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**Outside The Box / Re: Convergence theory**

« **on:**Nov 28, 05:09 PM 2019 »

Windows

Thanks

Thanks

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I would like to continue watching...

Please let me in

Thanks

Please let me in

Thanks

3

I also noticed that sometimes 3 lines are not enough to complete the game, for example

1 LLLWLL

2 LLW0WL

3 LL000W

probably it makes sense to add more lines ...

Hi Person, in your example, at least you should have played another 6 more spins. According to Kali, it must be close to equilibrium before moving on to line 2.

40% -60%. if you bet on red, before moving on to line 2, the number of reds must be close to at least 40%. Although I think Kali said his system wins with 30%. in your example you only have 16,6%

Regards!!

4

Hey Kali ! I think I am understanding but I have some doubts regarding line 2.

spin R/B Result

1 Black next spin we bet Black

2 Black +1 next spin we bet Black

3 Red -1 next spin we bet Red

4 Red +1 next spin we bet Red

5 Red +1 next spin we bet Red

6 Red +1 next spin we bet Red

7 Black -1 Here we have 4 hits and 2 failures. Your method only needs 30% delivery to profit and in this case

we have 66%. It is right?

Here and at this moment is where I have doubts. At first I thought we would have to play another 6 spins and only bet on spins 2 and 6 of line 2, but from what I see it is not so. you only play 2 more spins (spins 8 and 9). It is right?

for spin 8 (belonging to line 2) do you play the same color as in spin 7? that is, would we bet on black?

for spin 9 (belonging to line 2) do you play the same color as in spin 8? that is, would we bet on black?

Thanks in advance!!!

spin R/B Result

1 Black next spin we bet Black

2 Black +1 next spin we bet Black

3 Red -1 next spin we bet Red

4 Red +1 next spin we bet Red

5 Red +1 next spin we bet Red

6 Red +1 next spin we bet Red

7 Black -1 Here we have 4 hits and 2 failures. Your method only needs 30% delivery to profit and in this case

we have 66%. It is right?

Here and at this moment is where I have doubts. At first I thought we would have to play another 6 spins and only bet on spins 2 and 6 of line 2, but from what I see it is not so. you only play 2 more spins (spins 8 and 9). It is right?

for spin 8 (belonging to line 2) do you play the same color as in spin 7? that is, would we bet on black?

for spin 9 (belonging to line 2) do you play the same color as in spin 8? that is, would we bet on black?

Thanks in advance!!!

5

Hi Kali, thanks for your explanation. It is very interesting.

In this example:

1- RRRRRR RBBRRB

(-1-1-1-1-1-1 -1 +1 +1 -1 -1 +1)

Result is -6, Still not close to balance, would we play another 6 more spins in the 1 line?

What would be the balance to be achieved before moving to line 2?

Thanks!!!

In this example:

1- RRRRRR RBBRRB

(-1-1-1-1-1-1 -1 +1 +1 -1 -1 +1)

Result is -6, Still not close to balance, would we play another 6 more spins in the 1 line?

What would be the balance to be achieved before moving to line 2?

Thanks!!!

6

Hi Tesla, What is the name of the skype group?.

Thanks!

Thanks!

7

Please keep the topic going MoneyT101. it's very interesting.

Thanks!

Thanks!

8

Redhot, could you share your opinion about the last post please?

It's seem very interesting.

Thanks!!!

It's seem very interesting.

Thanks!!!

10

Does anyone know any website where to download RNG roulette permanences from online casinos?

Thanks in advance!!!

Thanks in advance!!!

11

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!

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!

12

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

Thanks!!

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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!!!

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!!!

14

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!!

15

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 rĂ©cord 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!!!

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