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    <p>I take it you tried the following:</p> <pre><code>private static boolean goForBiggerResource() { return false; }; </code></pre> <p>where none of the player try to go for the resource that is worth 2. They are hence guaranteed to each get a resource worth 1 every time hence:</p> <p><em>Payoff per player: 1.0</em></p> <p>I suppose also that if you ask this nice question is because you guess there's a better answer. </p> <p>The trick is that you need what is called a "mixed strategy".</p> <p><em>EDIT</em>: ok here I come with a mixed-strategy... I don't get how Patrick found the 20% that fast (when he commented, only minutes after you posted your question) but, yup, I found out basically that same value too:</p> <pre><code>private static final Random r = new Random( System.nanoTime() ); private static boolean goForBiggerResource() { return r.nextInt(100) &lt; 21; } </code></pre> <p>Which gives, for example:</p> <p><em>Payoff per player: 1.0951035</em></p> <p>Basically if I'm not mistaken you want to read the Wikipedia page on the "Nash equilibrium" and particularly this: </p> <p><em>"Nash Equilibrium is defined in terms of mixed strategies, where players choose a probability distribution over possible actions"</em></p> <p>Your question/simple example if I'm not mistaken also can be used to show why colluding players can do better average payoffs: if players could colude, they'd get 1.25 on average, which beats the 1.095 I got.</p> <p>Also note that my answers contains approximation errors (I only check random numbers from 0 to 99) and depends a bit on the Random PRNG but you should get the idea.</p>
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    1. POSolving a simple maximization game
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    1. COI don't see there's any strategy that, in the limit, is guaranteed to be better than "return false" here.
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    2. COSee Frigo's answer to see how to compute precisely the probability needed so that the mixed-strategy is optimal given the problem's constraints
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