Quote from: Gizmotron on January 01, 2013, 11:43:49 PM
Nothing more than absolute laziness. I know there can't be a negative result. I knew I needed a quick and easy result that had to be greater than 30. So I slapped the full quantity with the isolated group. So yes, I sloppied the equation. You should take notice that the full equation was understood. The communication process worked. The point was made. I expect syntax error in a development environment. SO GET OFF MY BACK!

There. That wasn't too hard, was it?

Thank you for this clear, understandable explanation.

You should have stated this in the first instance, instead of posting the first reply that I quite rightly asked you to clarify.


[I'M OFF IT....FOR NOW...]

Quote from: Ophis on December 29, 2012, 05:27:19 PM
Im looking for statistics or some sort of mathematical data about probability of how many spins can groups of numbers sleep.

groups example:
1,2,3,4,5,6
15,22,5,1,32,28,16,33
X,X,X...
(various sizes of groups)

how many spins can group of numbers with specific size sleep?
what should be average appearance rate?

could someone help me get that sort of data or formula how to calculate this?

Hi Ophis,

Here is a general formula which finds the length of sleep for any group of numbers from 1-36:

N = Z²(1 − p) / p

N is the number of spins that the number(s) will sleep.
p is the probability of the number(s) hitting.
Z is the Z-score (number of standard deviations from the mean).

Keep in mind that there is no theoretical maximum sleep, all you can do is calculate it for some number of standard deviations from the average. In practice, a z-score of 4-5 is the maximum.

You can easily set up a spreadsheet to calculate the sleeps for any group of numbers using Z = 3 to 5.

Example 1: Single number -

For a single number, the probability p is 1/37 and 1 − p = 36/37. Let's take Z to be 4. The formula gives -

N = 4² × 36/37 ÷ 1/37 = 576 spins. (note: 4² = 4 × 4 = 16)

Example 2: Double Street -

A double street covers 6 numbers, so p = 6/37 and 1 − p = 31/37. Use Z = 4 again.

N = 4² × 31/37 ÷ 6/37 = 83 spins (rounded up).

The average appearance rate is just the reciprocal of the probability (ie; turn the probability fraction upside down). e.g., a single number will appear (on average) once every 37 spins, a particular double street will appear every 37 / 6 = 6.17 spins, a particular street will appear every 37 / 3 = 12.33 spins etc.

Sorry Gizmotron,


Where can I find the charts etc and I will review them.


Thanks for your help.

Quote from: Gizmotron on December 29, 2012, 08:57:22 PM
Here's a basic formula

6 - 37 = 31

31 divided by 37 = .8378  -- 1 sleeper in a row

.8378 times .8378 = .7019  -- 2 sleepers in a row

.7019 times .8378 = .5880  -- 3 sleepers in a row

.5880 times .8378 = .4926  -- 4 sleepers in a row ...etc.

If this is confusing, you have a 83% chance of missing the sleeper on the first step.

Gizmo, I don't understand how you worked this out, and it doesn't tell you how long a particular group of numbers will sleep. What is meant by X sleepers in a row? 

Quote from: Bayes on January 02, 2013, 01:01:07 PM


The average appearance rate is just the reciprocal of the probability (ie; turn the probability fraction upside down). e.g., a single number will appear (on average) once every 37 spins, a particular double street will appear every 37 / 6 = 6.17 spins, a particular street will appear every 37 / 3 = 12.33 spins etc.

It is right, but in a very large sample.  I see the average very seldom (if ever in some cases).

Bayes, I'm just multiplying the fractions.

If you have a 50% chance of getting a heads then you have a 25% chance of getting two heads in a row.

.5 x .5 = .25 -- (notice the symmetry. can't screw this one up)

These are simple times-in-a-row of occurring results. The point is to deliberately miss the sleeping double street.  So you want to get the odds for missing them.

Quote from: AMK on January 02, 2013, 01:06:07 PM
Where can I find the charts etc and I will review them.

it's here:

Quote from: Gizmotron on November 30, 2012, 10:53:47 PM
On every spin I check to see if a new characteristic is forming or that any existing ones are changing. So I scan the chart. I can see a characteristic in approximately one third of a second.

Look for sleepers in the dozens and columns.
Look for singles, then doubles, triples, and larger in the dozens and columns.
Look for global effects for all this.
Look for perfect and almost perfect patterns in the dozens and columns.
Look for a perfect or almost perfect dominance in the dozens and columns.
Look for dominance in all of the even chance bets.
Look for sequences of singles, doubles, and triples and above in the even chance bets.
Look for perfect patterns in the even chance bets.
Look for sleeping zeros and wide awake zeros.
Look for any active attack bet ending.
Evaluate the effectiveness of the current state.
Repeat this process after every spin and before every bet.

and here:

Quote from: Gizmotron on December 04, 2012, 12:04:28 AM
My Chart:

| 1 2 3 | L M H | | B  R | L  H | O  E | -- ##
|     X |   X   |  | X    |    X | X    | -- 29 --  1
|   X   | X     |  | X    | X    | X    | -- 13 --  2
|   X   |   X   |  | X    |    X |    X | -- 20 --  3
| X     | X     |  | X    | X    |    X | -- 10 --  4
| X     |     X |  |    X | X    |    X | -- 12 --  5
|     X |     X |  |    X |    X | X    | -- 27 --  6
|   X   | X     |  |    X |    X | X    | -- 19 --  7
|   X   | X     |  | X    | X    | X    | -- 13 --  8
|     X |   X   |  | X    |    X | X    | -- 29 --  9
|   X   |     X |  | X    | X    | X    | -- 15 -- 10
| X     |     X |  |    X | X    | X    | --  3 -- 11
|     X |   X   |  |    X |    X |    X | -- 32 -- 12
| X     | X     |  |    X | X    | X    | --  1 -- 13
|   X   | X     |  | X    |    X |    X | -- 22 -- 14
|   X   |     X |  |    X | X    |    X | -- 18 -- 15
|   X   | X     |  |    X | X    |    X | -- 16 -- 16
| X     |     X |  | X    | X    |    X | --  6 -- 17
| X     | X     |  |    X | X    | X    | --  7 -- 18
| X     | X     |  | X    | X    |    X | -- 10 -- 19
|     X |     X |  |    X |    X |    X | -- 36 -- 20
| X     |   X   |  | X    | X    |    X | --  2 -- 21
|   X   |     X |  |    X | X    |    X | -- 18 -- 22
| X     | X     |  |    X | X    | X    | --  1 -- 23
|     X |     X |  |    X |    X |    X | -- 30 -- 24
|     X | X     |  | X    |    X |    X | -- 28 -- 25
|   X   | X     |  | X    |    X |    X | -- 22 -- 26
|     X |   X   |  | X    |    X | X    | -- 35 -- 27
| X     |     X |  |    X | X    |    X | -- 12 -- 28
|     X |     X |  |    X |    X |    X | -- 30 -- 29
|   X   |     X |  | X    | X    | X    | -- 15 -- 30
| X     |   X   |  | X    | X    |    X | --  2 -- 31
| X     |     X |  |    X | X    | X    | --  9 -- 32
| X     |     X |  |    X | X    | X    | --  3 -- 33
| X     | X     |  |    X | X    | X    | --  1 -- 34
|   X   | X     |  | X    | X    | X    | -- 13 -- 35
|     X |   X   |  | X    |    X |    X | -- 26 -- 36
| X     |     X |  |    X | X    |    X | -- 12 -- 37
|     X |     X |  |    X |    X |    X | -- 30 -- 38
| X     |     X |  |    X | X    |    X | -- 12 -- 39
|------------------------------------| -  0 -- 40


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