# Game Theory: determining the value of the foxhole game

"A soldier can hide in one of five foxholes, and a gunner can hide in four spots: A, B C, and D. The configuration looks like this: 1 (A) 2 (B) 3 (C) 4 (D) 5. If a shot is fired at a location and the soldier is in an adjacent foxhole (ex: shot is fired at B and soldier is in hole #2 or #3), the gunner received a reward of 1. Otherwise, the gunner receives a reward of 0. Assume that this is a zero-sum game

We are given that the optimal strategy for the soldier is to hide 1/3 of the time in foxholes 1, 3 and 5. For the gunner, an optimal strategy is to shoot 1.3 of the time at A, 1/3 of the time at D, and 1/3 of the time at B or C.

I have to determine the value of the game for the gunner. Honestly, I have no idea where to start. I know why the player should hide at holes one and five, but not three. How would I go about solving this>.

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Edited. Thank you –  user1764389 Feb 5 '13 at 3:11
Wow, edited again. Yes, one can assume that it is a zero sum game. Sorry for all of the mistakes. CS major taking a ST485 course :/ –  user1764389 Feb 5 '13 at 3:35
Without going into the details of mixed-strategy Nash equilibria (which is what they mean in this case when they say this is an "optimal strategy"), you can check that this pair of strategies is optimal for both players in the sense that neither player can gain from shifting some of the probability from one option to another: If the soldier takes the gunner's strategy as given, all the soldier's options that are assigned a non-zero probability ($1$, $3$ and $5$) yield the same expected payoff, and vice versa if the gunner takes the soldier's strategy as given. –  joriki Feb 5 '13 at 8:39
Note that, even if we accept the slightly inaccurate use of "optimal strategy" instead of "mixed-strategy Nash equilibrium", it's wrong and rather misleading to speak of "the optimal strategy for the soldier" -- a Nash equilibrium is an entire strategy profile, in this case a pair of strategies, one for each player, and the soldier's strategy is only "optimal" with respect to the gunner's strategy. There's no such thing as the optimal strategy for the soldier, since that depends on the gunner's strategy; there's only what might be called a mutually optimal pair of strategies. –  joriki Feb 5 '13 at 8:43

HINT: Look at the payoff matrix below. The top line shows the soldier’s options, and the lefthand column shows the gunner’s options. The numbers in the body of the table show the gunner’s payoff for each combination of options chosen by him and the soldier. You’ve been told that the soldier chooses each of his options with probability $1/3$, and that the gunner does the same. Thus, each of the nine outcomes is equally likely. The value of the game to the gunner is his expected payoff, which is simply the weighted average of the $9$ possible payoffs, each weighted according to its probability of occurrence.

$$\begin{array}{r|ccc} &1&3&5\\ \hline A&1&0&0\\ BC&0&1&0\\ D&0&0&1 \end{array}$$

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So, formally, the value of the game would be: 1/9(1) + 1/9(0) + 1/9(0) + 1/9(0) + 1/9(1) + 1/9(0) + 1/9(0) + 1/9(0) + 1/9(1), which equals 1/3? –  user1764389 Feb 5 '13 at 3:24
@user1764389: Formally it would just be an algebraic version of what I said more understandably in words. Are you asking what it would be numerically? If so, let me ask first whether you know what either a weighted average or an expected value is. –  Brian M. Scott Feb 5 '13 at 3:27
Well, a weighted average is an average where some points contribute more to the average than others (in this case, the holes where the gunner actually scores points). But I guess I meant numerically, but what would it be algebraically? –  user1764389 Feb 5 '13 at 3:30
@user1764389: I wouldn’t mess about with algebra here: there are $9$ equally likely outcomes, and $3$ of them pay $1$ while the rest pay $0$, so on average the gunner wins how much? –  Brian M. Scott Feb 5 '13 at 3:33
It would just be 1/3, correct? Essentially, 1(1/9) + 1(1/9) + 1(1/9)? –  user1764389 Feb 5 '13 at 3:36