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As I mentioned in the previous article "The Good, the Bad and the Weird: Duels, Truels and Utility Functions", a classic probability puzzle involves a 3-way duel (called a "truel").
A, B and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unit. What should A's strategy be?
There's a subtle issue involved in these types of problems in that we don't know how each participant values each outcome. If we allow duelists to deliberately miss there are \(2^3-1=7\) possible outcomes; each person may or may not be shot and at least one person will not be shot. Even if deliberate missing isn't allowed, there are still 3 possible outcomes. A, for example, could conceivably value B winning more than C winning.
The classic solution concludes that with the given hit probabilities the optimal strategy for the first trueler is to deliberately miss. My contention is that this is an incomplete solution; for some sets of utility values this may not be the first trueler's optimal strategy. Let's examine duels using utility values as the first step towards addressing truels.
The classic solution concludes that with the given hit probabilities the optimal strategy for the first trueler is to deliberately miss. My contention is that this is an incomplete solution; for some sets of utility values this may not be the first trueler's optimal strategy. Let's examine duels using utility values as the first step towards addressing truels.
Let the two duelers have hit probabilities \( p_i \) for \( i=1,2\), and also let the values each assigns to a dueling win, loss or tie be \( W_i, L_i, T_i\). I'm assuming these values are all known to both duelers and that \( W_i > T_i > L_i\). Note that a strategy that optimizes expected utility is preserved under a positive affine transformation, so under the assumption that all values are finite, we may assume \(W_i=1\) and \(L_i=0\) for all \(i\).
We'll declare the duel a "gentleman's draw" if both duelers deliberately miss in a single round. Let the expected utility value for dueler \(i\) be \(U_i\).
Trivially, if the optimal strategy for dueler 1 is trying for a hit, the optimal strategy for dueler 2 must be to also try for a hit. Conversely, the optimal strategy for dueler 1 can be to deliberately miss if and only if the optimal strategy for dueler 2 is also to deliberately miss. Assume the draw is taken if there's indifference (they are, after all, gentlemen).
If dueler 1 deliberately misses, dueler 2 has two choices. If he deliberately misses, it's a gentleman's draw and \(U_2 = T_2\); if he tries for a hit, both duelers will subsequently try to hit each other and \(U_2 = p_2 + q_2 q_1 U_2 \). Solving, we get \[U_2 = \frac{p_2}{1-q_1 q_2}.\] It's therefore optimal for dueler 2 to take the gentleman's draw if and only if \[T_2 \geq \frac{p_2}{1-q_1 q_2}.\] As a consequence, dueler 1 will not deliberately miss if \[T_2 < \frac{p_2}{1-q_1 q_2}.\] If dueler 1 tries for a hit, both will subsequently try to hit each other and we have \( U_1 = p_1 + q_1 q_2 U_1\). Solving, we get \[ U_1 = \frac{p_1}{1-q_1 q_2}.\] Thus, the optimal strategy for dueler 1 is to deliberately miss if and only if \[ T_1 \geq \frac{p_1}{1-q_1 q_2}\] and \[T_2 \geq \frac{p_2}{1-q_1 q_2}.\] These are, as expected, precisely the conditions under which dueler 2 will deliberately miss.
For example, if \(p_1=p_2=1/2\), it'll be a gentleman's draw if and only if \(T_1,T_2 \geq 2/3\). Paradoxically, both will fire under these hit probabilities if \( T_1 = 4/5 \) and \( T_2 = 1/2 \), even though this results in lower expected utility for both duelers than if they had agreed to a draw. This is a type of prisoner's dilemma.
We'll declare the duel a "gentleman's draw" if both duelers deliberately miss in a single round. Let the expected utility value for dueler \(i\) be \(U_i\).
Trivially, if the optimal strategy for dueler 1 is trying for a hit, the optimal strategy for dueler 2 must be to also try for a hit. Conversely, the optimal strategy for dueler 1 can be to deliberately miss if and only if the optimal strategy for dueler 2 is also to deliberately miss. Assume the draw is taken if there's indifference (they are, after all, gentlemen).
If dueler 1 deliberately misses, dueler 2 has two choices. If he deliberately misses, it's a gentleman's draw and \(U_2 = T_2\); if he tries for a hit, both duelers will subsequently try to hit each other and \(U_2 = p_2 + q_2 q_1 U_2 \). Solving, we get \[U_2 = \frac{p_2}{1-q_1 q_2}.\] It's therefore optimal for dueler 2 to take the gentleman's draw if and only if \[T_2 \geq \frac{p_2}{1-q_1 q_2}.\] As a consequence, dueler 1 will not deliberately miss if \[T_2 < \frac{p_2}{1-q_1 q_2}.\] If dueler 1 tries for a hit, both will subsequently try to hit each other and we have \( U_1 = p_1 + q_1 q_2 U_1\). Solving, we get \[ U_1 = \frac{p_1}{1-q_1 q_2}.\] Thus, the optimal strategy for dueler 1 is to deliberately miss if and only if \[ T_1 \geq \frac{p_1}{1-q_1 q_2}\] and \[T_2 \geq \frac{p_2}{1-q_1 q_2}.\] These are, as expected, precisely the conditions under which dueler 2 will deliberately miss.
For example, if \(p_1=p_2=1/2\), it'll be a gentleman's draw if and only if \(T_1,T_2 \geq 2/3\). Paradoxically, both will fire under these hit probabilities if \( T_1 = 4/5 \) and \( T_2 = 1/2 \), even though this results in lower expected utility for both duelers than if they had agreed to a draw. This is a type of prisoner's dilemma.
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