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?
(In reply to
Incorrect assumption by FrankM)
Only ed bottemiller continues to believe one should refuse to play or stop playing when there is an equality in the number of + and - cards remaining. I initally thought that, but leming corrected me. My extended table below,
b: 0 1 2 3 4 5 6 7 8 9 10 11 12
1.00 0.50 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
2.00 1.33 0.67 0.20 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
3.00 2.25 1.50 0.85 0.34 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
4.00 3.20 2.40 1.66 1.00 0.44 0.07 0.00 0.00 0.00 0.00 0.00 0.00
5.00 4.17 3.33 2.54 1.79 1.12 0.55 0.15 0.00 0.00 0.00 0.00 0.00
6.00 5.14 4.29 3.45 2.66 1.91 1.23 0.66 0.23 0.00 0.00 0.00 0.00
7.00 6.13 5.25 4.39 3.56 2.76 2.01 1.34 0.75 0.30 0.00 0.00 0.00
8.00 7.11 6.22 5.35 4.49 3.66 2.86 2.11 1.43 0.84 0.36 0.05 0.00
9.00 8.10 7.20 6.31 5.43 4.58 3.75 2.95 2.21 1.52 0.92 0.43 0.10
10.00 9.09 8.18 7.28 6.39 5.52 4.66 3.83 3.04 2.30 1.61 1.00 0.50
11.00 10.08 9.17 8.26 7.35 6.46 5.59 4.74 3.91 3.12 2.38 1.69 1.08
12.00 11.08 10.15 9.24 8.32 7.42 6.54 5.66 4.81 3.99 3.20 2.46 1.77
shows that if there is 1 +, play should stop if there are 2 or more -'s. If there are 2 +'s, stop if there are 4 or more -'s (where the zeros begin in the expected values). If there are 3 +'s, stop if there are 5 or more -'s. If there are 4 +'s stop if there are 7 or more -'s.
The table is of course recursive, rather than closed-form, but does give the expected value for a given A and B, such as 8.32 when A=12 and B=4.
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Posted by Charlie
on 2008-03-12 11:18:43 |