Royal Panda roulette

Author Topic: The longest EC ZigZag ever?  (Read 1029 times)

0 Members and 1 Guest are viewing this topic.

Stratege

Re: The longest EC ZigZag ever?
« Reply #30 on: February 04, 2019, 07:25:36 PM »
In fact I spoke in general, for your game or that of other players, it is essential to find a critical situation for chance. Then, the particular case can be played with few ships. Lanky Divisor is very old (it seems like an exercise that was offered to American students in the 1970s and the best solution was to divide losses into 3). The problem is that you attack the game directly and it's a mistake to avoid, even if you then divide your losses.
 
The following users thanked this post: Crazy K

Re: The longest EC ZigZag ever?
« Reply #31 on: February 04, 2019, 07:46:54 PM »
In fact I spoke in general, for your game or that of other players, it is essential to find a critical situation for chance. Then, the particular case can be played with few ships. Lanky Divisor is very old (it seems like an exercise that was offered to American students in the 1970s and the best solution was to divide losses into 3). The problem is that you attack the game directly and it's a mistake to avoid, even if you then divide your losses.
Some great advices here... Thanks. Let's get into details. Are you saying that watching wheel's outcomes gives us a clue (edge) about the upcoming results? I'm not new to this approach, heard a lot of cons and pros, but couldn't come to an acceptable conclusion.
 

Sputnik

Re: The longest EC ZigZag ever?
« Reply #32 on: February 04, 2019, 07:51:25 PM »


 Crazy K i have a version of the Parachute that made me 3K.
 It begins with a Marty and i wanted to keep my bets down when moving on into other locations in the Parachute.
 My solution to keep bets down was to use +1 unit every second step and every second step break even.

 For example original Marty eight steps

 1 2 4 8 16 32 64 128

 Total cost 258 units

 My solution

 1 1 3 5 11 21 43 85

 Total cost 170

 Difference 88 units
 
The following users thanked this post: rimsky, Third, Crazy K

Re: The longest EC ZigZag ever?
« Reply #33 on: February 04, 2019, 08:17:38 PM »

 Crazy K i have a version of the Parachute that made me 3K.
 It begins with a Marty and i wanted to keep my bets down when moving on into other locations in the Parachute.
 My solution to keep bets down was to use +1 unit every second step and every second step break even.

 For example original Marty eight steps

 1 2 4 8 16 32 64 128

 Total cost 258 units

 My solution

 1 1 3 5 11 21 43 85

 Total cost 170

 Difference 88 units

Interesting progression. However this cannot be the answer we are missing here. This performs way better than Classic Marty, but still can't improve the routine for more than one extra level with the same bankroll as before. It still needs 1,431,657,131 units to stretch to the worst case scenario (32 steps), where there is still the chance of 33rd unlucky outcome and losing 1.5 million units  ;D

I believe the answer lies behind a perfect debt management system. Still researching... Hope to hear more of your ideas. Thanks.
 
The following users thanked this post: Third

Re: The longest EC ZigZag ever?
« Reply #34 on: February 04, 2019, 11:25:05 PM »

VERY IMPORTANT UPDATE:

I developed a piece of program that generates 50 million spins and tests both simple EC (Always betting on RED) and my ZigZag method (changing to the other EC bet when lost). I did the test 50 times (250 million spins!) for each method and here are the results:

The longest bad run against my ZigZag method was 30.
The average of "longest bad runs" in 50 tests (50 million spins for each test) was [25.82].

The longest bad run against the simple EC betting (Always betting on RED) was 35.
The average of "longest bad runs" in 50 tests (50 million spins for each test) was [27.06].

This shows that my ZigZag strategy performs at least 4.58% better than the simple EC betting!

This also means that any method using EC bets can perform 4.58% better by using my ZigZag method instead of simply betting on one color! I don't have any statistical/mathematical reasoning for this yet, but the results are more than obvious!

Any thoughts?
 
The following users thanked this post: Third

MickyP

Re: The longest EC ZigZag ever?
« Reply #35 on: February 05, 2019, 02:51:23 AM »
Interesting.
So your zig zag system preforms better than normal  or simple EC betting. Very good from a simplicity point of play. Your Zig Zag pattern is static so it may do you well to test it against a non-static pattern.
Take say the last 5 spins and make each game 5 spins long.Lets say the spins came out RRRBR. now you can either bet for or against this pattern repeating. With each complete game count back 5 spins and use that pattern for the next game.

How do you think this will preform against the zig zag pattern?
 
The following users thanked this post: Third

Sputnik

Re: The longest EC ZigZag ever?
« Reply #36 on: February 05, 2019, 10:31:10 AM »


Crazy K I want to elaborate about a theory.
Maybe you can code this.

a) Singles versus larger series
b) Series of two versus larger series
c) Series of three versus larger series
d) Series of four versus larger series

Marigny De Grilleau talk about overrepresented and underrepresented events, imbalance and correction.
The deviation from the norm.

The values are as follows

Singles has the value of 1
Series of two has the value of 0
Series of three has the value of 1
Series of four has the value of 2
Series of five has the value of 3
And so it continues

Same for a, b, c, d,

Series of two has the value of 1
Series of three has the value of 0
Series of four has the value of 1
Series of five has the value of 2
Series of six has the value of 3
And so it continues

Now, look at this theory that has not been tested.
Assume we would live in a perfect world and get a perfect sequence.
Where the deviation of the norm increase step by step with no present regression/correction for each level.

