Hi, I would like to know more about the options Regular and Laggard. I want to bet 5 chances (6 lines each) but I would like to modify the choice of the lines in each step based upon the previous results. The idea is to bet each time in the 5 most frequent 6 lines. With the option Regular I only can choose the most frequent chance. Is there any possibility of having the options Regular 1, Regular 2, ...., Regular 5 (the 5 most frequent 6 lines)? or is there any way to set it up?

Thanks!

Thanks!

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chilote:

The idea is to bet each time in the 5 most frequent 6 lines.

Hello chilote,

did you try the option "All - Favorite - Six line"?

Depending on the initial coups and the

*threshold*for this option,

you can set this function to work very similar to "the 5 most frequent 6 lines".

Maybe even better, because it uses more or less chances, depending on the statistical spread.

You can use the setting

*Runtime statistics in bet log for: Six line*in the test configuration,

to display the statistics in the bet log.

This can help you to find a good value for the threshold.

As much as i understand your question, it is currently not possible to simulate it 1:1.

I also asume logical problems, because it won't be always clear, which 5 six lines are the most frequent.

What if there are 6 or 7 six lines with the same occurance count within your initial coups?

Its the curse and blessing of totally deterministic strategies, to define them. :-)

Have fun with strategy,

trizero

Hello Trizero,

Sorry I think I wasn't very clear. I hope to simplify the problem with the following example. Lets assume that we want a strategy which always bets on the 2 columns with lower frequency of occurrences. Lets use in Initial coups the following setting: Columns=25 and Floating.

Suppose that in certain spin the frequency of the columns 1,2 and 3 are 8 9 7 respectively. Lets use following setup:

column; one ; Regular (Favorite>8 & 8>Laggard)

column; one ; Laggard ( 8>Laggard)

with this configuration we can bet according with the objective of the strategy.

But if in certain spin the frequency of the columns 1,2 and 3 are 8 8 9 respectively, the program will proceed by betting only on one column, decreasing the probability of winning and not following the objective of the strategy.

In order to avoid this problem we could think about using the setup:

column; one ; Regular (Favorite>8 & 8>Laggard)

column; one ; Laggard ( 9>Laggard)

But still there is the possibility of that the program is choosing the same column (for instance, the column 1) in both bets,decreasing the probability of winning and not following the objective of the strategy.

Also we could think about the setup:

column; all ; Laggard (9>Laggard)

which works with the spin of frequencies 8 8 9 but doesn't work with the spin of frequencies 7 9 9.

Effectively sometimes the frequency of different columns are the same, but in that case I think it wouldn't be a problem for the program to choose randomly 2 different columns between those of lowest frequency of occurrence (For instance, when we are dealing with the option

If I missing something please let me know.

Chilote.

Sorry I think I wasn't very clear. I hope to simplify the problem with the following example. Lets assume that we want a strategy which always bets on the 2 columns with lower frequency of occurrences. Lets use in Initial coups the following setting: Columns=25 and Floating.

Suppose that in certain spin the frequency of the columns 1,2 and 3 are 8 9 7 respectively. Lets use following setup:

column; one ; Regular (Favorite>8 & 8>Laggard)

column; one ; Laggard ( 8>Laggard)

with this configuration we can bet according with the objective of the strategy.

But if in certain spin the frequency of the columns 1,2 and 3 are 8 8 9 respectively, the program will proceed by betting only on one column, decreasing the probability of winning and not following the objective of the strategy.

In order to avoid this problem we could think about using the setup:

column; one ; Regular (Favorite>8 & 8>Laggard)

column; one ; Laggard ( 9>Laggard)

But still there is the possibility of that the program is choosing the same column (for instance, the column 1) in both bets,decreasing the probability of winning and not following the objective of the strategy.

Also we could think about the setup:

column; all ; Laggard (9>Laggard)

which works with the spin of frequencies 8 8 9 but doesn't work with the spin of frequencies 7 9 9.

Effectively sometimes the frequency of different columns are the same, but in that case I think it wouldn't be a problem for the program to choose randomly 2 different columns between those of lowest frequency of occurrence (For instance, when we are dealing with the option

**Favorite**,if we choose the option**one**, the program can choose a column randomly when there are more than one column satisfying the requirements).If I missing something please let me know.

