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Two players are playing a game. The first player has unlimited gold coins of 2 types, $C_1=2\$ and $C_2=5\$. Each turn he chooses one of these coins and hides it in his hand. If the second player guesses correctly which type of coin the first player is hiding in his hand, he gets this coin; otherwise he loses $x$ cents. Find the largest integer $x$ for which the game is beneficial to the second player.

I knew the answer, but I forget how we got it. I would appreciate it if someone would explain it for me. Thank you.

Answer is x=316

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    Imho the payoff of the first player is not clearly specified. Does he receive $x$ when the second player loses $x$? Does he receive the coin when when the second player loses? It is important to determine the strategy of the first player.2012-08-26

2 Answers 2

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Since the first player has an unlimited number of coins he shouldn't care how much money he loses and there aren't any predictions on his strategy possible. But lets just assume he wants to loose as little money as possible.

Let player 1 choose the coin $C_1$ with probability $p$ and let player 2 choose the coin $C_1$ with probability $q$.

Then the expected value for player 2 is $2pq+5(1-p)(1-q)-(1-pq-(1-p)(1-q))x$ $=7pq-5p-5q+5-x(p+q-2pq)$

Player 2 wants to maximise his profit depending on the first player's strategy. So we derive wrt $q$:

$7p-5-x(1-2p)$

If the dervative is positive, player 2 will choose $q$ maximal i.e. $q=1$ with profit $2p-x(1-p)$, if it is negative, player 2 will choose $q$ minimal i.e. $q=0$ with profit $5-5p-xp$, and if it is zero, his choice doesn't matter and for all that it's worth we can assume he takes $q=0$ as well.

So player 1 wants to minimise the maximum of $2p-x+px$ and $5-5p-xp$. Since one is decreasing as a function of $p$ and the other one is increasing the equilibrium is the point where they are equal:

$2p-x+px=5-5p-xp\Leftrightarrow p=\frac{5+x}{7+2x}$

The profit of player 2 is then

$(2+x)\frac{5+x}{7+2x}-x$

This is positive if and only if $x\leq \sqrt{10}\simeq 3.16$

Edit I assume we might be able to streamline the solution a bit since this $x$ is precisely the value for which the derivative above vanishes. I suspect this is no coincidence. Unfortunatly I have no idea of the general theory and based my solution solely on common sense.

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    @Ilmari: That's true if by "component pure strategies" you mean only those pure strategies which have non-zero probability in the player's mixed strategy. The ones with zero probability may have lower payoff.2012-08-26
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This is a zero-sum game, so the solution boils down to finding the minimax strategies for the two players, calculating their expected payoffs for those strategies and finding the threshold value of $x$ for which the expected payoff to each player is zero.

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    Even if interpreted your way, it's still a constant-sum game: the sum of the two players' payoffs is independent of their strategies. As you correctly note, the theory of constant-sum games is essentially equivalent to that of zero-sum games, as the addition of a constant to all payoffs does not affect the optimal strategies. Indeed, it's a common abuse of terminology (which I may be guilty of above) to use the term "zero-sum game" even for games with a _non_-zero constant total payoff.2012-08-26