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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(3): What's the catch? - it's not that simple | Comment 9 of 37 |
(In reply to re(2): What's the catch? - it's not that simple by Charlie)

I like the following way of presenting your positive expected value (and therefore fair amount to pay to play) with only 2 plus against 3 minus. Below are the 10 sequences in which the five cards could be located within the deck, and the strategy is to quit when one is either ahead or even:

`---++  lose 1  (stopped when out of cards)--+-+  lose 1  (stopped when out of cards)--++-  even--0 (stopped when even after 4 cards)-+--+  even--0 (stopped when even after 2 cards)-+-+-  even--0 (stopped when even after 2 cards)-++--  even--0 (stopped when even after 2 cards)+---+  gain 1 (stopped when ahead at 1st card)+--+-  gain 1 (stopped when ahead at 1st card)+-+--  gain 1 (stopped when ahead at 1st card)++---  gain 1 (stopped when ahead at 1st card)`

So the overall question is still unresolved in the general case. And even when one starts with more +'s than -'s, it still could be advantageous to continue playing even when the pluses are outnumbered.

 Posted by Charlie on 2008-03-11 16:09:16

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