For example, if I play against singles my expectation is to get a series of three.
Assume I use a benchmark or a certain value for each level where the overrepresented events have to reach for example 2.07 Z-Score.

I have a 36 step progression winning two in a row.

That would be six singles with no series of three, first level reaching 2.07 Z-Score
Second, if we lose, would be to wait for a series of two to aim for a series of four, if we lose again we reach 2.07 Z-Score
The third time, if we lose, we wait for a series of three to become a series of five.
The fourth time, if we lose, we wait for a series of four to become series of six.
The fifth time, if we lose, we wait for a series of five to become series of seven.
Time six, if we lose, we wait for a series of six to become series of eight.

Do you see the magic of the deviation of the norm only need to regress with two events in any given situation?

Cheers

Code: [Select]
1. Z-Score 0,02 -  1 singles contra 1 serie of three
 2. Z-Score 0,73 -  2 singles contra 1 serie of three
 3. Z-Score 1,18 -  3 singles contra 1 serie of three
 4. Z-Score 1,53 -  4 singles contra 1 serie of three
 5. Z-Score 1,82 -  5 singles contra 1 serie of three
 6. Z-Score 2,07 -  6 singles contra 1 serie of three
 7. Z-Score 2,30 -  7 singles contra 1 serie of three
 8. Z-Score 2,51 -  8 singles contra 1 serie of three
 9. Z-Score 2.58    9 singles contra 1 serie of three
10. Z-Score 2,70 - 10 singles contra 1 serie of three
11. Z-Score 2,89 - 11 singles contra 1 serie of three
12. Z-Score 3,06 - 12 singles contra 1 serie of three
13. Z-Score 3,22 - 13 singles contra 1 serie of three
14. Z-Score 3,37 - 14 singles contra 1 serie of three
15. Z-Score 3,52 - 15 singles contra 1 serie of three
16. Z-Score 3,66 - 16 singles contra 1 serie of three
17. Z-Score 3,80 - 17 singles contra 1 serie of three
18. Z-Score 3,93 - 18 singles contra 1 serie of three
19. Z-Score 4,06 - 19 singles contra 1 serie of three
20. Z-Score 4,18 - 20 singles contra 1 serie of three
21. Z-Score 4,30 - 21 singles contra 1 serie of three
22. Z-Score 4,42 - 22 singles contra 1 serie of three
23. Z-Score 4,53 - 23 singles contra 1 serie of three
24. Z-Score 4,64 - 24 singles contra 1 serie of three
25. Z-Score 4,75 - 25 singles contra 1 serie of three
26. Z-Score 4,86 - 26 singles contra 1 serie of three
27. Z-Score 4,96 - 27 singles contra 1 serie of three
28. Z-Score 5,07 - 28 singles contra 1 serie of three
29. Z-Score 5,17 - 29 singles contra 1 serie of three
30. Z-Score 5,26 - 30 singles contra 1 serie of three
31. Z-Score 5,36 - 31 singles contra 1 serie of three
32. Z-Score 5,45 - 32 singles contra 1 serie of three
33. Z-Score 5,55 - 33 singles contra 1 serie of three
34. Z-Score 5,64 - 34 singles contra 1 serie of three
35. Z-Score 5,73 - 35 singles contra 1 serie of three
36. Z-Score 5,82 - 36 singles contra 1 serie of three
37. Z-Score 5,90 - 37 singles contra 1 serie of three
38. Z-Score 5,99 - 38 singles contra 1 serie of three
39. Z-Score 6,07 - 39 singles contra 1 serie of three

« Last Edit: February 05, 2019, 10:33:08 AM by Sputnik »
 

dobbelsteen

Re: The longest EC ZigZag ever?
« Reply #37 on: February 05, 2019, 11:10:48 AM »
All EC systems with a Martingale kind of bet selection have a lot in common. SSB uses after every hit  a new trigger of 10 random sequence.  In my challenge I use a 10 step Martingale or grand Martingale because  the table limit is reached by about 10 steps. On average the profit is about 45% of the number of throws. The profit percentage varies between 2% and 25 % . In the blog you can find the reports and examples of the Excel program.
 

Stratege

Re: The longest EC ZigZag ever?
« Reply #38 on: February 05, 2019, 06:51:14 PM »

Crazy K : "Are you saying that watching wheel's outcomes gives us a clue (edge) about the upcoming results? I'm not new to this approach, heard a lot of cons and pros, but couldn't come to an acceptable conclusion."

If the past could not give clues to future results, it would be impossible to win in the long run. You did not get an acceptable conclusion on this because chance is almost perfect. But he has some flaws. Maybe one day you will find one of these flaws? To not leave you alone with your hopes here are some ideas. Sputnik speaks of Marigny's figures. Zig-Zag is a suite of Singles (B-R-B-R-B). Marigny ranked the Singles vs. Series. In 1024 spins, there are 256 Singles (128 single or units, 64 groups of 2, 32 groups of 3... 1 group of 8 (or more). We also know that Marigny has ranked the Series. In 1024 spins 256 series (128 series of 2, 64 series of 3 ... 1 series of 10 or more!). So, there is a difference between Series and "Zig-Zag". The longest series on 1024 spins is longer than the largest group of Zig-Zag. Crazy K, your statistics presented today show that there is a difference in length. You are absolutely right. Another observation. If you follow the Singles vs. series waves, you will see a lower deviance than with B / R. It's all a set of little things like that that allow you to find flaws in chance.
« Last Edit: February 05, 2019, 06:56:22 PM by Stratege »