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: .. what went wrong .. by Steve Herman)
Hi Steve,
Let's see here.
The domain of W is pairs of nonnegative integers. Hence, to use the recurson relation it is necessary that A, B (the arguments on the left hand side) are both positive. It is thus nonsense to apply the recursion relation to cases like W(0,1).
What about a case like A=1,B=3, which leads to a negative payoff (so that W is reset to 0)?. Here we'd have
4 W(1,3) = (1 + 0) + 3 (1 + 0) = 3
which indeed would be unacceptable. So you have a point.
It appears we need to modify the recursion relation to:
(A+B) W(A,B) = Max[ 0, AB + AW(A1,B) + BW(A,B1) ]
____________________
Aside: I was just reading about various approaches for calculating close form solutions to recursion relations. There are effective tools available, but everything I've read deals with functions of a single variable. Thought is needed to see how to adapt these procedures to our case..

Posted by FrankM
on 20080325 14:51:20 