3 Cups Probability Problem

Question

Person 1 convinces Person 2 to play a game. He has to pay 1 dollar to play. There are 3 cups on a table. Person 2 hides 2 dollars inside one of the three cups. He swaps them around at speed and Person 1 has to keep up. If he identifies the correct cup again, he is given the 2 dollars, thus gaining 1 dollar. Unfortunately, Person 1 does not know where the 2 dollars end up and has to make a guess at random. Person 2, who knows the location of the money, lifts up one of the two cups that Person 1 didn't select, showing him that there is nothing beneath it. He asks him whether he wants to stick with his chosen cup or switch. Should he stick, switch or does it make any difference? Assuming Person 1 guesses, but plays with the optimal strategy, would it be worth playing? Person 2 then offers Person 1 to play again if he gives 3 dollars and he hides 4.5 dollars inside one of the cups. Should Person 1 accept?

Answer 1

As in the solution to the Monte Hall problem, Person 2 should always switch, in which case Person 2 wins with probability 2/3. Thus, assume Person 2 always switches.

In the $(\$1,\$2)$ version of the game, Person 2 wins $\$1$ with probability $2/3$ and loses $\$1$ with probability $1/3$, so Person 2 has a per-game expectation of $1/3$ of a dollar. Since Person 2's expectation is positive, Person 2 should opt to play.

In the $(\$3,\$4.5)$ version of the game, Person 2 wins $\$1.5$ with probability $2/3$ and loses $\$3$ with probability $1/3$, so Person 2 has a per-game expectation of zero. Since Person 2's expectation is zero, Person 2 has no incentive to play, but also no disincentive, so it doesn't matter -- either way, Player 2's expectation is zero.