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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(2): A fair amount? | Comment 30 of 37 |
(In reply to re: A fair amount? by ed bottemiller)

To "see the logic of a system of play which would continue drawing cards if the remaining B are greater than the remaining A", consider the simplest case, of A=2,B=3.

`Prob. 2/5 of getting +, if so stop with                    +1, probability 2/5`
`In the 3/5 prob of getting - there are     2 + and 2 - remaining, givingprob. 1/2 of getting +, if so stop with prob. 1/2 * 3/5    with total winnings of zero (-1+1)         ... stopped                                         0,  prob. 3/10         in the overall prob. of 3/5 * 1/2 of having drawn   two -'s in a row you are now down by 2, but with    2 +'s and 1 - remaining   now new conditional prob of  1/3 of getting another -,      leaving only +'s in which case you continue on      to take all 5 cards for net -1      ... overall prob. is 3/5 * 1/2 * 1/3 = 3/30 = 1/10    -1, prob. 1/10         alternate conditional prob of 2/3 getting + and     leaving 1 + and 1 - means you are now down by 1,     but continue playing with       that 1+ and 1-, so you then have              1/2 prob of getting + and being even and stopping:              overall prob of 3/5 * 1/2 * 2/3 * 1/2 = 6/60=1/10    0, prob. 1/10                1/2 prob of getting - then the remaing +         to be down by 1 still.                              -1, prob 1/10         `

By multiplying the outcomes by the probabilities, in the right margin, you get 2/5 - 2/10 = 1/5 as the expected value when starting out with A=2,B=3.

 Posted by Charlie on 2008-03-13 21:33:33

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