Recent posts
#51
Multiple / Re: bgGES Roulette ver.Inna
Last post by exe11 - Oct 23, 2024, 10:47 PMBIG WIN ROULETTE
#52
Money management / fitting money management with...
Last post by Random - Oct 15, 2024, 08:41 PMhello.
I always choose bet selection first. Then when I observed thousands of loss win strings I try to fit in a progression witch I believe is suitable for.
I have been using a star carsh variant:
10
10
20
20
50
50
110
110
220
220
and the aim is to win to bets in a row and reset the progression.
There are recovery levels for it, but it does not matter now.
I use it to catch regression to the mean with success. Only latter I have learned about Victor's 50% and 75% bankroll management method and I must say it works amazingly.
I always choose bet selection first. Then when I observed thousands of loss win strings I try to fit in a progression witch I believe is suitable for.
I have been using a star carsh variant:
10
10
20
20
50
50
110
110
220
220
and the aim is to win to bets in a row and reset the progression.
There are recovery levels for it, but it does not matter now.
I use it to catch regression to the mean with success. Only latter I have learned about Victor's 50% and 75% bankroll management method and I must say it works amazingly.
#53
FreeRoulette / Re: Kind of breaks my brain
Last post by winkel - Oct 12, 2024, 04:36 PMTake it easier:
Take the last 37 spins and selct the unhit
You will get a hit very soon
Take the last 37 spins and selct the unhit
You will get a hit very soon
#54
FreeRoulette / Re: Kind of breaks my brain
Last post by TwoUp - 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.
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.
#55
FreeRoulette / Kind of breaks my brain
Last post by FreeRoulette - Oct 01, 2024, 07:29 AMLets make the assumption that picking 10 numbers will guarenteed hit in 100 spins.
So we look backwards through the number history removing numbers from the board until only 10 remain and we see that it took 60 spins.
So we select those 10 numbers, and 60 spins came up without hitting it, then it should hit within 40 spins, because we have a guaranteed hit in 100 spins based on billions of trials.
Here is where it gets funky. Had you picked those 10 numbers before the 60 spins, then at this point, you would hit within 40 spins. But because you looked back through the numbers, you kept moving the target around and those numbers had zero chance of hitting. Since they had zero chance of hitting, they don't count towards the 100 spins.
So we look backwards through the number history removing numbers from the board until only 10 remain and we see that it took 60 spins.
So we select those 10 numbers, and 60 spins came up without hitting it, then it should hit within 40 spins, because we have a guaranteed hit in 100 spins based on billions of trials.
Here is where it gets funky. Had you picked those 10 numbers before the 60 spins, then at this point, you would hit within 40 spins. But because you looked back through the numbers, you kept moving the target around and those numbers had zero chance of hitting. Since they had zero chance of hitting, they don't count towards the 100 spins.
#56
Coding requests / Re: RIBOT WEB - Actuals conver...
Last post by VLS - Sep 23, 2024, 06:06 AMThe core template for the foundational modules is complete. Right now, the Red/Black/Green actuals converter is online and available! 😊
Link: https://rouletteideas.com/ribot/temp/ribot-web-actuals-converter-red-black-green.html

The other mentioned foundational modules are coming up based on it (sharing code). 👍
Link: https://rouletteideas.com/ribot/temp/ribot-web-actuals-converter-red-black-green.html
The other mentioned foundational modules are coming up based on it (sharing code). 👍
#57
Albalaha / Re: base URL is being taken ov...
Last post by VLS - Sep 20, 2024, 11:15 AMNote: The base URL "https://rouletteideas.com/ribot/" is being transitioned to a new back-end service.
If you prefer the current version, feel free to save this temporary module as it is.
It will return with a new look and feel, aligned with the new engine's style, at its permanent URL.
If you prefer the current version, feel free to save this temporary module as it is.
It will return with a new look and feel, aligned with the new engine's style, at its permanent URL.
#58
Coding requests / Re: Foundational modules
Last post by VLS - Sep 20, 2024, 10:44 AMQuote from: VLS on Sep 12, 2024, 07:41 PMActuals transformers (from numbers to betting locations: RB, EO, HL, D1/D2/D3, C1/C2/C3, etc.).
These are definitely the opening modules. Can't have a Roulette software without them!
#59
Multiple / Re: bgGES Roulette ver.Inna
Last post by exe11 - Sep 17, 2024, 06:58 PMformula change. adding a new chart. generator testing
#60
TwoUp / Optimal stopping for positive ...
Last post by TwoUp - Sep 17, 2024, 03:25 PMWhen playing even chance we are presented with a predicament of should we stop taking our wins or accepting our losses or continuing.
This is known as "The optimal stopping Problem" and more commonly the dating problem or the 37% rule.
The optimal stopping problem has many variations and is an ongoing area of research, but it provides some precise insights in obtaining an advantage in an even chance game through a process of deciding when it is optimal to stop.
This video introduces the concept of optimal stopping in an entertaining way:
Now there is a special variation of the stopping problem that is solved for a fixed number of events and is partly solved for an indefinite number of events. These problems are known as the Robbins Problem and the related Chow-Robbins game where one tosses a coin and after each toss, you decide if you take the fraction of heads up to now as a payoff, otherwise you continue.
It provides an answer to betting optimally on even chance outcomes and providss a positive expectation.
