Flip the Coin!

There is a $\dfrac 1 2$ chance that you win one dollar, a $\dfrac 1 4$ chance that you win two dollars, a $\dfrac 1 8$ chance that you win three dollars, etc.

Therefore, the average value of your winnings is $\dfrac 1 2 + \dfrac 2 4 + \dfrac 3 8 + \dfrac 4 {16} ........ (1)$

This may be written as

$(\dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + ....) + (\dfrac 1 4 + \dfrac 1 8 + \dfrac 1 {16}) + (\dfrac 1 8 + \dfrac 1 {16} ....)+............ (2)$

which equals

$1 + \dfrac 1 2 + \dfrac 1 4 + ......... = 2$

So you expect to win an average of two dollars each time you play this game.

An interesting variation of this problem presents a counterintuitive answer. If, after n flips, you receive $2^n$ dollars instead of n dollars, you should on average expect to win an infinite amount of money. If given the opportunity to play this game, you should, on average, pay any amount of money in order to play. Theoretically, at least, your very small chance of gaining money is made up for by the amount you would gain if you did make a net profit.