Author Topic: A new way of analyzing your coverage/chances of winning based on wheel layout?  (Read 251 times)

heatmap

  • New
  • **
  • Posts: 35
  • Thanked: 19 times
I bring you my theory, or way of calculating your chances of winning at roulette. I am not a math wizard of any kind so this might not even be correct in any way.

 The old way goes just as you might expect. I take the amount of numbers covered and divide it by 37 or 38. Get a percentage.

So this is based on the amount of numbers you have covered. Its also assuming you want to cover the maximum numbers possible with the least amount of money, but still getting the greatest odds. You might ask yourself, how might there be any different way of calculating something that only has 2 variables - the amount covered and the amount uncovered - in order to get the percentage you possibly have to win?

It is technically not new, but a new way of looking at your chances of winning. You take the amount of numbers covered and divide it by 38. Wait that's not new? Well let me explain it like this. Let's say you only cover one number. Your chances are 1/38 or about a 2% of the numbers covered. Still nothing new. The new part is how you describe the way you come to get the percentage of winning.

We need to go a step further in this case. We need to look at the wheels number layout. 

Say we choose 17 as our single number.

 American Wheel Layout :

          1                                x                                                3                                          4 

0-32-15-19-4-21-2-25-17-34-6-27-13-36-11-30-8-23-10-5-24-16-33-1-20-14-31-9-22-18-29-7-28-12-35-3-26

European Wheel Layout :

          1                                     x                                            3                                         4

0-28-9-26-30-11-7-20-32-17-5-22-34-15-3-24-36-13-1-00-27-10-25-29-12-8-19-31-18-6-21-33-16-4-23-35-14-2

The "x" is where are bet lies.

The ball can only land in one location. No matter what your chances of winning are ONLY 2%, because it can only land in one location. But the way you increase your chances of winning are to increase the amount of numbers you have covered. But thats not completely true either. Once again we have to look at the layout and where the numbers we have chosen are located.

The next concept is nothing new. Neighbors. This is how we actually increase our chances of winning. We can have more than one number chosen but lets just say the numbers are on opposite sections of the wheel, and lets just say the ball is about to land in a section completely foreign to the numbers we have chosen. Do we still have the same chances of winning now? No because the ball is not going to land in the section or the numbers we have chosen. The only chances we have at winning are if the ball is going toward the numbers we have picked. Therefore, the only thing that actually matters when it comes to winning is if the ball is going toward one or all of the numbers we have chosen, and we have less of a chance of winning if the ball goes to the opposite ends.

When we want to calculate the "actual" percentage of a win, we need to iterate through each number in order. While going through each number, I take N numbers at a time and average those N numbers together, in the first case, we take one number at a time. When we go through each number, we assume the ball is going to land in that number. Next we ask if the number that we are currently on is covered. If it isn't covered, there is a 0 percent chance of winning, if it is covered you count that into the current average.

To take you through the system ill use the american wheel layout. Like I said, I take one number at a time, but we need to go through the numbers next to it if we decide we want to factor in the neighbors around it. Lets just say we are factoring in the next neighbor to our current number we are iterating on.

Going through each number along with one other neighbor, brings our "N" variable to 2 possible numbers we are looking at, at a time then.

A = Amount of numbers we go through in one iteration

N = Amount of numbers we possibly have covered

CASE 1 where  A = 1 and N = 2

A[1] : N[1]0, N[2]32 - 0/2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it (0/2)=0
A[2] : N[1]32, N[2]15 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it (0/2)=0
15, 19 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
19-4 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
4-21 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
21-2 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
2-25 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
25-17 - 1 out of 2 numbers equating to a fifty percent chance of winning IF THE BALL goes toward it = .5 +
17-34 - 1 out of 2 numbers equating to a fifty percent chance of winning IF THE BALL goes toward it  = .5 +
34-6 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
6-27 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
27-13 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
13-36 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
36-11 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
11-30 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
30-8 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
8-23 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
23-10 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
10-5 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
5-24 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
24-16 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
16-33 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
33-1 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
1-20 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
20-14 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
14-31 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
31-9 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
9-22 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
22-18 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
18-29 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
29-7 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
7-28 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
28-12 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
12-35 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
35-3 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +
3-26 - 0 out of 2 numbers equating to a zero percent chance of winning IF THE BALL goes toward it = 0 +

CASE 1 PROOF :

 0 + 0 + 0 + 0 + 0 + 0 + 0 + .5 + .5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 1/38 or 2%

In all there are 38 iterations each with 2 numbers that were averaged out. With these 38 separate averages, we then again average those out. Surprisingly the number acquired by doing this is very close if not more or less than if you were to simply take the covered and uncovered amounts and divide them. The thing is that this is slightly more accurate, as well as telling you if you have less or more of a chance of winning because these averages rely on the concentrations of bets on the wheel rather than the amount of numbers in any arbitrary order.
« Last Edit: April 06, 2017, 04:46:12 PM by heatmap »


 
The following users thanked this post: kav, Reyth

Harryj

  • Veteran Member
  • ****
  • Posts: 359
  • Thanked: 164 times
  • Gender: Male
Hi heatmap, welcome to the forum. I have done considerable research around this concept in my attempts to find the best partners for DS groups.
   
      Where you have sought to find the  best quantity of pockets covered to give the best results. Whether the ball was approaching your group or receeding from it. I sought to establish a series of small groups scattered around the wheel. So that as the ball receeded from one group it would always be approaching another.
         This approach has always worked well for me.
           Regards.
                          Harry
 
The following users thanked this post: kav, Reyth

heatmap

  • New
  • **
  • Posts: 35
  • Thanked: 19 times
@Harryj, Yes that is the exact concept of what I am talking about. The strategy is usually setup 3 numbers at a time then skip one, 3 more numbers covered, skip one, and do that for 9 numbers on one part of the wheel and cover the exacty other half with the same bet. The double-dozens and double column bets actually mimic exactly what i have just described.
 
The following users thanked this post: Reyth