quasi
Posts:
10,208
Registered:
7/15/05
|
|
Re: Beating the Odds?
Posted:
Feb 3, 2013 3:04 PM
|
|
pepstein5@gmail.com 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.
Right.
>If player 2 picks n, player 2 should guess that the other is
>smaller.
Right.
>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.
OK.
>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.
Right.
The problem is to find the minmax value of the game and to
specify optimal strategies for each player.
>Here's a wild guess at the solution. Player 1 should pick all
>possible choices with probability 1/(n * (n-1) /2 ).
No.
For example, player 1 should never choose the pair (1,n).
>Player 2 should switch if player 2's first choice <= n/2.
No, that's not correct.
>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.
Try the case n=3 and you'll see your proposed optimal strategies
are not correct -- not even close.
quasi
|
|
