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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): .. what went wrong .. - to S Herman | Comment 36 of 37 |
(In reply to re: .. what went wrong .. by Steve Herman)

Hi Steve,

Let's see here.

The domain of W is pairs of non-negative 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, A-B + AW(A-1,B) + BW(A,B-1) ]

____________________

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 2008-03-25 14:51:20

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