Roulette Ideas

By combining 3 different betting principles we can select a few numbers which overlap into all 3 of those betting principles:

1. There are 3 pairs of ECs, High/Low, Odd/Even and Black/Red, when you combine them by betting 3 ECs it's like betting 4 to 5 numbers depending on the combination of EC groups.
On average 1 out of 3 ECs which are associated to the last spun number will come again in the next spin, while the other 2 will be the opposite ECs in the next spin.
But you may wonder "which one?", in order to determine this we have to compare it in conjuction with 2 other criteria (betting principles).

2. The average cycle for dozens and columns is: 1 of them is repeated, 1 of them comes only once and the last is not coming at all for a cycle of 3 spins, why 3 spins? Because there are 3 dozens and 3 columns of 12 numbers each, therefore 3 x 12 = 36

3. By reversing the cycle of 3 spins by 12 numbers in order to make it like 12 cycles of 3 spins we would find a certain way to replicate the average probability of 1 cycle of 3 spins into the 12 cycles of 3 spins.
Such implementation is simpler than what may seem, we have to convert the result of the last spun column into the associated dozen and vice versa.
For example, let's say that the last dozen was the 2nd and the last column was the 3rd when the number 24 hit.
So, by reverting I mean that the next selection should indicate the 3rd dozen and 2nd column, the combination of these 2 betting elements (dozen + column) ends up with four numbers, different sets of 4 numbers.
The 2nd dozen became 2nd column and the 3rd column became 3rd dozen with mutual exchange.

From the 4-5 numbers' groups of ECs we are aplying the other 2 principles related to the dozens and columns in order to filter out the rest of the numbers and conclude on a number(s) which belongs to ECs (4 to 5 numbers each) + dozens/columns groups of numbers (4 numbers each), in other words, it should satisfy all given criteria (3 betting principles).

Sometimes there will not be any qualified number(s) to bet, this is normal, in such cases we could bet the 3 ECs and the dozen and column optionaly.
However, the main target is straight up numbers and not outside bets because they merely utilized as betting filters/triggers.

No progression whatsoever is being used, just flat bets.
What about the 0?
You might prefer to bet it on every spin with just 1 unit flat, I prefer to entirely ignore it.

Would you like some practical example?
Please provide me with some sequences of numbers, but please not from "Random.org", preferably neither RNGs.

Thanks for your participation, dear Angelo. I'm carefully reviewing this to begin coding it as a new module.

Quote from: BlueAngel on Dec 03, 2024, 11:03 AMNo progression whatsoever is being used, just flat bets.

We don't see much flat betting threads around the forums, so you've certainly caught my attention (and I'm sure, the community's as well).

In case more clarification is needed, I'll "ping" with a message here.

Cheers!
Vic

Yes sure, although I feel that a small sample will demonstrate these betting principles in a specific way, thus leaving less room for vague inerpretations.

1     This is the first number, oldest in the list.
23   The 2nd number is both different dozen and column, on the same time is 2 out of 3 ECs the same.
      Therefore we are looking for a repeat in dozen and column in combination with 2 different out of 3 EC pairs.
      Finally the revertion principle between column and dozen in this case we have 2nd dozen and 2nd column from the last spun number 23, thus we would check the nubers: 14,17,20,23 if any of them could be combined with 2 different EC pairs out of 3. 14 is qualified, 17 is qualified and 20 is qualified in regards with 23, the last spun number.

34  Bet 14,17,20 and we are losing 3 units. From 23 to 34 there are 2 different ECs and 1 of the same, also 2 different dozens and 2 different columns within 2 last spins. The revertion principle indicates that from 34 which belongs to 3rd dozen and 1st column should become 1st dozen and 3rd column, but since the previous number was 23 which belongs to 2nd dozen and 2nd column, there wouldn't be any repeat for dozens and coluns within the last 3 spins, we wouldn't bet on the next spin.

29  No bet, here there is a change of 2 ECs out of 3, 29 belongs to the 3rd dozen and the 2nd column, different column and same dozen from the previous number, by reverting we have 2nd dozen and 3rd column, but it doesn't match the principle for the columns because it would be 3 different within the 3 last spins. Therefore no bet.

19  2 of the same ECs and 2 different dozens and columns within the last 2 spins. By reverting the number 19 from 2nd dozen/1st column into 1st dozen/2nd column we find out that it would be 3 different dozens in the last 3 spins, thus we would not bet.

18  No bet. There's a repeat of dozen, but not for column, additionally there are 2 different ECs. From 2nd dozen/3rd column we revert to 3rd dozen/2nd column, but we find out that there would be 3 different columns within the last 3 spins. Conclusion, discard.

33  Here we have a repeat for columns and 2 different dozens, we should expect a repeat for a dozen and another column (different). We revert 3rd dozen/3rd column into 3rd dozen/3rd column, there wouldn't be another column, thus discard.

20  2 of the same ECs and different dozen and column within the 2 last spins. From 2nd dozen/2nd column we revert to 2nd dozen/2nd column and this time it satisfies our betting criteria in regards with the dozens & columns.
From 14,17,20,23 we would select the number(s) with 2 different ECs in regards with number 20, in this case all 3, except number 20, are qualified.

1  Bet 14,17,23. We would lose 3 units with an overall balance of -6. Here there are 3 different ECs in combination with different dozen and column. By reverting from 1st dozen/1st column into 1st dozen/1st column we would have a repeaticion for dozen and column, thus it'd satisfy our dozens & columns betting principle.
From the numbers: 1,4,7,10 we should select only those which would be 2 different ECs/1 EC same in comparison with the last spun number, 1. Two numbers are qualified: 4 and 10.

4  Bet 4 and 10. Win 35 units - 6 = 29 units net
27
3
30
17
5
5
32
19
31
10
35
8
15
11
16
29
20
5
7
26
22
10