Chilote.

chilote:

Effectively sometimes the frequency of different columns are the same, but in that case I think it wouldn't be a problem for the program to choose randomly 2 different columns between those of lowest frequency of occurrence

Hello chilote,

i think this is the same logical problem that i asumed above.

There is one more option: "Column - !m - Favorite" - which comes most close to your demand.

If you can define a logical statement that matches your wish, i will implement it.

But right now i can't see a solution to this (logical) problem:

There are no two laggard columns if they occured like 8-9-9.

Did you already think about constructing your condition without runtime chances?

In principle, you can do

**everything**with a step tree, but 25 initial coups would be a mass of steps... ;-)

In general there is no difference in the result if you test a "pattern" on every coup or on every 2nd or 10th coup.

Have fun with strategy,

trizero

Hello Trizero,

ok, I'm going to explain the tools that I have in mind and I'll try to show the way they should work. To avoid conflicts with the options Favorite, Regular and Laggard and their original objectives lets create 3 new options which only will be available when we are using a positive number of initial coups.

Using either the option Floating or Growing, after n inital coups any spin has a ternary associated, lets say (x1,x2,x3), where xi is the frequency of occurrences of the column i. In this context we want to define the following bet options:

infimum= k-th column such that xk is the

supremum= k-th column such that xk is the

mean= k-th column such that is not the infimum neither the supermum.

In order to solve the problem of how to define

Consider the ternary (x1,x2,x3). The definition of

Case 1: x1,x2 and x3 are pairwise different. (The set {x1,x2,x3} has 3 different elements).

In this case there is no problem by defining

and

Case 2: The set {x1,x2,x3} has 2 different elements. Here we have 3 subcases.

Case 2.1: x1=x2 and different than x3.

If x1<x3, then

If x3<x1, then

Case 2.2: x1=x3 and different than x2.

If x1<x2, then

If x2<x1, then

Case 2.3: x2=x3 and different than x1.

If x2<x1, then

If x1<x2, then

Case 3: The set {x1,x2,x3} has 1 element (x1=x2=x3).

Here

Then

This idea can be applied not only for columns, but streets, singles and all the non-overlapping bets. The main idea is always the same: to give an order to the components of the k-th vector (x1,x2,...,xk).

Let me know what you think.

Chilote.

ok, I'm going to explain the tools that I have in mind and I'll try to show the way they should work. To avoid conflicts with the options Favorite, Regular and Laggard and their original objectives lets create 3 new options which only will be available when we are using a positive number of initial coups.

Using either the option Floating or Growing, after n inital coups any spin has a ternary associated, lets say (x1,x2,x3), where xi is the frequency of occurrences of the column i. In this context we want to define the following bet options:

infimum= k-th column such that xk is the

**"minimum number"**in the ternary (x1,x2,x3)supremum= k-th column such that xk is the

**"maximum number"**in the ternary (x1,x2,x3)mean= k-th column such that is not the infimum neither the supermum.

In order to solve the problem of how to define

**infimum**and**supremum**in cases like the ternary (8,8,9) or (8,8,8), lets see the following idea.Consider the ternary (x1,x2,x3). The definition of

**infimum**and**supremum**of this ternary will depend on different cases.Case 1: x1,x2 and x3 are pairwise different. (The set {x1,x2,x3} has 3 different elements).

In this case there is no problem by defining

**minimum number**as the minimum between x1,x2 and x3,and

**maximum number**as the maximum between x1,x2 and x3. Then**infimum**and**supremum**will be well defined.Case 2: The set {x1,x2,x3} has 2 different elements. Here we have 3 subcases.

Case 2.1: x1=x2 and different than x3.

If x1<x3, then

**supremum**will be column 3, and**infimum**will be the k-th column where k is randomly chosen between 1 and 2.If x3<x1, then

**supremum**will be the k-th column where k is randomly chosen between 1 and 2, and**infimum**will be column 3.Case 2.2: x1=x3 and different than x2.

If x1<x2, then

**supremum**will be column 2, and**infimum**will be the k-th column where k is randomly chosen between 1 and 3.If x2<x1, then

**supremum**will be the k-th column where k is randomly chosen between 1 and 3, and**infimum**will be column 2.Case 2.3: x2=x3 and different than x1.

If x2<x1, then

**supremum**will be column 1, and**infimum**will be the k-th column where k is randomly chosen between 2 and 3.If x1<x2, then

**supremum**will be the k-th column where k is randomly chosen between 2 and 3, and**infimum**will be column 1.Case 3: The set {x1,x2,x3} has 1 element (x1=x2=x3).

