Bob and Charlie are playing a game, where they count “one, two, three” and then
ID: 2922716 • Letter: B
Question
Bob and Charlie are playing a game, where they count “one, two, three” and then simultaneously show either one or two fingers each. If the total number of fingers shown is odd (i.e. 3), then Charlie pays Bob 3$. If the sum is even (i.e. 2 or 4), then Bob pays Charlie the value of the sum in dollars. Suppose Charlie plays one finger with probability q, and two fingers with probability 1 q. Let X = Bob’s payoff, i.e. the amount of money he wins or loses. (a) Calculate Bob’s average payoff, if he plays one finger with probability r and two fingers with probability 1 r. 1 (b) Which value of r (call this rc) should Bob choose, in order to maximize his expected payoff for a given q? What are the average payoffs to Bob in this case? (c) Suppose Charlie knows that Bob will choose his optimal strategy rc. Which value of q should he choose, in order to minimize the payoff to Bob (i.e. in order to maximize his own average payoff)? What is the average payoff to Bob in this case? (d) Whom would you rather be in this game?
Explanation / Answer
Pr( Charlie - 1 finger) = q
Pr( Charlie - 2 finger) = 1- q
X = Bob's payoff
When Charlies shows 1 finger and Bob's shows 2 figer
X = 3 ; Pr(X = 3) = q(1-r) + r (1-q) = r + q - 2rq
X =2 ; Pr(X =2) when both shows 1 fingers = qr
X = 4; Pr( X = 4) when both shows 2 fingers = (1-q) (1-r) = 1 - q - r - qr
Average Payoff = X * P(X) = -2 * rq + 3 * ( r + q - 2rq) - 4 * ( 1 - q - r + qr)
= -2rq + 3r + 3q - 6rq - 4 + 4q + 4r - 4rq
= -12rq + 7q + 7r - 4
(b) to maximise its average payoff, which is (-12rq + 7q + 7r -4).
so, E(r) = -12rq + 7q + 7r -4
dE(q)/ dq = - 12r + 7
dE(q)/ dq = 0
-12r + 7 = 0
r = 7/12
rc = 7/12
Average payoff in this case = (-12rq + 7q + 7r -4)rc = 7/12 = 1/12
(c) Bob has choosen his optimal strategy rc .
Expected payoff for Bob =(-12rq + 7q + 7r -4)rc = 7/12
= -12 * (7/12) q + 7q + 7 * (7/12) - 4
= 1/12
so it doesn't matter that what q we will choose it will be minimum payoff to Charlie when it is maximum for Bob.
the average payoff to Bob in this case = 1/12
(d) I would rather be Bob as minimum payoff for Bob is equal to 1/12 in worst case.