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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.)
re(2): A fair amount? | Comment 30 of 37 |
(In reply to re: A fair amount? by ed bottemiller)

To "see the logic of a system of play which would continue drawing cards if the remaining B are greater than the remaining A", consider the simplest case, of A=2,B=3.

Prob. 2/5 of getting +, if so stop with                    +1, probability 2/5

In the 3/5 prob of getting - there are
    2 + and 2 - remaining, giving
prob. 1/2 of getting +, if so stop with prob. 1/2 * 3/5
    with total winnings of zero (-1+1)
        ... stopped                                         0,  prob. 3/10
       
in the overall prob. of 3/5 * 1/2 of having drawn
   two -'s in a row you are now down by 2, but with
    2 +'s and 1 - remaining
   now new conditional prob of  1/3 of getting another -,
      leaving only +'s in which case you continue on
      to take all 5 cards for net -1
      ... overall prob. is 3/5 * 1/2 * 1/3 = 3/30 = 1/10    -1, prob. 1/10
     
   alternate conditional prob of 2/3 getting + and
    leaving 1 + and 1 - means you are now down by 1,
     but continue playing with
       that 1+ and 1-, so you then have
      
       1/2 prob of getting + and being even and stopping:    
         overall prob of 3/5 * 1/2 * 2/3 * 1/2 = 6/60=1/10    0, prob. 1/10
        
       1/2 prob of getting - then the remaing +
         to be down by 1 still.                              -1, prob 1/10
        


By multiplying the outcomes by the probabilities, in the right margin, you get 2/5 - 2/10 = 1/5 as the expected value when starting out with A=2,B=3.        
   

 


  Posted by Charlie on 2008-03-13 21:33:33
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