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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