Mathematically, roulette can't be beaten. This is a fact.
However, this is under the assumption that roulette will behave in a mathematically expected way defined by probability. If it doesn't, the above statement no longer holds true.
If we could find a situation where roulette results do not conform to the mathematical expectation, we may have a chance to beat the game.
A simple example is a biased wheel, certain numbers/sectors hit above the mathematical expectation. Betting on these numbers gives an advantage.
Another example is what I call the 'Repeat Window' -
In a 38 spin cycle, a repeat is guaranteed however we don't know when it will occur.
Mathematically, the repeat can occur on any spin up to and including spin 38. Mathematically it is possible to see all 37 numbers appear with a repeat coming on spin 38. If this is true, then roulette can't be beaten. It's impossible to benefit from a repeat at spin 38.
Interesting fact 1 - we all know this isn't true! All 37 numbers appearing without a repeat has NEVER and will NEVER happen. Mathematically this is possible but in reality it's not possible, so here we have an example of roulette results that do not conform to mathematical expectation.
Can this give an advantage? Well, let's take 1,000,000 cycles of 38 spins and see when the repeat occurs:
Cycle Length 1: 26839 --- 2.6839%
Cycle Length 2: 52544 --- 5.2544%
Cycle Length 3: 74880 --- 7.488%
Cycle Length 4: 91570 --- 9.157%
Cycle Length 5: 101780 --- 10.178%
Cycle Length 6: 105320 --- 10.532%
Cycle Length 7: 103471 --- 10.3471%
Cycle Length 8: 96449 --- 9.6449%
Cycle Length 9: 84391 --- 8.4391%
Cycle Length 10: 70998 --- 7.0998%
Cycle Length 11: 57243 --- 5.7243%
Cycle Length 12: 43701 --- 4.3701%
Cycle Length 13: 31892 --- 3.1892%
Cycle Length 14: 22293 --- 2.2293%
Cycle Length 15: 14934 --- 1.4934%
Cycle Length 16: 9430 --- 0.943%
Cycle Length 17: 5640 --- 0.564%
Cycle Length 18: 3245 --- 0.3245%
Cycle Length 19: 1788 --- 0.1788%
Cycle Length 20: 874 --- 0.0874%
Cycle Length 21: 411 --- 0.0411%
Cycle Length 22: 177 --- 0.0177%
Cycle Length 23: 84 --- 0.0084%
Cycle Length 24: 29 --- 0.0029%
Cycle Length 25: 12 --- 0.0012%
Cycle Length 26: 2 --- 0.0002%
Cycle Length 27: 2 --- 0.0002%
Cycle Length 28: 1 --- 0.0001%
Cycle Length 29: 0 --- 0%
Cycle Length 30: 0 --- 0%
Cycle Length 31: 0 --- 0%
Cycle Length 32: 0 --- 0%
Cycle Length 33: 0 --- 0%
Cycle Length 34: 0 --- 0%
Cycle Length 35: 0 --- 0%
Cycle Length 36: 0 --- 0%
Cycle Length 37: 0 --- 0%
As expected, a repeat on spin 38 (cycle length 37) never happened. In fact, the furthest it went to was spin 29 (cycle length 28) which happened once in 1 million cycles.
Mathematically the repeat can appear at spin 38, but in reality the limit is spin 29.
The mathematical 'repeat window' is 38/38 = 100% of the cycle - the repeat can appear on all spins
The real 'repeat window' is 29/38 = the repeat can only occur in 76% of the cycle.
Unfortunately, 76% is still quite a large proportion of the cycle and when adding in table limits, this is still difficult to create an advantage from.
We need to make the 'repeat window' smaller so we're only betting in a small section of the cycle.
Interesting fact 2 - Increasing the number of options, decreases the repeat window! Try it out for yourself - imagine if roulette had 74 numbers, what happens to the repeat window?
