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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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 re: Incorrect assumption | Comment 20 of 37 |
(In reply to Incorrect assumption by FrankM)

FrankM: Since this is your puzzle, would you please clarify two points: (1) does the putative player know the values of A and B at the outset? and (2) is the fee to play (re "a fair amount to pay for the right to play") a one-time up-front fee, or is it before each draw?  If the answer to (1) were "no" then I think there is no puzzle to solve. I would suppose the answer to (2) should be the one-time fee (your W), but that isn't explicit (i.e. the "expected payoff" would be for the entire sequence until the player would decide not to draw more).

I see I was too hasty, but want to be sure I know what the task is.  For the case of A=1 and B=1, I agree the fair amount W (if onetime upfront) would be 50 cents, since one is assured with equal probability of either 1.00 (if first draw is A, and you stop), or 0.00 (if first draw is B, and hence second draw is A).  I assume what Charlie has called the "general case" would be a formula for W(A,B) for any positive values of A and B, and hence also a rule for when to stop drawing.  (I shall change my rating of the problem!)

If I now understand the task, the value of W(2,1) would be 1.33.. since AA/B gives 2.00, and ABA and BAA would give 1.00 each-- three equally likely cases.

 Posted by ed bottemiller on 2008-03-12 11:59:35

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