associated payoff per each outcome and summing the two numbers:

\[\begin{array}{cccc} \ & \textbf{IL} & \textbf{IM} & \textbf{IC} \\ \textbf{DL} & (3, -5) & (3, -10) & (3, -10)\\ \textbf{DM} & (10, -5) & (8, -6) & (8, -10)\\ \textbf{DC} & (10, -5) & (10, -6) & (10, -10)\\ \end{array}\]

Summing these values and completing the movement diagram yields

\[\begin{array}{cccccc} \ & \textbf{IL} & \ & \textbf{IM} & \ & \textbf{IC} \\ \textbf{DL} & -2 & \Rightarrow & -7 & \Leftrightarrow & -7\\ \ & \Downarrow & \ & \Downarrow & \ & \Downarrow \\ \textbf{DM} & 5 & \ & 2 \ & & -2\\ \ & \Updownarrow & \ & \Downarrow & \ & \Downarrow \\ \textbf{DC} & 5 & \Rightarrow & 4 & \Rightarrow & 0\\ \end{array}\]

There is a nash equilibrium at (DC, IC) – here the game has a value of 0 and neither player can unilaterally improve.