To get an idea following diagram, any repeat position in %:
37 numbers higher maximum
74 numbers lower maximum
calculated results
Quote from: redhot on May 09, 10:19 AM 2019
If we could find a situation where roulette results do not conform to the mathematical expectation, we may have a chance to beat the game.
True. So create automated software to perform analysis, use cross referencing to reduce the data needed for sufficient confidence, and determine in what conditions the anomalies occur.
Its not rocket science.
Quote from: Steve on May 09, 06:56 PM 2019
True. So create automated software to perform analysis, use cross referencing to reduce the data needed for sufficient confidence, and determine in what conditions the anomalies occur.
Its not rocket science.
There's a fundamental difference between your approach to that and what redhot is talking about. You gather data on specific wheels, anticlockwise and clockwise spins, and scatter distances etc. He seems to work with only the pure results. Trying to work with cycles in numbersets. Such numbersets as I understand it can come from anywhere, even RNG.
Maybe the two can meet somewhere, but as I see it, you're barking up different trees at the moment.
Yes his parameters are as relevant as how many birds fly past the window.
:girl_to:Bet from spin 18 to 30 with a long progression 1unit x 5 bets 2 units x 5 bets 3 units x 5 bets ... Hope then u eventually ahead lol
unless u also organise the repeats into categorys like even chance and choose the dominant one to have a longer progression
Redhot
Out of curiousity how many 6 unique bets were lost in a row in ur million spins
Quote from: Firefox on May 09, 07:46 PM 2019
There's a fundamental difference between your approach to that and what redhot is talking about. You gather data on specific wheels, anticlockwise and clockwise spins, and scatter distances etc. He seems to work with only the pure results. Trying to work with cycles in numbersets. Such numbersets as I understand it can come from anywhere, even RNG.
Maybe the two can meet somewhere, but as I see it, you're barking up different trees at the moment.
Correct. The stats presented above are true for any random numberset consisting of 37 unique possible outcomes.
Those results do not need to come from a wheel, they can be from RNG, random.org, picked out of a hat etc...
There's no concern for any physical variables because the physical wheel is not required.
Back to the topic...
Results for "74 number roulette" over 1,000,000 cycles below:
Cycle Length 1: 13304 --- 1.3304%
Cycle Length 2: 26682 --- 2.6682%
Cycle Length 3: 38595 --- 3.8595%
Cycle Length 4: 49786 --- 4.9786%
Cycle Length 5: 58584 --- 5.8584%
Cycle Length 6: 66219 --- 6.6219%
Cycle Length 7: 70545 --- 7.0545%
Cycle Length 8: 72921 --- 7.2921%
Cycle Length 9: 73418 --- 7.3418%
Cycle Length 10: 72189 --- 7.2189%
Cycle Length 11: 68070 --- 6.807%
Cycle Length 12: 62735 --- 6.2735%
Cycle Length 13: 57523 --- 5.7523%
Cycle Length 14: 50872 --- 5.0872%
Cycle Length 15: 44516 --- 4.4516%
Cycle Length 16: 37691 --- 3.7691%
Cycle Length 17: 31290 --- 3.129%
Cycle Length 18: 25650 --- 2.565%
Cycle Length 19: 20263 --- 2.0263%
Cycle Length 20: 15896 --- 1.5896%
Cycle Length 21: 12362 --- 1.2362%
Cycle Length 22: 9256 --- 0.9256%
Cycle Length 23: 6752 --- 0.6752%
Cycle Length 24: 4870 --- 0.487%
Cycle Length 25: 3325 --- 0.3325%
Cycle Length 26: 2345 --- 0.2345%
Cycle Length 27: 1529 --- 0.1529%
