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The 'repeat window' - Math vs Reality

Started by redhot, May 09, 10:19 AM 2019

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0 Members and 3 Guests are viewing this topic.

andreib1986

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.

RayManZ

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...

andreib1986

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?

andreib1986

I mean there are just more numbers to bet on...

quos

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!!!

Firefox

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 ?

zhone

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.

Anastasius

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.

Boom boom sir

zhone

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.

Anastasius

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.
Boom boom sir

redhot

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.

quos

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!!


quos

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!!!

quos


Hi redhot!, Did you read the previous two messages ?. I also wrote you a PM

Thanks!!

Ricky

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

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