Here

**infimum**will be the k-th column where k is randomly chosen between 1,2 and 3.Then

**supremum**will be the n-th column where n is randomly chosen between the elements of the set {1,2,3} - {k}.This idea can be applied not only for columns, but streets, singles and all the non-overlapping bets. The main idea is always the same: to give an order to the components of the k-th vector (x1,x2,...,xk).

Let me know what you think.

Chilote.

Hello chilote,

thank you very much for the clear and detailed explanation.

Did you still see the "old runtime chances"? They had a similar random-component built-in! :-)

But what would be the resulting bet, that you want to make possible with such a function?

Can you give a short explanation on how this would work for single-chances, please?

I read your suggestion like this:

Case 1: take the highest frequent as supremum, the lowest as infimum. mean is the rest.

Case 2: take the highest frequent as supremum, and choose one other randomly for "infimum". no mean?

Case 3: all random ;-)

I am curious, where you gonna take me to! :-)

trizero

PS: you inspired me already: i'll build in a "random" option for all chances ;-)

thank you very much for the clear and detailed explanation.

Did you still see the "old runtime chances"? They had a similar random-component built-in! :-)

But what would be the resulting bet, that you want to make possible with such a function?

Can you give a short explanation on how this would work for single-chances, please?

I read your suggestion like this:

Case 1: take the highest frequent as supremum, the lowest as infimum. mean is the rest.

Case 2: take the highest frequent as supremum, and choose one other randomly for "infimum". no mean?

Case 3: all random ;-)

**but**the Column.sup != Column.inf ? mean is rest?I am curious, where you gonna take me to! :-)

trizero

PS: you inspired me already: i'll build in a "random" option for all chances ;-)

update:

The random option is ready now!

Hello Trizero:

I think this webpage has a huge potential, spending some time here totally worth it. Even more if I can help.

Sorry, I don't fully understand what you mean.

Among the strategies that I've tested so far, the martingale-style ones are the more promising. The problem that I found so far is that we only can elaborate "static" or "deterministic" strategies, for example my "column martingala (4steps)" strategy use

There is one property of the wheel that we haven't used yet and could optimize our strategies and making them more "dynamic": the chances of the roulette come back in balance on non long term. Let me explain this. When I was testing "column martingala (4steps)" I did it with the following options:

Initial Coups: Column=25 ; Floating

Checking the tests I realized that after the spin 25, the ternaries of frequencies of occurrences of the columns had numbers very close to each other (i.e. always something like (8,7,9), (8,8,10) and never like (1,1,20) using the configuration Initial Coups: Column=25;

You got the case 1. In case 2, after defining the supremum and infimum, the mean always will be the rest.

In case 3, all random, but all different! Independent of the case always we will have that Column sup, Column inf and Column mean are different names for column 1, 2 and 3 (no necessarily in the same order). To do this one option would be:

step 1: define supremum (it has 3 possibilities)

step 2: define infimum between the rest (2 possibilities since supremum had already used one)

step 3: define mean as the last and only possible option.

In this way of defining the columns, mean and infimum cannot be defined before the supremum column.

Sure! But I cannot guarantee that is going to be a short one :-)

In the case of columns and dozens we proceed giving them an order a<b<c regarding of the frequency of occurrences and then defining infimum=a, mean =b, supremum =c.

With streets is the same idea. but the number of variables is bigger (3 columns< 12 streets

ordered street 1, ordered street 2, ordered street 3, ... ,ordered street 12; where they are ordered in a decreasing way according with their frequency of occurrences. For example,

...

...

The same idea with single-chances: ordered single 1, ordered single 2,..., ordered single 37; where

Now with these options we could think about a strategy of betting on the so called "hot numbers" and "cold numbers" if we use the option Growing instead of Floating.

I'm glad of that! This is a very exciting topic!

Chilote

trizero:

thank you very much for the clear and detailed explanation.

I think this webpage has a huge potential, spending some time here totally worth it. Even more if I can help.

trizero:

Did you still see the "old runtime chances"? They had a similar random-component built-in! :-)

Sorry, I don't fully understand what you mean.

trizero:

But what would be the resulting bet, that you want to make possible with such a function?

