The recalcitrant James and Dean are playing their more dangerous variant of chic

Question

The recalcitrant James and Dean are playing their more dangerous variant of chicken again (see Exercise S6). They've noticed that their payoff for being perceived as "tough" varies with the size of the crowd. The larger the crowd, the more glory and praise each receives from driving straight when his opponent swerves. Smaller crowds, of course, have the opposite effect. Let k > 0 be the payoff for appearing "tough." The game may now be represented as: (a) Expressed in terms of k, with what probability does each driver play Swerve in the mixed-strategy Nash equilibrium? Do James and Dean play Swerve more or less often as k increases? (b) In terms of k, what is the expected value of the game to each player in the mixed-strategy Nash equilibrium found in part (a)? (c) At what value of k do both James and Dean mix 50-50 in the mixed-strategy equilibrium? (d) How large must k be for the average payoff to be positive under the alternating scheme discussed in part (c) of Exercise S6?

Explanation / Answer

It is as we can see symmetric, so we'd be calculating for James and same would be valid for Dean.

Say probability of Dean Swerving = p,

probability of him going straight = 1-p.

Expected Payoff for James swerving = p*0 + (1-p)*(-1)

Expected Payoff for James going straight = p*k + (1-p)*(-10)

Then for mixed strategy nash equilibrium, Excpected Payoff for swerving = Expected Payoff for straight

=> (1-p)*9 = p*k

=> p = 9/(9+k)

and, 1-p = k/(9+k)

a) Swerve prob. = 9/9+k. So, as k increases, swerving probability decreases.

Straight prob. = 1-p = k/(9+k) and with increase in k, p decreases so 1-p increases, hence going straight increases

b) Now, Expected payoff for swerving(E1) = -(1-p) = -k/(9+k)

Expected payoff for going straight(E2) = p*k - 10*(1-p) = 9k/(9+k) - 10k/(9+k) = -k/(9+k)

Expected payoff = p*E1 + (1-p)*E2 = -9k/(9+k)2 - k2/(9+k)2 = -k(k+9)/(k+9)2 = -k/(k+9)

c) p = k/(9+k) = 0.5

=> k = 9/2 + k/2 => k = 9

d) In this given example Expected payoff = -k/(k+9) which is always negative for k > 0. and the alternating scheme has not been described here. So, without that, whatever k be Expected Payoff cannot be positive.

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