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?
After Charlie's initial post, I had been working on this problem looking for an equation that might provide the fair amount if given A and B. During my initial foray, I had arrived at the same results as had Leming, e.g., recognizing that there may be positive return even with A < B.
I have not found the single equation, yet I do not believe FrankM's equation, given in his soluton is correct:
A(A+1B)/(A+1) when A >= B. It provides a close approximation, but it is not precise.
I am not guaranteeing the following will work for all nonnegative integer values of A given B where A >= B, but the following seem to provide the values in Leming's tables for the given values of B:
Let D = (A+B)!/(A!*B!)
Where B = 0: (A^1)/D
Where B = 1: (A^2)/D
Where B = 2: (A^3 + A^2  2*A)/(2*D)
Edited on March 14, 2008, 12:19 am

Posted by Dej Mar
on 20080313 17:49:43 