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Topic: Beating the Odds?
Replies: 35   Last Post: Feb 6, 2013 3:44 PM

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Posts: 12,067
Registered: 7/15/05
Re: Beating the Odds?
Posted: Feb 3, 2013 3:19 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply wrote:
> 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
>>If player 2 picks n, player 2 should guess that the other is
>>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

>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.


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