Cycle Length 28: 1098 --- 0.1098%
Cycle Length 29: 677 --- 0.0677%
Cycle Length 30: 432 --- 0.0432%
Cycle Length 31: 258 --- 0.0258%
Cycle Length 32: 163 --- 0.0163%
Cycle Length 33: 87 --- 0.0087%
Cycle Length 34: 43 --- 0.0043%
Cycle Length 35: 30 --- 0.003%
Cycle Length 36: 8 --- 0.0008%
Cycle Length 37: 6 --- 0.0006%
Cycle Length 38: 5 --- 0.0005%
Cycle Length 39: 2 --- 0.0002%
Cycle Length 40: 2 --- 0.0002%
Cycle Length 41: 1 --- 0.0001%
Cycle Length 42: 0 --- 0%
Cycle Length 43: 0 --- 0%
Cycle Length 44: 0 --- 0%
Cycle Length 45: 0 --- 0%
Cycle Length 46: 0 --- 0%
Cycle Length 47: 0 --- 0%
Cycle Length 48: 0 --- 0%
Cycle Length 49: 0 --- 0%
Cycle Length 50: 0 --- 0%
Cycle Length 51: 0 --- 0%
Cycle Length 52: 0 --- 0%
Cycle Length 53: 0 --- 0%
Cycle Length 54: 0 --- 0%
Cycle Length 55: 0 --- 0%
Cycle Length 56: 0 --- 0%
Cycle Length 57: 0 --- 0%
Cycle Length 58: 0 --- 0%
Cycle Length 59: 0 --- 0%
Cycle Length 60: 0 --- 0%
Cycle Length 61: 0 --- 0%
Cycle Length 62: 0 --- 0%
Cycle Length 63: 0 --- 0%
Cycle Length 64: 0 --- 0%
Cycle Length 65: 0 --- 0%
Cycle Length 66: 0 --- 0%
Cycle Length 67: 0 --- 0%
Cycle Length 68: 0 --- 0%
Cycle Length 69: 0 --- 0%
Cycle Length 70: 0 --- 0%
Cycle Length 71: 0 --- 0%
Cycle Length 72: 0 --- 0%
Cycle Length 73: 0 --- 0%
Cycle Length 74: 0 --- 0%
The limit for a repeat was spin 42 however mathematically the limit is spin 75.
The repeat window is 42/75 = 56% of the cycle.
By increasing the options, the repeat window has decreased and is moving further and further away from the mathematical expectation.
Hi Redhot, very interesting. It is clear that as the options increase, the repeat window decreases. But going back to European roulette for example. Would it be possible in some way to decrease that repeat window ?.
Thanks!
How do u use 74 spins?
Does it decrease again at 111 spins
Quote from: Anastasius on May 10, 07:09 AM 2019
Does it decrease again at 111 spins
It's not spins, it's options but - Yes, increasing the options decreases the repeat window:
"111 number roulette"
The repeat window is 50/112 = 44% of the cycle.
Quote from: quos on May 10, 07:02 AM 2019
Hi Redhot, very interesting. It is clear that as the options increase, the repeat window decreases. But going back to European roulette for example. Would it be possible in some way to decrease that repeat window ?.
Thanks!
The roulette wheel is just a random number generator.
Use the outcomes to create your own game. Create "74 number roulette" or "111 number roulette" or "1000 number roulette" whatever you like to get your desired repeat window.
Quote from: redhot on May 10, 08:51 AM 2019
The roulette wheel is just a random number generator.
Use the outcomes to create your own game. Create "74 number roulette" or "111 number roulette" or "1000 number roulette" whatever you like to get your desired repeat window.
But how is that possible? We only have 37 numbers. That is, only 37 options. It is not possible to create a window of 1000 roulette numbers.
Thanks for continuing explaining!!!
Quote from: quos on May 10, 09:07 AM 2019
But how is that possible? We only have 37 numbers. That is, only 37 options. It is not possible to create a window of 1000 roulette numbers.
Thanks for continuing explaining!!!
You can create any amount of options you like.
For example, "74 number roulette":
Take the result of 2 spins, for the first spin record if it's red or black, then for the second spin record the straight number.
