Is the way to win at roulette to bet on green? I apologize for the juvenile, and probably incorrect, math you are about to see. I have no intention of gambling. This post is made solely from curiosity. Anyways, I am underage, and thought of the idea merely while reading The Gambler.
Some roulette tables have 15 slots:

*

*1 green, which multiplies your bet by 14x.


*7 red and 7 black, which multiply your bet by 2x.
The idea is to wait till there is a long streak without a single green, say, 20 rolls from the previous green, and then start continuously betting on green until it lands. Repeat this process. Slowly increasing your bet by 1/15 per unsuccessful roll to ensure profit. A 'roll' is one round (1 spin of the roulette wheel).
The crux of this theory is the law of big numbers. As you execute more unsuccessful bets, the likelihood of a green bet occurring increases to match the average probability of getting green, even if rolls are independent of one another. I will plug in some calculations.
Assuming that I have 1000 dollars in total, and start off betting 1 dollar. I increase my bet by 1/15x (estimated to 0.07) every unsuccessful roll. Then we can calculate the total number of rolls my 1000 dollars will permit me (assuming all rolls are unsuccessful).
$y = x(1.07)^n$, y = total capital, x = starting bet, n = number of unsuccessful bets
$1000 = 1(1.07)^n, n=102$
We can execute 102 bets before running out of money. So, plugging in the binomial formula to calculate the probability of a string of unsuccessful bets, we use the formula:
$P(L)=(a | b)(x)^b (y)^(a-b)$, where a = number of unsuccessful rolls, b = number of successful rolls we want, x = probability of a successful roll, y = probability of an unsuccessful roll
$P(L)=(102+20 | 1)(1/15)^1(14/15)^(102+20-1)$, therefore P(L) is 0.0019 (0.19%), and therefore the probability of profit in the above mentioned case is 99.81%.
I use the number 102+20 as the number of unsuccessful rolls, this is because 20 accounts for the 20 unsuccessful rolls we observe before placing continuous bets on green. Unsuccessful rolls mean rolls that do not land on green, and vice versa.
Now, I am absolutely sure I made an error in that (if you spot it please do point it out), but otherwise, why is this theory incorrect? I am sure it is, but I am not smart enough to figure out why.
Of course, it is far likelier that you will hit a green in your first 14 bets, thus making a notable profit. But that is not the point of the post, which is to see whether this theory is likely to fail and bankrupt in practice.
I apologize for the length of this post and poor formatting of the equations, as I do not frequent this forum often. Have a nice day :)
 A: If anything, you have it backwards:
If we assume that the roulette wheel is perfect, then (as many others have pointed out, but you refuse to believe) the chance of a green after 20 non-greens is exactly 1/15, the same as it always is. So your strategy is pointless.
But the possibility remains that the roulette wheel is affected by local fluctuations in temperature of humidity or whatever, so that if you see a string of non-greens, the probability of a green is less than 1/15.
In either case, you may be assured that you are a very welcome customer at the casino!
A: I believe the main flaw in your analysis is confusing conditional probability with full probability. The full probability of observing  a priori a set of $N+M$ losses is indeed much smaller that $N$ losses, and then at least one  out of the $M$ winning. That does not imply that if you wait out the $N$ loosing tries, you have improved your odds in the following $M$ trials. You need to ask: given that I have observed $N$ losses, what is the conditional probability that the there will be at least one win in the next $M$. That is, $P(\text{win at least once in M} | \text{no win in previous N})$. This conditional probability is what is relevant, and for a memoryless roullete, this is the same as the full probability of   $P(\text{win at least once in M})$, regardless of the past.
