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
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, giving
prob. 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.
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Posted by Charlie
on 2008-03-13 21:33:33 |
re(2): A fair amount?
