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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(2): A fair amount? | Comment 31 of 37 |
(In reply to re: A fair amount? by ed bottemiller)

Ed, I hope this is a more clear explanation:

Looking at the A=2 & B=3 example. If we list all the possible arrangements we get the following:

AABBB
ABABB
ABBAB
ABBBA
BAABB
BABAB
BABBA
BBAAB
BBABA
BBBAA

By replacing each A and B with the running total of gains/losses with drawing each card in the permutated sequence, we get the ten possible arrangements:

1  2  1  0 -1
1  0  1  0 -1
1  0 -1  0 -1
1  0 -1 -2 -1
-1  0  1  0 -1
-1  0 -1  0 -1
-1  0 -1 -2 -1
-1 -2 -1  0 -1
-1 -2 -1 -2 -1
-1 -2 -3 -2 -1

As shown, the ending result in each row (drawing all five cards) gives a loss of \$1, yet we can choose to stop at anytime thus our probable outcome is different than that.

If we draw when we have an even or greater chance of improving our total, in this example of ten possibilities, five we end with a \$1 gain, two we break even, and three we lose \$1. This is what gives us the 1/5 probability when A=2 & B=3 that we should win more than we lose; and it is an example of how we can have a probable positive outcome when there are more B (-) than A (+).

1  2  1  0 -1
1  0  1  0 -1
1  0 -1  0 -1
1  0 -1 -2 -1
-1  0  1  0 -1
-1  0 -1  0 -1
-1  0 -1 -2 -1
-1 -2 -1  0 -1
-1 -2 -1 -2 -1
-1 -2 -3 -2 -1

 Posted by Dej Mar on 2008-03-14 02:16:21

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