Been reading on this forum and elsewhere a bunch about how martingale doesn't work, and I get why with a game like craps or roulette. But I was hoping to have a discussion around a particular game.

Say you have a very large bankroll of 1,000,000usd in this example. You bet .000001 mBTC set on autopilot. You say, less than 80 is a win. So you have about a 79% chance of winning each spin. You win about a fifth of your bet on each spin. On each loss you increase your bet by 6, which covers the loss and earns you a small amount. After a win, the bet returns to original.

What I'm trying to wrap my head around is what the odds are of rolling enough losing spins in a row to lose your massive bankroll. Let's take the million in the example, we increase the bet after each consecutive loss by 6:
.000001
.000006
.000036
.000216
.001296
.007776
.046656
.279936
1.679616
10.077
60.46
362.79
2176.78
13k
78k
470k


So what are the odds then of losing 16 times in a row when any singular spin has about a 79% chance of winning? I get that each spin is independent of the next and is referred to as the "gamblers fallacy" but it seems like losing 16 times in a row is a very small probabaility when singular spins are 79% in your favor.

In experiments, I see three losses in a row with some regularity, 4 too. 5 and 6 are very rare.

Someone tell me what I'm missing here. Seems like you could win if your bankroll was large enough to basically work outside of the very unlikely trend of losing so many times in a row. Sure maybe after a long enough time of doing this you would lose but shouldn't this work if you just slam the game with a big bankroll and then stop?

I see this as materially different than playing a hi/lo 50/50 game since you have no odds in your favor but rather a coin flip.