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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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 Incorrect assumption | Comment 17 of 37 |

Many people have offered "solutions" based on the assumption that a player should stop drawing as soon as there is an equality between the number of + and - cards remaining.

No one has attempted to justify this assumption, and it is easily seen to be wrong by considering the trivial case A=B=1. Following a strategy of drawing until you reach the + card, a player has a 50% chance of winning \$1 and a 50% chance of breaking even.

We can already say some things about the "fair price to play function", W(A,B). For instance

W(1,1) = .5

W(A,0) = A

W(0,B) = 0

The solution would express W in terms of A and B. Cases like these are useful as "sanity checks" on the form of the function W.

Edited on March 12, 2008, 10:23 am
 Posted by FrankM on 2008-03-12 10:10:48

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