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 on the previous results.
The idea is to bet each time on the 5 most frequent 6-lines. With the option Regular I can only 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!

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 fewer chances, depending on the statistical spread.

You can use the setting
Run time statistics in bet log for: Six line in the test configuration,
to display the statistics in the bet log.
This can help you 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 assume logical problems, because it won't always be clear which 5 six-lines are the most frequent.
What if there are 6 or 7 six-lines with the same occurrence count within your initial coups?
It's the curse and blessing of totally deterministic strategies, to define them. :-)

Have fun with the strategy, trizero

Hello trizero,

Sorry, I think I wasn't very clear. I hope to simplify the problem with the following example.
Let's assume that we want a strategy which always bets on the 2 columns with the lower frequency of occurrences.
Let's use in Initial coups the following setting: Columns=25 and Floating.
Suppose that in a certain spin the frequency of columns 1, 2 and 3 are 8, 9, 7 respectively. Let's use the following setup:
column; one; Regular (Favorite>8 & 8>Laggard)
column; one; Laggard (8>Laggard)
With this configuration we can bet according to the objective of the strategy.

But if in a certain spin the frequency of 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 there is still the possibility that the program is choosing the same column (for instance, 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 is the same,
but in that case I think it wouldn't be a problem for the program to randomly choose 2 different columns among those of the lowest frequency of occurrence
(for instance, when dealing with the option Favorite, if we choose the option one, the program can choose a column randomly when there is more than one column satisfying the requirements).
If I am missing something please let me know.

Chilote

chilote:

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

Hello chilote,
I think this is the same logical problem that I assumed above.

There is one more option: "Column - !m - Favorite" - which comes closest 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 occurred like 8-9-9.

Did you already think about constructing your condition without run time 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 the 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, let's create 3 new options which will only be available when we are using a positive number of initial coups.

Using either the option Floating or Growing, after n initial coups any spin has a ternary associated, let's say (x1,x2,x3), where xi is the frequency of occurrences of 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 that is neither the infimum nor the supremum.

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), let's 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 in 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 sub-cases.

Case 2.1: x1=x2 and different from 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 from 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 from 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 from the set {1,2,3} - {k}.

This idea can be applied not only for columns, but also for streets, singles and all 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

update:

The random option is ready now!

Hello chilote,
thank you very much for the clear and detailed explanation.
Did you still see the "old run time 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 will take me! :-)
trizero

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

Hello trizero:

trizero:

thank you very much for the clear and detailed explanation.

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

trizero:

Did you still see the "old run time 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 can only elaborate "static" or "deterministic" strategies.
For example my "column martingala (4steps)" strategy uses 2 fixed columns and your "Martingale on two Dozens" strategy,
although it changes the dozens during the time, the choice of the dozens is 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 make 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 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 be able to predict the column of the next spin. So answering 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 of "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 case 1. In case 2, after defining the supremum and infimum, the mean will always be the rest.
In case 3, all random, but all different! Independent of the case, we will always have that Column sup, Column inf and Column mean are different names for column 1, 2 and 3 (not 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 it's going to be a short one. :-)
In the case of columns and dozens we proceed by giving them an order a<b<c regarding the frequency of occurrences and then defining infimum=a, mean=b, supremum=c.
With streets it's the same idea. But the number of variables is bigger (3 columns < 12 streets (let's 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 3 different tags: infimum, mean and supremum. Now with streets we will need 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 to their frequency of occurrences.
For example, ordered street 1 will be the street of the highest frequency of occurrences while ordered street 12 will be the street of the lowest frequency of occurrences.
To define them we need to proceed recursively:

step 1: define ordered street 1 as the street of the highest frequency of occurrences among the 12 streets, using randomness if necessary.
step 2: define ordered street 2 as the street of the highest frequency of occurrences among the 11 remainder streets, using randomness if necessary.
step 3: define ordered street 3 as the street of the highest frequency of occurrences among the 10 remainder streets, using randomness if necessary.
...
step k: define ordered street k as the street of the highest frequency of occurrences among the 13-k remainder streets, using randomness if necessary.
...
step 12: define ordered street 12 as the only remainder street.

The same idea applies to single-chances: ordered single 1, ordered single 2,..., ordered single 37; where ordered single 1 will be the single-chance of the highest frequency of occurrences and ordered street 12 will be the single-chance of the 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:

chilote:

trizero:

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

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

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.

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.

chilote:

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

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

quarters lined up

What about this functionality:
I will use an artificial chance "quarters" defined as (Zero+) 36/4 = 9 singles.
Let's 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 it's 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 on 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 on 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 "mathematical" terms.
Of course this will be based on the "initial coups".

I am happy that we found a solution for your needs! It's 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 the strategy,
trizero

trizero:

There is only one little problem: you need to be patient,
it will take some weeks, as there are other things in progress currently...

We were very patient - when will the option Regular be implemented?
I am still waiting...

Just a little while... ;-)

wheel:

I am still waiting...

Me too!

Spielpirat:

wheel:

I am still waiting...

Me too!

Me too! :-)

chilote:

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.

hot and cold numbers

The options hot and cold numbers are implemented in between.
I would like to complete this topic, even if it took a little longer than I thought... ;-)

trizero:

it took a little longer than I thought... ;-)

Days, months, years, decades... what's the difference!? In result? ;-)

trizero:

Spielpirat:

wheel:

I am still waiting...

Me too!

Me too! :-)

But not me!

banksy:

But not me!

Then you really missed out! ;-)