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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)

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 solution--I think | Comment 1 of 37

If the best strategy is to keep drawing until the minuses equal or exceed the pluses, then, as one plays, the number of pluses drawn will exceed the number of minuses drawn by A - B.  This is true even if the number of minuses goes to zero before the number of pluses does so, and eventually you are assured that at some point the pluses will get down to the level of the minuses.

So the expected payoff (fair amount to pay to play) is A - B.

Admittedly I did not realize this until I was looking to explain the results of a computer program, which produced this payoff matrix:

` B= 1  2  3  4  5  6  7  8  9`
`A1   0  0  0  0  0  0  0  0  02   1  0  0  0  0  0  0  0  03   2  1  0  0  0  0  0  0  04   3  2  1  0  0  0  0  0  05   4  3  2  1  0  0  0  0  06   5  4  3  2  1  0  0  0  07   6  5  4  3  2  1  0  0  08   7  6  5  4  3  2  1  0  09   8  7  6  5  4  3  2  1  0`

based in the recursion E(a,b) = (a/(a+b))*(E(a-1,b) + 1) + (b/(a+b))*(E(a,b-1) - 1), with E(x,x) = 0 and e(x,0)=x.

DECLARE FUNCTION e! (p!, m!)
FOR p = 1 TO 9
FOR m = 1 TO 9
PRINT e(p, m);
NEXT
PRINT
NEXT

FUNCTION e (p, m)
IF m >= p THEN e = 0: EXIT FUNCTION
IF m = 0 THEN e = p: EXIT FUNCTION
e = (p / (p + m)) * (e(p - 1, m) + 1) + (m / (p + m)) * (e(p, m - 1) - 1)
END FUNCTION

 Posted by Charlie on 2008-03-11 14:41:49

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