Now you have 74 possible outcomes - R0, R1, R2....R36 and B0, B1, B2... B36
Quote from: redhot on May 10, 09:59 AM 2019
You can create any amount of options you like.
For example, "74 number roulette":
Take the result of 2 spins, for the first spin record if it's red or black, then for the second spin record the straight number.
Now you have 74 possible outcomes - R0, R1, R2....R36 and B0, B1, B2... B36
But you did not provide the statistics for that option. That should be the same as 1/37. Because now you do 2/74
He means you treat 7R or 7B as distinct numbers so you need 2 spins to generate a number. And if u have 74 numbers it will take 148 spins to create a cycle of the 74 roulette numbers.
Quote from: andreib1986 on May 10, 11:40 AM 2019
He means you treat 7R or 7B as distinct numbers so you need 2 spins to generate a number. And if u have 74 numbers it will take 148 spins to create a cycle of the 74 roulette numbers.
I know what he means. But the stats won't be the same as he posted earlier right. Because you dont have 37 numbers and 74 spins but 74 option and 74 spins. I would only be usefull if the stats would atleast be better then 37 numbers in 37 spins.
But maybe he has his reason to not post those stats because they may reveal too much...
I dont quite understand the million test spin of the cycle. The probability having repeaters of spin 37 is the same, so how can it help? I ve seen tests on repeaters and eventualy always get cold. Maybe some steady progressions?
I mean there are just more numbers to bet on...
Quote from: redhot on May 10, 09:59 AM 2019
You can create any amount of options you like.
For example, "74 number roulette":
Take the result of 2 spins, for the first spin record if it's red or black, then for the second spin record the straight number.
Now you have 74 possible outcomes - R0, R1, R2....R36 and B0, B1, B2... B36
Hello redhot, thanks.
Then, we can make a roulette of 1369 numbers.
We take the resul of 2 spins. In the first spin we récord the straight number and in the second spin as well.
Then we have 1369 options.
0-0, 0-1, ..... 0-36. the last number would be 36-36.
With 1369 options the repeat window should be a very low percentage in relation to the cycle.
What percentage (window of repetition) would give us this scenario ?.
Many thanks in advance Redhot.
Regards!!!
Quote from: quos on May 10, 05:33 PM 2019
Hello redhot, thanks.
Then, we can make a roulette of 1369 numbers.
We take the resul of 2 spins. In the first spin we récord the straight number and in the second spin as well.
Then we have 1369 options.
0-0, 0-1, ..... 0-36. the last number would be 36-36.
With 1369 options the repeat window should be a very low percentage in relation to the cycle.
What percentage (window of repetition) would give us this scenario ?.
Many thanks in advance Redhot.
Regards!!!
More to the point, how is this exploited to gain an advantage betting on 37 numbers with a pay off of only 35-1 ?
Quote from: Firefox on May 10, 08:45 PM 2019
More to the point, how is this exploited to gain an advantage betting on 37 numbers with a pay off of only 35-1 ?
Not sure if this is considered the player edge: wait for 18 numbers (out of 37) to be covered, and bet for a repeat. The results show +ve earning even after thousands of attempts. NOTE: I've only tested with 400,000 RNG numbers.
Zhone do u mean
1. 18 numbers unique
2. 18 number with repeater or
3. 18 number that hit just 1 time and exclude repeats
I thought of this today :if u collect 1 hitters and exclude repeats . How many spins until a hit of 18 single hits
Surely. At some point of collecting 1 hitters u would guarantee a quick win.
I think i played this way once before and when it gets to a certain amount of 1 hitters it doesnt go over . It should be looked at
Test could be : how many max 1 hitters at a single time until all 37 numbers become 2 hitters.