Among the strategies that I've tested so far, the martingale-style ones are the more promising. The problem that I found so far is that we only can elaborate "static" or "deterministic" strategies, for example my "column martingala (4steps)" strategy use

**2 fixed columns**and your "Martingale on two Dozens" strategy although changes the dozens during the time, the choice of the dozens are not related with the "tendencies" of the wheel (which I'll explain).There is one property of the wheel that we haven't used yet and could optimize our strategies and making them more "dynamic": the chances of the roulette come back in balance on non long term. Let me explain this. When I was testing "column martingala (4steps)" I did it with the following options:

Initial Coups: Column=25 ; Floating

Checking the tests I realized that after the spin 25, the ternaries of frequencies of occurrences of the columns had numbers very close to each other (i.e. always something like (8,7,9), (8,8,10) and never like (1,1,20) using the configuration Initial Coups: Column=25;

**Floating**). Also I could notice that, if the ternary of a spin is (x1,x2,x3) and the lower number between x1,x2, and x3 is x1, generally in the**next 2 spins or so**the column 1 will show up at least once. So (in theory) we'll able to predict the column of the next spin. So answering to your question, the strategy would be a martingale of 2 columns and in each step we'll use the columns infimum and supremum defined in my previous post. Also we could use the columns infimum and mean. But currently we don't have the tools to elaborate such a strategy. That is why I would love to have the opportunity of using the options infimum, mean and supremum for columns and dozens, in order to test this theory "balance of the wheel" and if we can use this in our favor.trizero:

I read your suggestion like this:

Case 1: take the highest frequent as supremum, the lowest as infimum. mean is the rest.

Case 2: take the highest frequent as supremum, and choose one other randomly for "infimum". no mean?

Case 3: all random ;-) but the Column.sup != Column.inf ? mean is rest?

You got the case 1. In case 2, after defining the supremum and infimum, the mean always will be the rest.

In case 3, all random, but all different! Independent of the case always we will have that Column sup, Column inf and Column mean are different names for column 1, 2 and 3 (no necessarily in the same order). To do this one option would be:

step 1: define supremum (it has 3 possibilities)

step 2: define infimum between the rest (2 possibilities since supremum had already used one)

step 3: define mean as the last and only possible option.

In this way of defining the columns, mean and infimum cannot be defined before the supremum column.

trizero:

Can you give a short explanation on how this would work for single-chances, please?

Sure! But I cannot guarantee that is going to be a short one :-)

In the case of columns and dozens we proceed giving them an order a<b<c regarding of the frequency of occurrences and then defining infimum=a, mean =b, supremum =c.

With streets is the same idea. but the number of variables is bigger (3 columns< 12 streets

**(lets forget for a moment about the streets involving 0 because they overlap, just consider the streets involving the numbers 1,2,3,...,36)**). When we worked with columns we needed of 3 different tags infimum, mean and supremum. Now with streets we will need of 12 tags and we could think about the following alternative:ordered street 1, ordered street 2, ordered street 3, ... ,ordered street 12; where they are ordered in a decreasing way according with their frequency of occurrences. For example,

**ordered street 1**will be THE STREET of highest frequency of occurrences while**ordered street 12**will be the STREET of lowest frequency of occurrences. To define them we need to proceed in a recursively way:**step 1:**define**ordered street 1**as the street of highest frequency of occurrences among the**12**streets, using randomness if it is necessary.**step 2:**define**ordered street 2**as the street of highest frequency of occurrences among the**11**remainder streets ,using randomness if it is necessary.**step 3:**define**ordered street 3**as the street of highest frequency of occurrences among the**10**remainder streets ,using randomness if it is necessary....

**step k:**define**ordered street k**as the street of highest frequency of occurrences among the**13-k**remainder streets ,using randomness if it is necessary....

**step 12:**define**ordered street 12**as the only remainder street.The same idea with single-chances: ordered single 1, ordered single 2,..., ordered single 37; where

**ordered single 1**will be the single-chance of highest frequency of occurrences and**ordered street 12**will be the single-chance of lowest frequency of occurrences.Now with these options we could think about a strategy of betting on the so called "hot numbers" and "cold numbers" if we use the option Growing instead of Floating.

trizero:

PS: you inspired me already: i'll build in a "random" option for all chances ;-)

I'm glad of that! This is a very exciting topic!

Chilote

Hello chilote:

My english language might not be perfect... ;-) As you may have noticed.

I meant: the previous (old) version of the runtime chances had a component that used randomness.

But for a straight traceability i changed it to the present ones.

