I tried to wrap my head around this. I got 55% edge on my oponent every single time (probably more, but just being a little conservative). So its a 55/45 bet in my favor. If I use the Martingale method, how many times should I double to expect to never go broke (as in less than 5% over a lifetime)?. Thinking about keeping a roll so I cap if I ever reach 50%. I expect to do this bet 50-100 times a day at max. Is the Martingale method applicable to this situation or should I go for a different system?

Been a while since I was on here but I cant go to bed without answers.

Tried to do the maths myself but my brain just melted.

This is a 10% edge, not 55.

Use a risk of ruin calculator and solve for 5% RoR. It will give a bankroll fraction to wager. Doubling won't be part of it. And note that this won't be the way to maximize bankroll growth. That would be Kelly betting.

Yea, I know its 10%. Sorry for the spelling error. I was thinking 55% of 100 and wrote 55% edge lol. I tried to find calculators online but I didnt find any I like and I have no idea how to do the maths to come up with a number :/

This seems to be amenable to analytical methods. Let me give it a try.

Let P = prob of win each wager (55% in your example)

Let Q = 1-P be the prob of losing each wager (45% in your example)

Let I = bettor's initial bankroll in betting units (where the size of the first wager of every round is 1 betting unit).

Let a "round" be the series of Martingale wagers which ends in a Win; that is, after each round either the bettor's bankroll has grown by 1 or she has gone broke.

Let L(B) = prob of going broke in a round starting with a bankroll of B

Let N(B) = the number of consecutive losses needed to go broke in a round starting with a bankroll of B.

Let H = your horizon, the maximum number of rounds in your "lifetime"

Then it is straightforward to show:

N = Ceiling[(ln(B+1))/(ln(2))]

L = Q^N

Prob(Ruin in H rounds) = Prob(Ruin in Round 1) + Prob(Ruin in Round 2 | bettor reaches Round 2) + Prob(Ruin in Round 3 | bettor reaches Round 3) + ... + Prob(Ruin in Round H | bettor reaches Round H)

= L(I) + Sum (R goes from 2 to H) L(I+R-1) * Product (S goes from 1 to R-1) [1-L(I+S-1)]

This formula is fairly easy to code up in any programming language.

The idea is that, for a given Wager Win Prob (P) and Lifetime Horizon (H), you can try different Initial Bankroll values (I) to get your desired Risk of Ruin.

I doubt there's a good reason to Martingale, but I'll take a crack at this nonetheless.

We have to define ruin. Let's say it's when your bankroll falls to 1% of what it started. There are infinite ways to keep your RoR under 5% while martingaling, so I'll find the martingale which maximizes growth rate.

Say you start with a base unit of 1% bankroll and double 3 times (4 bets total, up to 8 units). Then with each series you're risking 15% of your bankroll (combined) for a 1-.45^4 chance to win 1%. Let's plug that combined bet into Kelly:

f = p - (1-p)/y where y is payout odds (as a ratio of potential $won to $risked)

f(p,y) = f(.959, 1/15) = 34.39%

To maximize growth, increase your base unit so that your risk equals f: base unit = 34.39% / 15% = 2.29%
With that base unit, your RoR in an infinite timeline is 1% (assuming you adjust your dollar base unit after each winning bet or 4-bet losing streak).

The growth rate of that strategy is (.959^1.0229)(.041^.9707) = 4.3%

What happens if we double to 16 units? That's risking 31% to win 1%.
f(p,y) = f(.9815, 1/31) = 40.95%
Bump the base unit to f / 31% = 1.321%

g = (.9815^1.01321)(.01845^.9868) = 1.91%

That's less growth for the same RoR, so let's try the other direction: only double to 4 units (a total risk of 7 units).
p = 1-.45^3 = .908875
f = p - 7(1-p) = .271
Increase base unit to 3.87%

g = (p^1.0387)(1-p)^.9613 = 9.05%

If we only double once, f=19%, base unit = 6.33% and g=17.6%

If we don't martingale at all, then to maximize growth is to simply Kelly bet. We're risking 10% on each wager and g = 25.25% (the highest attainable growth rate, and with the same RoR as all the martingales above).

It makes sense that the optimal Martingale is the one most similar to Kelly betting: only double once and use a base unit of 6.33% bankroll.

Disclaimer: all of this assumes you know your exact edge, and aren't talking about something like sports betting. If that assumption is false, then my solution above is too aggressive.

Sorry, in the growth formula I had the bases and exponents reversed. The probabilities should be the exponents. And if your bankroll is growing, the growth should be >1, so I should have known I screwed up when I got results <1. When working with log growth the result only needs to be positive to produce profit.

The strategy will be the same, but I'll fix the growth numbers below. Lucky coincidence that bad numbers had the correct ranking.

Oh weird, kelly-martingaling results in a higher growth than plain Kelly betting? That can't be right, can it? I think I still have mistakes in my math. I'll return to this tomorrow.

Gahhh I see what I did wrong: in the growth formulas, I was treating the payout and risk as equal, when they're not (except in the non-martingale case).
Bold is wrong and should be 1-f, and the same goes for the other examples. In this one, f=.3439, so g = 1.0044

The next ones should be 1.0032, 1.0057 and 1.0063

Those last two may look higher than Kelly but they're not. Kelly's growth rate is per a single bet whereas those are per martingale series. To see how the growths compare, we have to compare apples to apples.

The shortest martingale (with g=1.0063) has an average series length of .55+2*.45 = 1.45.

One way to compare apples to apples: 1.005^1.45 = 1.0073 (>1.0063)
Another way: find the per-bet growth of the best martingale = 1.45th root of 1.0063 = 1.00434 (<1.005)

I hope my struggles have provided amusement