Doubling a roulette bet

A very simple question has me very confused..

Assume that a roulette table has 38 numbers 0, 00, 1, 2 ..., 35, 36. 0 & 00 are green, and the remaining numbers are split evenly between red and black. If a player bets black, the odds of winning and doubling his money is 18/38 = 0.473%.

Consider the following strategy. The player initially bets $2 on black. On each turn, if the player wins, he stops playing. If the player loses, he doubles his bet and keeps playing. Calculate the expected value of the the player's earnings.

E[Game] = E[First Turn] + E[Second Turn] + ... + E[nth Turn] = [2(0.473) - 2(.527)] + [4(0.473) - 4(.527)] + ... = some negative number

Now, assume that the roulette table has 38 numbers evenly split between red and black. How does this affect the expected earnings?

I would have assumed that you are expected to break even.. but when I think about it, I don't understand why you have not doubled your initial bet by the end. It seems inevitable that black will eventually hit, and if you double your bet each time, then when you decide to finish, you will have only lost $2 less than you won...

Can someone help me work through this conceptually?