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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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 re: .. what went wrong .. | Comment 34 of 37 |
(In reply to .. and on the eighth day, he began to scratch his head and wonder what went wrong .. by FrankM)

Sorry, Frank.

The published recursion relation only has one solution.  What went wrong is that the published recursion relation is incorrect (your confidence notwithstanding).  The published recursion relation would be correct if the player is forced to draw a next card, but the player is not forced to draw a next card.  For instance W(0,0) = 0.  The recursion formula says W(0,1) is -1, but it is actually 0 (since a player will stop before drawing the single remaining B card).

See eigenray's solution for the correct recursion relation.
 Posted by Steve Herman on 2008-03-25 11:08:25

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