A simple rule of thumb is if you are up 8 vs 5 or better then stop, otherwise continue until you break even, but we can do much better using a decision table I provide below for the first 1000 spins.
The math shows that when considering a horizon of up to 268 million spins has an upper bound the optimal stopping advantage for a perfect even chance game of 0.79295350640 or 79.2953% which is pretty spectacular for a 50/50 outcome which in simple terms has no long term advantage.
Now practically we cannot play that many outcomes but it is clear that deciding when to stop based on math provides an advantage to the player which I sfar better than 50/50.
The table below covers the opening decisions of the Chow-Robbins game, showing when to stop and when to continue, through computational analysis of the theory and shows all decision points a player has to make up to 1000 spins computed using a 10 million spin horizon.
So for starters let's consider a case where we have won on 19 reds against losing on 14 blacks, (19-14) with the difference therefore being 5.
According to the table, stopping with a difference of 5 is best even up to 23 reds vs 18 blacks "23–18", so we must therefore stop.
As can be seen in the table, the mathematical theory is complete up to a difference of 11, while for a difference of 12 the status of what to do for the outcome 116 reds vs 104 blacks "116–104" is currently unknown to science but a 12 unit win is still a win so stopping is a prudent option whether or not it is known to be optimal.
For the outcome of 16 reds vs 12 blacks (16–12), the decision is extremely close. Whilst for a run of 1 million spins the math fails to determine the optimal decision, whilst a run of 10 million spins shows that stopping is optimal.
The following table was computed for all possible outcomes up to 10 million spins.
The 3 columns are:
difference stop-with continue-with
1 1–0 2–1
2 5–3 6–4
3 9–6 10–7
4 16–12 17–13
5 23–18 24–19
6 32–26 33–27
7 42–35 43–36
8 54–46 55–47
9 67–58 68–59
10 82–72 83–73
11 98–87 99–88
12 115–103 117–105
13 134–121 135–122
14 155–141 156–142
15 176–161 177–162
16 199–183 201–185
17 224–207 225–208
18 250–232 251–233
19 277–258 279–260
20 306–286 307–287
21 336–315 338–317
22 368–346 369–347
23 401–378 402–379
24 435–411 437–413
25 471–446 473–448
26 508–482 510–484
≥ 27 stop
The above table and analysis from Rigorous computer analysis of the Chow-Robbins game
This is known as "The optimal stopping Problem" and more commonly the dating problem or the 37% rule.
The optimal stopping problem has many variations and is an ongoing area of research, but it provides some precise insights in obtaining an advantage in an even chance game through a process of deciding when it is optimal to stop.
This video introduces the concept of optimal stopping in an entertaining way:
Now there is a special variation of the stopping problem that is solved for a fixed number of events and is partly solved for an indefinite number of events. These problems are known as the Robbins Problem and the related Chow-Robbins game where one tosses a coin and after each toss, you decide if you take the fraction of heads up to now as a payoff, otherwise you continue.
It provides an answer to betting optimally on even chance outcomes and providss a positive expectation.
A simple rule of thumb is if you are up 8 vs 5 or better then stop, otherwise continue until you break even, but we can do much better using a decision table I provide below for the first 1000 spins.
The math shows that when considering a horizon of up to 268 million spins has an upper bound the optimal stopping advantage for a perfect even chance game of 0.79295350640 or 79.2953% which is pretty spectacular for a 50/50 outcome which in simple terms has no long term advantage.
Now practically we cannot play that many outcomes but it is clear that deciding when to stop based on math provides an advantage to the player which I sfar better than 50/50.
The table below covers the opening decisions of the Chow-Robbins game, showing when to stop and when to continue, through computational analysis of the theory and shows all decision points a player has to make up to 1000 spins computed using a 10 million spin horizon.
So for starters let's consider a case where we have won on 19 reds against losing on 14 blacks, (19-14) with the difference therefore being 5.
According to the table, stopping with a difference of 5 is best even up to 23 reds vs 18 blacks "23–18", so we must therefore stop.
As can be seen in the table, the mathematical theory is complete up to a difference of 11, while for a difference of 12 the status of what to do for the outcome 116 reds vs 104 blacks "116–104" is currently unknown to science but a 12 unit win is still a win so stopping is a prudent option whether or not it is known to be optimal.
For the outcome of 16 reds vs 12 blacks (16–12), the decision is extremely close. Whilst for a run of 1 million spins the math fails to determine the optimal decision, whilst a run of 10 million spins shows that stopping is optimal.
The following table was computed for all possible outcomes up to 10 million spins.
The 3 columns are:
difference stop-with continue-with
1 1–0 2–1
2 5–3 6–4
3 9–6 10–7
4 16–12 17–13
5 23–18 24–19
6 32–26 33–27
7 42–35 43–36
8 54–46 55–47
9 67–58 68–59
10 82–72 83–73
11 98–87 99–88
12 115–103 117–105
13 134–121 135–122
14 155–141 156–142
15 176–161 177–162
16 199–183 201–185
17 224–207 225–208
18 250–232 251–233
19 277–258 279–260
20 306–286 307–287
21 336–315 338–317
22 368–346 369–347
23 401–378 402–379
24 435–411 437–413
25 471–446 473–448
26 508–482 510–484
≥ 27 stop
The above table and analysis from Rigorous computer analysis of the Chow-Robbins game