is it possible to transform each old Martingale from lost to Progress Gain?

[French_cunning_Fox]

Hello everyone!

A Martingale, are a betting form, if player have lost Bet or any time hes Bet.
A progression, so as he name explain, progress with Gain so with money win from Casino.

Basis Martingale D'Alembert (old Mathematicien) which have found: 1-2-3-4-5...
So if lost 1 unit, player should next net 2 unit, then 3 units, then 4, and so long time each bet is lost. And by winning one bet, as example: -1-2-3-4-5 +6, the next Bet, the player Go down to 5 units then 4 units...
So here the player have lost -1-2-3-4-5 = -15 unit, by winning +6 (-15 -+6 = -9) here player need to win +5 +4 +3 +2 +1, so to win a serial of 6 to coming out, after lose a serial 5, or in multiple other way, so as example to 9x bet to win 6x, or to 12x bet win 9x, and each time, the player win +1 unit, by still standing on win phases. So player can win only +8 unit, by a serial from 8 which coming out.

And to wining progression.
here the player should win +1 +1 before betting 2 unit, so +1+1+2+3+4 = +11 from serial 5 Winning hands.

If, for example, a player wins 70 out of 100 bets and loses 30 or 28, what is their actual net gain?
here with lose 30, is integrated the coming of 0.
on 100 Game hands, the Croupier, can reach 2-3x to coming out 0.

To Verify this, is useful to use my old number from Bad-Homburg, from date 06-06-2026. (French Roulette Table)

[French_cunning_Fox]

Attachment is awaiting approval.

Well, even when I play, I like to know, what came out as luck or chance before launching into a game.
And here the eternal players all ask the same question, just seeing his little series.
How to win your series on both sides. As much red as black to can play it?.

This is called Figure. As example:
R-R-R-R-B-B Figure from 6 Spin.
6 spin = 64 Combination of different figure; so it have 1/64 chance to find the figure which coming out, before it start it.

I like more 3 spin Figure, this have only 8 combination of different figure. so as: R-R-R; R-B-R; R-B-B  and R-R-B + same wo start with first spin as Black color.

[French_cunning_Fox]

So I follow my explain:

It is also clear that, no matter what system a player uses on the even-money bets in roulette, the chances of winning or losing will remain 48.7% or 51.3%—and here I am specifically referring to the percentage chance of winning or losing when the ball lands on zero.
Because with the d'Alembert betting progression, at some point, as we bet larger amounts, we'll be quite happy that the 0, when it comes up, gives us a little bonus.

Translated with DeepL.com (free version)