I have been wondering about the following question for a while.
Consider the martingale strategy for roulette, where you bet on a color, such as red. If you put \$1 on red and win. You walk away, or bet another on \$1. If you lose, you double your bet and put it on red. That way if you win, you win back your previous investment, plus the \$1 payout from the original wager. If you lose, you double your bet and try again. Obviously with a finite amount of money this strategy will eventually bankrupt you.
Now consider the augmented strategy, where after a certain number of losses, you accept the loss and rather than keep doubling your bet you start over at \$1. For example, after 4 losses in a row, you would have \$15 Invested ;1+2+4+8. Now rather than keep doubling you start the system over.
So the question becomes are you able to hit on 15 reds before you hit on 4 blacks in a row. Runs of black less than 4 are obviously allowed.
In roulette there are 38 equally likely outcomes, 18 red, 18 black, 2 green.
Best Answer
Each bet is a losing proposition. No sum of negative numbers is positive. The strategy as a whole is losing.
We can compute the expected value of one run of the strategy, which is playing until you get either one red or four blacks in a row. On each spin you win with probability $\frac {18}{38}$ and lose with probability $\frac {20}{38}$. You lose $4$ in a row with probability $\left(\frac{20}{38}\right)^4$ and lose $15$ in that case. In all other cases you win $1$. The expected value is then $$1\cdot\left(1-\left(\frac{20}{38}\right)^4\right)-15\left(\frac{20}{38}\right)^4=-\frac{29679}{130321}\approx -0.2277$$ So the expected value of each trial is about $-23\%$ of the initial bet. On each run you have $1-\left(\frac{20}{38}\right)^4\approx 92.3\%$ chance of winning, but because the losses are so much larger than the wins the expected value is negative. The chance of winning $15$ times in a row is $\left(1-\left(\frac{20}{38}\right)^4\right)^{15}\approx 0.302$