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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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 Table of Fair Amount | Comment 8 of 37 |
(In reply to re(2): What's the catch? - it's not that simple by Charlie)

I built a spreadsheet for calculating the Fair Amount. From that point on, the itterative equation is:

f[A,B]= (Probability of drawing A) * (profit of drawing A plus profitability of f[A-1,B]) + (Probability of drawing B) * (Loss of drawing B plus profitability of f[A,B-1])

f[A,B] = A/(A+B)*(1+f[A-1,B]) + B/(A+B)*(-1+f[A,B-1])

where f[0,B] = 0 and f[A,0] = A

A / B   0        1        2       3        4       5

0       0.00   0.00   0.00   0.00   0.00   0.00

1       1.00   0.50   0.00   0.00   0.00   0.00

2       2.00   1.33   0.67   0.20   0.00   0.00

3       3.00   2.25   1.50   0.85   0.34   0.00

4       4.00   3.20   2.40   1.66   1.00   0.44

5       5.00   4.17   3.33   2.54   1.79   1.12

Edited for clarity and format and content

Edited on March 11, 2008, 4:29 pm
 Posted by Leming on 2008-03-11 16:04:47

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