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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?

No Solution Yet Submitted by FrankM    
Rating: 2.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts .. and on the eighth day, he began to scratch his head and wonder what went wrong .. | Comment 33 of 37 |

Remarks:

1. Gee, it's exciting to see the problem generate so much interest and good thinking!

2. I feel badly to have lost sight for so long, and even to have missed important requests for clarification. Maybe the site could have some facility for author notification. Perhaps a theme for the forums..

Substance:

1. There is more to this problem than met my eye! As has been pointed out, the proposed solution is at fault, and will have to be corrected.

2. I would much prefer to replace it with a closed form solution. Dej Mar (comment 27) may have pointed out a way. I'll need to research this, or perhaps someone may get there before me. I would only be willing to accept a spreadsheet solution of last resort. 

3. I continue to have confidence in the published recursion relation. We can also believe that the published (and erroneous) formula fulfills the recursion relation condition. What I had overlooked was the possibility that the recursion relation could have multiple solutions. Presumably this is cause of the trouble.


  Posted by FrankM on 2008-03-24 17:06:48
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