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
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.