email@example.com wrote: >firstname.lastname@example.org wrote: >>quasi wrote: >>> >>> Here's a finite version of the problem, cast as a game. >>> >>> Perhaps the game is well known, I'm not sure, however the >>> resolution of the game is nice, and not immediately obvious, >>> >>> so I'll pose it here as a challenge. >>> >>> Two players play a game, win or lose, for 1 dollar, based on a >>> fixed positive integer n > 1, known in advance to both players. >>> >>> Player 1 chooses two distinct integers from 1 to n inclusive, >>> writes them on separate index cards, and places them face down >>> on the table. >>> >>> Player 2 then selects one of the cards and turns it face up, >>> exposing the hidden value. Player 2 can then either "stay", >>> yielding the value on the chosen card, or "switch", yielding >>> the value on the other card instead. Player 2 wins (and >>> player 1 loses) if player 2's final value is the higher of >>> the 2 values, otherwise player 2 loses (and player 1 wins). >>> >>> In terms of n, find the value of the game for player 2, and >>> specify optimal strategies for both players. >> >>I don't have a solution but I have an outlined approach. >> >>Player 1 has n(n-1)/2 choices for pairs. Each of these >>n(n-1)/2 choices should be picked with a probability p >>(where p is different for each pair). >> >>If player 2 picks 1, player 2 should guess that the other is >>larger. >> >>If player 2 picks n, player 2 should guess that the other is >>smaller. >> >>Each of the n-2 non-endpoints is assigned a probability q_x >>where x varies from 2 to n-1. Player 2 switches with >>probability q_x. >> >>Player 1 chooses the probability for each pair according to the >>min of the max of the expected gain from all the possible choices >>for the set of q_x. >> >>It's a Nash equilibrium minmax problem. >> >>Here's a wild guess at the solution. Player 1 should pick all >>possible choices with probability 1/(n * (n-1) /2 ). >> >>Player 2 should switch if player 2's first choice <= n/2. >> >>If this is right, then the expected value is trivial to >>calculate. Furthermore, if I'm right about player 1's strategy >>(not at all confident about this) then player 2's strategy is >>trivially correct. However, (again assuming I'm right), >>player 2's strategy is not unique if n is odd and (n+1)/2 is >>selected. > >Sorry, this isn't as clear as it could be. I meant that if >player 2 picks a non-endpoint x, player 2 switches with >probability q_x. Player 1 minmaxes to get player 1's selection >strategy. My solution-guess makes each q_x either 0 or 1 >depending on whether x <= n/2.
You proposed solution strategy for player 2 is clear enough, but as I indicated in my previous reply, for n > 2, it's not correct.