Roulette Ideas

CARPANTA'S SYSTEM FOR DOUBLE-STREET REPETITION WITH A SOFT PROGRESSION:
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At Edgardo's request, I'm going to share a system that has worked well for me..

This system belongs to the forum member Carpanta and was published in the "Arcane forum" and tested by several players with positive results in all cases.

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I know that the cutting point for taking a profit is "the crux" of winning at roulette.

Lately, I set the cut based on being able to close a micro-game in positive, even if it's just +1 chip.

My proposal is the following, you'll see!...

I wait for a double street repetition, a DS from the table layout; when this happens, I start playing 1 chip. If I lose, I continue with 1 until I lose five, then I continue with 2, etc. But if that double street has repeated once, and in the course of the next few spins another pattern repeats, I start playing it too. So, playing two double streets, I will play like this:

1.1
1.1
2.1
2.2
3.2
etc.

It's somewhat an "Atila betting" application, applied to this case.

If while I'm playing the 2 double streets that repeated, a third repetition occurs, then I immediately play this last double street that appeared + the previous one, discarding the oldest, and always like that. The only thing I don't cut and do maintain is the progression I was on previously.

If I have 1 win and I'm still in negative, then I continue with the progression until I win and end up in positive, recovering the passive of the previous losses + some profit chips.

You don't "parachute" on a win; instead, you return to the initial betting chip.

You'll be surprised to see that sooner or later, 2 or 3 consecutive wins will occur, allowing us to close the micro-game in positive.

In some cases, the passive (losses) can shoot up quite a bit, but it eventually closes as I said.

You need 500 chips to play this way, and you can aim for a daily profit of 100 chips. If you study it, you'll tell me. It's the best "system" I know to date, and it's enough to go eat coconuts under a palm tree in the Caribbean.
Regards,

CarlosGa.

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Well, I've adjusted the game to my liking and introduced some changes:

1) A 100-chip bankroll to aim for 30 to 50 chips of profit.

2) Carpanta talks about following the progression until recovering the passive and being at least 1 chip up. When I have a win, even if I'm still in negative, I look at the balance and the progression step, and if I can lower it, I do so to avoid the risk of a climb in the progression.

3) When the progression is at 3 or more, I always place one chip on the square/cormer formed by the last number that appeared in that line.

4) When I look for a double street repetition, I do it within the last 6 or 7 spins.

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I tested the software, recorded 300 spins+ no prediction..

Quote from: FreeRoulette on May 22, 2025, 04:24 PMLooking forward to seeing the web version.

RIBOTX WEB is active at: https://rouletteideas.com/ribotx/

Simply click on the module name to select it, then hit the "Launch modules" button.





Very streamlined!

amazing work, im trying the software side now, will let you know how I find it.

Looking forward to seeing the web version.

This system if for even bets.

Bankroll Needed: 85 units

Bet Progression: 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 22

The idea of this system is win twice. If you win once, then leave the money on the table (parlay) for the second win, the reset. 

One a loss, move to the next bet unit.

One a win, you have two options:

• Option 1: Keep the win on the table and try for the second win to complete the session.
• Option 2: Move back 1 in the progression.

This bet sequence has been calculated out to win at least 1 unit on second win.


You can test it out free.

gamble-galaxy.com/sim_wes/sim_martingale_parlay/index.html


It also has a calculator to create your own bet sequence.

gamble-galaxy.com/sim_wes/sim_martingale_parlay/martingaleProgression.html

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Quote from: TwoUp on Oct 08, 2024, 09:53 AMTo hit ten chosen numbers on a 37 pocket wheel, the probability is:

10/37 × 9/37 × 8/37 × 7/37 × 6/37 × 5/37 × 4/37 × 3/37 × 2/37 × 1/37 =

The number of spins on average to hit all ten numbers can be calculated by flipping the probability of each outcome and adding them up, so you get:

37/10 + 37/9 + 37/8 + 37/7 + 37/6 + 37/5 + 37/4 + 37/3 + 37/2 + 37/1 = 108.37

So basically 109 spins on average to hit all ten chosen numbers at least once.

Not sure where you got your numbers and assumptions from however there is no guarantee of any outcome, there is always the probability that the wheel doesn't hit all of the numbers or it takes hundreds of spins to hit the last or last few missing numbers.

My assumptions come from running python scripts to test how many spins it takes to hit all the numbers. Supose x spins is the most it took to hit y numbers. It seems tgat if I wait for it to get close to x spins and start betting, I should hit within those remaining spins. If I wait to start betting it is not the same as betting from spin 1. On other words, it doesn't work to pretend bet, you have to actually bet for the odds to work out.