Quote from: Anastasius on May 11, 06:39 AM 2019
Zhone do u mean
1. 18 numbers unique
2. 18 number with repeater or
3. 18 number that hit just 1 time and exclude repeats
I thought of this today :if u collect 1 hitters and exclude repeats . How many spins until a hit of 18 single hits
Surely. At some point of collecting 1 hitters u would guarantee a quick win.
I think i played this way once before and when it gets to a certain amount of 1 hitters it doesnt go over . It should be looked at
Test could be : how many max 1 hitters at a single time until all 37 numbers become 2 hitters.
It could be first or second, but you will come to see mostly second scenario.
Zhone in thr first 37 spins it can go to like 37 singles but as more have hit 3 times or more hopefully the amount of singles before a hit reduces.. like u could test always betting on 12 singles no matter what thr other numbers have hit.... all singles is too much wait cus some numbers dont hit 500 spin.
Quote from: quos on May 10, 05:33 PM 2019
Hello redhot, thanks.
Then, we can make a roulette of 1369 numbers.
We take the resul of 2 spins. In the first spin we récord the straight number and in the second spin as well.
Then we have 1369 options.
0-0, 0-1, ..... 0-36. the last number would be 36-36.
With 1369 options the repeat window should be a very low percentage in relation to the cycle.
What percentage (window of repetition) would give us this scenario ?.
Many thanks in advance Redhot.
Regards!!!
The repeat window is 189/1370 = 13.8% of the cycle.
Quote from: redhot on May 13, 04:34 AM 2019
The repeat window is 189/1370 = 13.8% of the cycle.
Than we have the next:
37 numbers = 76% of the cycle
74 numbers = 56% of the cycle
111 numbers = 44% of the cycle
.
.
1369 numbers = 13.8% of the cycle. In 189 * 2 = 378 spin max. (only 189 numbers of our roulette of 1369 numbers) appears always a repeat.
As we increase the options, our repeat window decreases, but each time it decreases less.
Maybe we have to find the balance in some intermediate scenario.
I would like to know what the repetition window is for the following scenarios:
Double Street/ Straight.......---> 222 numbers.
Street/ Straight.....................---> 444 numbers.
split / Straight.......................---> 666 numebrs. (Only 18 splits)
Red hot, for example in a roulette of 74 numbers R0......R36 and B0....B36 if looking for the repetition of a number land zero in 1st spin (we need two spins to make each number of our roulette of 74 numbers), Do we spin 1 more time?.
Thanks so much again!!
We could also have a 216-number roulette:
If we take 3 spins for each number of our roulette and write down in each of those 3 spins the number of the double street, we would have a roulette of 216 numbers.
111,211,121, ....... up to 666
redhot, did you calculate which is the best option (roulette with X numbers) possible (which is the most balanced in terms of window size and options)?
Thanks so much!!!
Hi redhot!, Did you read the previous two messages ?. I also wrote you a PM
Thanks!!
Quote from: redhot on May 09, 10:19 AM 2019It's impossible to benefit from a repeat at spin 38.
Not true. Ask Dysexlic
There is a way
Cheers,
Ricky
Quote from: redhot on May 09, 10:19 AM 2019Mathematically the repeat can appear at spin 38, but in reality the limit is spin 29.
:thumbsup:
Quote from: redhot on May 09, 10:19 AM 2019Unfortunately, 76% is still quite a large proportion of the cycle and when adding in table limits, this is still difficult to create an advantage from.
There are two paths to solve this problem. You have assumed one path - progression.
The other path will get you there with much less risk and more successful cycles then losses.
cheers,
Ricky
You need to take into account he was full of shit.
Quote from: redhot on May 09, 10:19 AM 2019We need to make the 'repeat window' smaller so we're only betting in a small section of the cycle.
Or somehow overcome the lost spins using some other means. What if it did not cost you anything to see the next number or you invented a 'time machine'?
Cheers,
Ricky
Quote from: Steve on May 16, 05:03 AM 2019
You need to take into account he was full of shit.
