8
$\begingroup$

I am really confused about how to think about this question. It was presented as a challenge by a peer.

Two people seek to kill a duck at a location $Y$ meters from their origin. They walk from $x=0$ to $x=Y$ together. At any time, one of the two may pull out their gun and shoot at the duck, however, the probability that person A hits is $P_{A}(x)$ and the probability that person B hits is $P_{B}(x)$. It is also known that $P_A(0)=P_B(0)=0$ and $P_A(Y)=P_B(Y)=1$ and both functions are increasing functions.

What is the optimal strategy for each player?

  • 0
    @Roah, as mentioned, it was presented as a challenge by a peer. I don't know its origins. I can try and find o$u$t.2011-11-05

4 Answers 4

9

I believe both should shoot at $P_A(x)+P_B(x)=1$. If either shoots earlier, the chance of winning is reduced. If either shoots later, the other could wait half as much later and have a better chance of winning. But what happens if they both hit or both miss?

  • 0
    @Craig: There is no unstable equilibrium in this game. There is no hidden information, so no need for any kind of mixed strategy. But you might find that you need an itchy trigger finger $-$ see my answer.2011-11-04
6

Suppose player $A$ takes a shot at distance $x$, before player $B$. He collects the price with $P_1 = p_A(x)$ probability, while player $B$ collects the price with probability $P_2 = 1-p_A(x)$.

If the player $B$ shoots first, then $A$ wins with probability $Q_1 = 1-p_B(x)$ and $B$ wins with $Q_2 = p_B(x)$.

The optimal strategy for $A$ is to shoot at the point minimizing $B$'s win, i.e. $x_A = \operatorname{argmin}_x \max(p_B(x), 1-p_A(X))$, while the optimal strategy of $B$ is to shoot at $x_B = \operatorname{argmin}_x \max(p_A(x), 1-p_B(X))$.


Here is a visualization, assuming duck is located at $Y=1$, and $p_A(x)$ and $p_B(x)$ are beta distribution cumulative distribution functions:

enter image description here

  • 2
    @picakhu Actually, my answer is equivalent to Ross's. Given that $p_A(x)$ and $p_B(x)$ are monotonic, it is easy to demonstrate that $x_A=x_B$, and at that point $p_A(x_A) + p_B(x_A) = p_A(x_B) + p_B(x_B) = 1$.2011-11-04
2

Suppose the probability functions $P_A, P_B$ are continuous, and that "increasing" means "non-decreasing". Then there is a unique maximal closed interval in which $P_A(x) + P_B(x) = 1$. Each player's strategy is identical: shoot at any time in this interval. (Ross Millikan simul-posted this answer.)

It gets a bit more complicated if $P_A$ or $P_B$ is not continuous. This is a realistic scenario $-$ for instance, the brow of a hill might obscure the duck up to a certain point (which might be different for each player). Then there might be a point $x$ before which $P_A + P_B < 1 - a$, and after which $P_A + P_B > 1 + b$, for some strictly positive $a,b$. There are two cases:

  1. $P_A$ is continuous at $x$, and $P_B$ is not. Then $P_A$ must shoot before $x$, but as shortly before $x$ as possible. Likewise if $A$ and $B$ are swapped.

  2. Neither $P_A$ nor $P_B$ is continuous at $x$. Then neither player wants to shoot before $x$, and neither player wants to allow the other to shoot after $x$. The situation becomes tense, and mathematics has little to say; in fact, the game should perhaps be called "chicken" in this case, rather than "duck".

  • 1
    Frankly, picakhu, I don't give a damn.2011-11-04
0

I think that we should be explicit about the payoffs in the case they both hit simultaneously and about the concept of optimality. I suppose that we search for the Nash equilibrium.

Let's consider 3 versions:

Fair-hunters version. The first who hits the duck gets +1. If both hit simultaneously they get +1/2 each. No Nash equilibrium in pure strategies. Simultaneous shot can not be an equilibrium because both players want to deviate and shoot $\varepsilon$ earlier.

Hunters-enemies. The first who hits the duck gets +1. If both hit simultaneously they get 0 because of a quarrel. No Nash equilibrium as in fair-hunters version.

Brothers-hunters. The first who hits the duck gets +1. If both hit simultaneously they get +1 each (they are very surprised and happy). In this case any pair of strategy $(x,x)$ such that $P_A(x)+P_B(x)\geq 1$ is a Nash equilibrium.