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?
(In reply to
re(4): What's the catch?  it's not that simple by ed bottemiller)
"If you do punt, I would think you should ALWAYS stop if the remaining B exceed the remaining A (assuming a fair shuffle, etc.) . Hence that is the GENERAL case, nicht wahr? (If you have exhausted all the A cards, a fort. you should stop since B > zero."
As Leming pointed out, if there are 2 pluses and 3 minuses, there's still a strategy for coming out ahead. As the diagram below summarizes, of the 10 ways the cards could be arranged, two result in the loss of 1 dollar and 4 result in the gain of a dollar, netting +2 over the course of 10 games on average, for a value of the game of 0.20, which would make 0.20 the fair amount to pay for entry if there were 2 pluses and 3 minuses.
The strategy is to stop when either you're even, or ahead, or the cards have run out:
++ lose 1 (stopped when out of cards)
++ lose 1 (stopped when out of cards)
++ even0 (stopped when even after 4 cards)
++ even0 (stopped when even after 2 cards)
++ even0 (stopped when even after 2 cards)
++ even0 (stopped when even after 2 cards)
++ gain 1 (stopped when ahead at 1st card)
++ gain 1 (stopped when ahead at 1st card)
++ gain 1 (stopped when ahead at 1st card)
++ gain 1 (stopped when ahead at 1st card)
By "the general case" I meant all conditions, not just those where B exceeded A, which is the seemingly paradoxical case of coming out ahead.

Posted by Charlie
on 20080311 22:48:18 