You mean 'Dyskexlic'? Never met the guy. But he may not have found the ideal solution but there is another way to solve his problem!
Cheers,
Ricky
Quote from: redhot on May 10, 08:47 AM 2019The repeat window is 50/112 = 44% of the cycle.
Are you using the same number of spin cycles to get your results?
I would have thought if you doubled your number set you should double your spin cycle to get an equivalent max uniques before a repeat. This is due to the more combinations in the larger spin set we have not exhausted all possible paths. This may be a reason why your ratios are dropping. You may also find if you double the 37 spin cycles you may get more than 29 uniques.
Cheers,
Ricky
Quote from: redhot on May 10, 09:59 AM 2019Take the result of 2 spins, for the first spin record if it's red or black, then for the second spin record the straight number.
Ignore my previous responses . Now I see how you are creating your 74 options from the 37 numbers so your analysis is correct. I should have read the entire thread before responding, I thought your were using a wheel with 74 numbers.
Cheers,
Ricky
Quote from: quos on May 13, 05:14 AM 2019Red hot, for example in a roulette of 74 numbers R0......R36 and B0....B36 if looking for the repetition of a number land zero in 1st spin (we need two spins to make each number of our roulette of 74 numbers), Do we spin 1 more time?.
Actually, if you include the Zero, would you not have RBG as the color? So there will be more than 74 options. This is where the house edge comes and your stats then become skewed as some results will be as you say G0 - G36
Quote from: Ricky on May 16, 05:54 AM 2019
Actually, if you include the Zero, would you not have RBG as the color? So there will be more than 74 options. This is where the house edge comes and your stats then become skewed as some results will be as you say G0 - G36
I'd ignore the zero for now, we need to create a method that can beat non-zero roulette first (1-36).
Once that is established, then think about how to handle zero.
Hi Redhot,the best method so far may be that of a roulette of 1369 numbers.
two spins to make each number of our roulette of 1369 numbers.
0-0, 0-1, 0-2, 0-3.......until 36-36
Our repeat window is 13,8% (189 numbers of the 1369)
Then I would start writing down numbers every two spins. Each time in the first spin, a number appears that previously appeared in another first spin, we will bet on its second turn.
Example:
sequence:
2-15
3-18
25-35
14-23
.
.
.
2- Here first spin corresponds to the first spin of a number that has already appeared. then we would bet for the number 15.
we would have to follow a progression that, in the event of a hit, leaves at least one unit up.
The question is that I do not know how far the progression could go and if it would reach the limits of the roulette.
My formula for calculating the bet is:
X = (1+Y) / (36-P)
X = bet to make in each Straight
Y = accumulated losses within the micro-game
P = Number of Straight to play in a certain spin
Thanks for continuing to explain Redhot!
Up
Looks complicated.
Redhot, could you share your opinion about the last post please?
It's seem very interesting.
Thanks!!!
Quote from: quos on Jun 02, 08:11 AM 2019
Redhot, could you share your opinion about the last post please?
It's seem very interesting.
Thanks!!!
Your method looks interesting have you tested it? How does it play out?
I don't have the answer, I'm not here to say what's right or wrong. I'm simply sharing my thoughts to hopefully inspire other ideas.
Quote from: redhot on May 09, 10:19 AM 2019Mathematically, roulette can't be beaten. This is a fact.
However, this is under the assumption that roulette will behave in a mathematically expected way defined by probability. If it doesn't, the above statement no longer holds true.
If we could find a situation where roulette results do not conform to the mathematical expectation, we may have a chance to beat the game.
A simple example is a biased wheel, certain numbers/sectors hit above the mathematical expectation. Betting on these numbers gives an advantage.
What an interesting proposition? Incorrect but interesting all the same. You are clearly mistaken about the math.
I agree with Orca, mathematically if it is possible to win at roulette ...
35 is lower than 37. Thats what he probably means, and he's correct.