You are right, but you can reach a lot with a tricky use of

The problem, i see is:

If you use e.g. "Laggards" for two thirds and "+ on win" (martingale function),

it comes up that you bet on only one third, when the frequency is like 8-9-9.

(betting on

This produces very high bet amounts (for the single third) and takes the chance of hitting two thirds.

You can already reach a lot, but there might be more!

At least its possible to test "static" vs "little tendency" ;-)

I'm happy that you can make your opinion that clear! :-)

I will use an artificial chance "quarters" defined as (Zero+) 36/4 = 9 singles.

Lets say those quarters came up like 1-3-3-6.

You can set this option in your strategy:

"quarters" "lined up" _1_ - which selects the 6 times quarter.

"quarters" "lined up" _4_ - selects the 1 times one

And the interesting point:

"quarters" "lined up" _2_ AND _3_ will both select one 3 times quarter but if you use both: never the same one.

This would mean: _2_ selects one random "quarter".

Thought in Columns, you could build a strategy:

"columns" "lined up" _2_ + "columns" "lined up" _3_ = don't bet on the favorite column, and even if its not cleared, take 2 chances

This "logic" can be derived to every other type of chance!

(For your information: this chance would break the: "no chance interferes with any other chance"-policy, but in an arguable manner)

What do you think?

trizero

chilote:trizero:

Did you still see the "old runtime chances"? They had a similar random-component built-in! :-)

Sorry, I don't fully understand what you mean.

My english language might not be perfect... ;-) As you may have noticed.

I meant: the previous (old) version of the runtime chances had a component that used randomness.

But for a straight traceability i changed it to the present ones.

chilote:

There is one property of the wheel that we haven't used yet and could optimize our strategies and making them more "dynamic":

the chances of the roulette come back in balance on non long term.

You are right, but you can reach a lot with a tricky use of

*runtime chances*.The problem, i see is:

If you use e.g. "Laggards" for two thirds and "+ on win" (martingale function),

it comes up that you bet on only one third, when the frequency is like 8-9-9.

(betting on

**none**is not that bad...)This produces very high bet amounts (for the single third) and takes the chance of hitting two thirds.

chilote:

That is why I would love to have the opportunity ..... to test this theory "balance of the wheel" and if we can use this in our favor.

You can already reach a lot, but there might be more!

At least its possible to test "static" vs "little tendency" ;-)

chilote:

I'm glad of that! This is a very exciting topic!

I'm happy that you can make your opinion that clear! :-)

__What about this functionality:__I will use an artificial chance "quarters" defined as (Zero+) 36/4 = 9 singles.

Lets say those quarters came up like 1-3-3-6.

You can set this option in your strategy:

"quarters" "lined up" _1_ - which selects the 6 times quarter.

"quarters" "lined up" _4_ - selects the 1 times one

And the interesting point:

"quarters" "lined up" _2_ AND _3_ will both select one 3 times quarter but if you use both: never the same one.

This would mean: _2_ selects one random "quarter".

Thought in Columns, you could build a strategy:

"columns" "lined up" _2_ + "columns" "lined up" _3_ = don't bet on the favorite column, and even if its not cleared, take 2 chances

This "logic" can be derived to every other type of chance!

(For your information: this chance would break the: "no chance interferes with any other chance"-policy, but in an arguable manner)

What do you think?

trizero

Hello Trizero,

If I understood your explanation correctly it's a faster explanation of what I tried to explain. Maybe I was too theoretical, sorry for that.

I have a couple of questions:

Is this going to work based upon initial coups? Is this going to be available with the options floating and growing?

I'll be very happy if I see these new options available!

Chilote.

If I understood your explanation correctly it's a faster explanation of what I tried to explain. Maybe I was too theoretical, sorry for that.

I have a couple of questions:

Is this going to work based upon initial coups? Is this going to be available with the options floating and growing?

I'll be very happy if I see these new options available!

Chilote.

chilote:

Maybe I was too theoretical, sorry for that.

I have a couple of questions:

Is this going to work based upon initial coups? Is this going to be available with the options floating and growing?

I'll be very happy if I see these new options available!

Hello chilote!

Your words were more "mathematically" terms.

Of course this will base on the "initial coups"

I am happy that we found a solution for your needs! Its a very good idea and a very useful functionality!

There is only one little problem: you need to be patient,

it will take some weeks, as there are other things in progress currently...

Have fun with strategy,

trizero