☄ alt2005 reveals the math behind roulette and answers questions

In my picture it is not normal

In what form exactly, I also have the distances sorted, only for one number

Alatissa wrote:
Here it all depends on what he takes for x (X).
For example: The player's goal is a certain number (let's say 27)
The number of "unlucky" spins before the first appearance of the "lucky" target number (27) is taken as X.
from 1 to n number.
no, no, no, 27 came up - game over (X1, X2, X3, X4)
27 came up right away - game over (X1)
etc.

The probabilities of these x's are distributed according to the geometriclaw.
(The number has a constant probability of appearing every spin of 0.027).

That's the point. Only personally I find it more convenient to present graphs in a different form. Just sort all the distances and build a graph.

It's a completely different matter when the number of "successful" spins is taken as X (X).
For example, over 2000+ spins #27 has appeared 64 times
This is the number x1, x2, x3, x4, x5......x64
.The probabilities of these x's are distributed according to the binomial law. (Geometric is not suitable)
because the value of x is different.
(The probability of a number appearing each spin is constant 0.027).

If you choose a number and bet on it, then the distance to the number and the distance to guessing the number are the same)) In other words, the number coming up and guessing the number are the same. It doesn't happen that "I guessed the number, but it didn't come up". So the number of successful spins has the same geometric distribution.
There is no binomial distribution for distances, nor is there a normal one. If it were possible to reduce distances to the binomial law, I would have become a millionaire long ago. And roulette would not exist as a business.

DLK wrote: alt2005 wrote:

As for the bell. It appears, for example, when there are deviations from this very average of -2.7. That is, the probability of losing -10% is much less than losing -3%. At one time, I tried a ton of betting schemes, both my own and others', and CGM won't let me lie. And they all gave approximately the same result from -4% to -1.5%. And the longer the sample, the closer this value is to -2.7.
Another bell can be by the frequency of numbers. The average frequency is 1/37, but in some segments it can be for some number, say, 1/36, and for another 1/38. I have also checked this many times. But the longer the sample, the closer the real frequency is to 1/37. Everything is in full accordance with the theory.

We have a number series - integers in the range 0-36. For example, 50 or 500 pieces long.
How can one scientifically determine whether this series is random or not?

Is there a method?

There is no strict definition of randomness. It is very difficult to say anything on 50 numbers, if at all possible. Although if all 37 numbers came up 50 times, you can already doubt randomness.

Personally, I would define it this way (it’s easier to take not 500 pieces, but 740 to make the count even).
1. By average frequency (should be close to 1/37). That is, each number will come up about 20 times (plus or minus).

2. Even if the average is 1/37, but something like 5-5-5-5-5-5-12-12-12-12-12-12-12-12-12 - it is already doubtful. Therefore, it is necessary to take all the repetition distances. If the picture is close to a geometric distribution, and the graph is more or less smooth - then the series is random.
What does geometric distribution mean?
The number of distances 1 (N1) should be approximately 20.
The number of distances 2 (N2) should be approximately 36/37 *N1
The number of distances 3 (N3) should be approximately 36/37 * N2 = (36/37)^2 * N1
Etc.

3. Finally, there is a 3 sigma criterion. I'm too lazy to give the formula.

alt2005 writes:

... And I consider the distances, there the distribution is geometric, hence the lack of memory, I have repeated this many times. And absolutely everything is within the norm. As for the bell. It occurs, for example, with deviations from this very average -2.7. That is, the probability of losing -10% is much less than losing -3%. At one time, I tried a ton of betting schemes, both my own and others', CGM will not let me lie. And they all gave approximately the same result from -4% to -1.5%. And the longer the sample, the closer this value to -2.7. The bell can also be based on the frequency of numbers. The average frequency is 1/37, but in some segments it can be for some number, say 1 / 36, and for another 1 / 38. I have also checked this many times. But the longer the sample, the closer the real frequency is to 1/37. Everything is in full accordance with the theory.

Here everything depends on what he takes for x (X).
For example: The player's goal is a certain number (let's say 27)
The number of "unlucky" spins before the first appearance of the "lucky" target number (27) is taken as X.
from 1 to n number.
no, no, no, 27 came up - game over (X1, X2, X3, X4)
27 came up right away - game over (X1)
etc.

The probabilities of these x's are distributed according to the geometriclaw.
(The number has a constant probability of appearing every spin of 0.027).


It's a completely different matter when the number of "successful" spins is taken as X (X).
For example, over 2000+ spins #27 has appeared 64 times
This is the number x1, x2, x3, x4, x5......x64
.The probabilities of these x's are distributed according to the binomial law. (Geometric is not suitable)
because the value of x is different.
(The probability of a number appearing each spin is constant 0.027).
In this example of 2000+ spins, with a huge N the binomial distribution of these probabilities will be very close to normal.
Provided that the number is truly random.
For a normal distribution, x is taken to be a continuous random variable.
In our game we are dealing with discrete (discontinuous) random variables, because the spins are recalculated (from 1 to nth), you can’t say one and a half spins.
All distribution laws are interconnected.

There is a method for determining the randomness of a series of at least 50 numbers, but such 50 current numbers need to be collected, 100 is 5 thousand spins...
However, there is a hypothesis that in such a study, only one out of 100 numbers will belong to the RNG.

In the case of a deck of cards (with replacement) - geometric distribution p.
In the case of a deck without card returns, the hypergeometric distribution is p.
Don't confuse me.

For example, let's say there is a basket with balls.
one ball with sticker #27, the remaining 36 are white.

The ball was taken out, looked at, returned, until it was caught with a sticker - game over.
(The probability of getting a ball with a sticker each time is constant 0.027)
All over again.
The number of unsuccessful hits is recalculated until the first successful one.
X is the number of these unsuccessful hits,
The probabilities of these x's are distributed according to a geometriclaw.

Second game.
one ball with a sticker, the other 36 are white.
The ball was taken out, looked at, returned(!), until it fell out with a sticker, but the game continues.
You are allowed to get the ball limited number of times.
For example, 100 times.
During these hundred attempts, the ball with the sticker can fall out:
1) never
2) once
3) twice
4) three times
etc.
The number of successful hits over a limited period is recalculated.
A completely different random variable is taken for x.
The probabilities of these x's are distributed according to the binominal law.

Third game.
one ball with a sticker, the other 36 are white.
They took out the ball, looked at it, but didn’t return it (!) and so on until they come across one with a sticker.
Here, another random variable is taken as x.
and the probability distribution of these x's is hypergeometric.

To solve each problem, carefully look at what is taken as x (X), and only then the distribution is determined.