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Pick a card, any card.. (Posted on 2008-03-11) Difficulty: 3 of 5
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?

See The Solution Submitted by FrankM    
Rating: 2.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): solution--I think | Comment 5 of 37 |
(In reply to re: solution--I think by Leming)

I don't follow Leming, and (after looking at Charlie's post only after posting mine) I think that Charlie has made this too complex rather than over simplified. I agree with Charlie that the (initial) fair cost would be A$ - B$ assuming A and B are known in advance (this would entail refusing to play at all if A is less than B -- as I recommend).

Leming: It seems that you are ignoring the upfront cost of the "right to play" altogether.

Never give a sucker an even break!  This sounds like an oversimplified version of the Connah Tiste gambit in a recent Enigma (NS).

Incidentally, what are the times associated with all the comments?  They seem to be two hours later than my time zone (in Madison, Wisconsin), hence one hour later than "East Coast time (NYC)" which seems to put the clock out to sea, or at least not in USA.

 


  Posted by ed bottemiller on 2008-03-11 15:35:21
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