Question
In a route-choice simulation, two rescue crews, A and B, each secretly choose either the narrow Pass (P) or a Detour (D). The two crews cannot both use the Pass safely. Their exercise scores, in points, are specified by the following game table; the first number is A's score and the second is B's:
| A's choice | B chooses P | B chooses D |
|---|---|---|
| P | \((−4, −5)\) | \((5, 1)\) |
| D | \((1, 4)\) | \((2, 2)\) |
Each crew wants to maximise its own expected score. First find the mixed Nash equilibrium in which each crew independently randomises its route. Then a dispatcher offers a different arrangement: they draw a joint pair of route recommendations, send each crew only its own recommendation, and announce the lottery's probabilities in advance. Neither crew sees the other message. Construct a joint lottery, using \((P,D)\), \((D,P)\) and \((D,D)\) with positive probabilities and never \((P,P)\), that makes both crews better off in expectation than the independent mixed equilibrium. Verify that neither crew gains by disobeying either recommendation, given what that message tells them.
Get hint
Hint 1
For independent mixed Nash, make each crew indifferent between P and D given the other's mixing probability. With a private recommendation, update what each crew believes about the other message.
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Solution
Let \(p\) be A's independent probability of P and \(q\) B's. A's payoff from P against B's mix is \(5(1-q)-4q=5-9q\); from D it is \(2(1-q)+q=2-q\). Indifference gives \(q=3/8\). B's payoff from P against A's mix is \(4(1-p)-5p=4-9p\); from D it is \(2(1-p)+p=2-p\). Hence \(p=1/4\). Both probabilities are interior, so this is the fully mixed Nash equilibrium. The corresponding expected scores are \(5-9(3/8)=13/8\) for A and \(4-9(1/4)=7/4\) for B.
Now consider the correlated recommendation distribution
\(\Pr(A=P,B=D)=2/5\), \(\Pr(A=D,B=P)=2/5\), \(\Pr(A=D,B=D)=1/5\), and \(\Pr(P,P)=0\).
A's expected score is \((2/5)5+(2/5)1+(1/5)2=14/5=2.8\); B's is \((2/5)1+(2/5)4+(1/5)2=12/5=2.4\). Both exceed their mixed Nash scores (1.625 and 1.75).
We must verify obedience, not just high average scores. If A is told P, B must have been told D; A earns 5 by obeying versus 2 by switching to D. If B is told P, A must have been told D; B earns 4 by obeying versus 2 by switching. If A is told D, conditional on that message B was told P with probability \((2/5)/(3/5)=2/3\), and D with probability \(1/3\). A's expected obeying score is \((2/3)1+(1/3)2=4/3\), versus \((2/3)(-4)+(1/3)5=-1\) if A deviates to P. If B is told D, A was told P with probability \(2/3\) and D with \(1/3\); B scores \(4/3\) by obeying and \((2/3)(-5)+(1/3)4=-2\) by deviating. All four inequalities favour obedience. This is a correlated equilibrium: a shared private lottery coordinates the crews in a way independent mixing cannot reproduce, without requiring any crew to observe the other's recommendation.
Assumptions
This is an explicitly scored simulation; scores are stipulated game points, not unexplained financial utility.
Each crew knows the game and recommendation lottery, observes only its private message, and may choose either route regardless of the message.
There is no communication after private recommendations and no repeated-game enforcement.
Final Answer
Independent mixed Nash: A uses P with probability \(1/4\), B with probability \(3/8\), for expected scores \(13/8\) (A) and \(7/4\) (B). One obedient correlated lottery is \(\Pr(P,D)=2/5\), \(\Pr(D,P)=2/5\), \(\Pr(D,D)=1/5\). It gives expected scores \(14/5\) and \(12/5\), improving both.