Posted by Diane on Sunday, November 2, 2008 at 3:32pm.
Suppose you toss a coin and will win $1 if it comes up heads. If it comes up tails, you toss again. This time you will receive $2 if it comes up heads. If it comes up tails, toss again. This time you will receive $4 if it comes up heads. Continue in this fashion for a total of 10 flips of the coin, after which you receive nothing if it comes up tails. What is the mathematical expectation for this game?

Math  economyst, Monday, November 3, 2008 at 10:08am
Ah, a variation on the good old St Petersburg paradox.
Prob of winning $1 is .5
Prob of winning $2 is .5*.5
Prob of winning $4 is .5*.5*.5
Prob of winning $8 is .5*.5*.5*.5
and so on,
Expected value is sum over all possible outcomes, the probability times the value of the prize.
E = .5*1 + .25*2 + .125*4 + ...
= .5 + .5 + .5
The expected value from each flip is $0.5 So, after 10 flips, the expected value is 5. (after 100 it would be 50)
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