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 Pick a card, any card.. (Posted on 2008-03-11)
You sit down with a well mixed deck containing A cards marked "+" and B cards marked "—". You may draw cards from this deck as long as you want, i.e., you can stop playing at any point. Each time you draw a + card you are given \$1 and each time you draw a — card you have to pay \$1. Cards are not replaced after having been drawn.

What would be a fair amount to pay for the right to play (i.e., what is the expected payoff) and under what circumstance should a player cease drawing?

 No Solution Yet Submitted by FrankM Rating: 2.7500 (4 votes)

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 Optimal strategy | Comment 24 of 37 |
As mentioned, the expected value of the game is actually

W(A,B) = Max(0, [1+W(A-1,B)]A/[A+B] + [-1+W(A,B-1)]B/[A+B]),
with W(A,0)=A, W(0,B)=0.

For example, W(26,26) = 41984711742427/15997372030584 ~ 2.62, not 26/27.

Here is the optimal strategy for a standard deck of cards (i.e., A<=26, B<=26): continue playing as long as B <= A + f(A), where f(A) is the number elements of the set S = {2,4,8,13,19} that are <= A.

For larger decks, take

S = {2,4,8,13,19,27,37,47,60,73,88,...}

Asymptotically, f(A) ~ 1.188 A^(1/2). There is a mathematical analysis here.
Edited on March 12, 2008, 10:21 pm
 Posted by Eigenray on 2008-03-12 22:07:34

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