A non-zero-sum game between Ortega and the dog

Text

Ortega sometimes pulls out a real gun, sometimes a fake one; the dog may bite or not. The payoffs (dog; Ortega) are:

	Dog bites	Dog doesn’t bite
Real gun	(+2; −7)	(−3; +4)
Fake gun	(−4; −2)	(+3; −5)

(a) Write the game table.

(b) The dog bites at random half the time ($50/50$): find Ortega’s best strategy.

(c) The dog copies Ortega, biting with the same probability $p$ with which Ortega shows the real gun: write the dog’s average payoff as a function of $p$ and show it is maximized at $p=0$; also determine the value of $p$ that maximizes Ortega’s payoff.

Solution

(a) The payoff table is the one given in the text.

(b) If the dog bites with probability $0.5$ independently of Ortega’s choice, Ortega’s average payoff with the real gun is $0.5\cdot(-7) + 0.5\cdot4 = -1.5$ while with the fake gun it is $0.5\cdot(-2) + 0.5\cdot(-5) = -3.5$ So it is better to always show the real gun (average payoff $-1.5$, less negative).

(c) Let $p$ be the probability that Ortega shows the real gun, with the dog biting with the same probability $p$. The dog’s average payoff is $E_{\text{dog}}(p) = p\big[2p - 3(1-p)\big] + (1-p)\big[-4p + 3(1-p)\big] = 12p^2 - 13p + 3$ This is an upward-opening parabola (positive coefficient of $p^2$), so the maximum occurs at one of the endpoints of $[0,1]$: $E_{\text{dog}}(0)=3$ and $E_{\text{dog}}(1)=2$, so the maximum is at $p=0$ (the dog never bites).

Ortega’s average payoff is $E_{\text{Ortega}}(p) = p\big[-7p + 4(1-p)\big] + (1-p)\big[-2p - 5(1-p)\big] = -14p^2 + 12p - 5$ This is a downward-opening parabola: the maximum is at the vertex $p = \frac{12}{2\cdot14} = \frac{12}{28} = \frac{3}{7} \approx 0.429$

$\boxed{\text{(b) real gun};\ p_{\text{dog}}=0;\ p_{\text{Ortega}}=\tfrac37\approx0.43}$