Discrete Distributions (Alatissa expert in probability theory)

Alatissa wrote:
What kind of coordinate system does Olynes have, where does he see thousands and millions of sigmas, I have no idea.

But I'll explain a little - you've probably heard such expressions, for example - "as high as a mountain" or - "as many as there are stars in the sky"... or the most common - "so... high".

Or something like this - when I was studying, there was some kind of assignment and the teacher explained what needed to be calculated and said "And take the data from the reference book of Vasily Potolkov (I'm trying to translate the last name)... well, there was one who went to look for such a reference book in the library.
I have a suspicion that you would be a match for this seeker

The fact that I wrote millions means... ...well, I didn't count how many there are, and what difference does it make, it's the same as with the data from Vasya Potolkov's encyclopedias... how could you do that...?

And finally, in the game I play, sigmas are not counted in order to find some rare coincidence - they are counted to make sure that what is constantly happening is not a coincidence.
Read again - "happens all the time" and since I'm talking about that, I mean that 10 out of 10 or close to that is always guessed.

You understand - I'm not interested in rarities because it's impossible to use them. I'm interested in certain constancies because my game is built on the fact that what was - will be in the future, because this is the shortest path to victory... again in the game I play.

klick wrote: Alt, is it necessary to say anything about your linguistics if D. London never wrote novels about casinos and roulette.
For the fact that you went to Google to look for De Steil's reference book, that's a credit! It's immediately obvious that you're far from the Ukrainian language, as well as from foreign literature.

I apologize in advance to the Admin and all other forum members. But I'll still answer Klick. Read, klick, about Smoke Bellew, how he won at the dried-up roulette, and then judge who wrote about what and who is far from what. Of course, I don't know Ukrainian to such an extent that I can immediately understand the joke about Steil, but the humor itself is somehow dubious ))

Now on to the topic. I don't actually count sigmas. Whether the graph is negative or not can be seen from the distribution of distances. But since we're talking about sigmas, I'll express some thoughts.

10 spins in my opinion is not enough to calculate sigma. But let's take 370 spins. The average number of number drops or, what is the same, the average number of guesses on this segment = 10. We calculate sigma:
1/37 * 36/37 * 370 = 9.73
Sigma = SQRT(9.73) = 3.12 so 3 sigma = 9.36
Then if guessing on 370 spins consistently occurs more than 20 times (10+9.36), then there is reason to consider this not a random guess, but a positive strategy.
But if you regularly guess less than 1-2 times (10 - 9.36) - then there is reason to suspect a casino setup. In any case, such numbers are no longer (most likely) random.

Alatissa wrote:
I will write my answers both for distances and for the 10-spin interval in a normal environment,

thanks for the detailed description
there are some things that I stubbornly can't get my head around... but with your help I'll definitely get my head around them )))

alt2005 replied to the topic ☄ alt2005 reveals the mathematics of roulette and answers questions Olynes writes: Here's a problem for you as a mathematician - let's say a player plays well and his average distance is 8.0 over a long distance. The player makes 10 bets on one number, what is his chance in sigmas that the result will be 9.4 (what Coyne got in the end, that is, 0.155 more than the average), yes, maybe not in sigmas, just what is the chance.. Can I try to solve it? Based on the average distance of 9.24. So, like this:
All calculations are for averages
1) 9.24 / 8 = 1.155 is how many times your probability is greater than the standard.
2) The distance 9.24 corresponds, as I understand it, to 19 numbers (0 + 9 to the right, 9 to the left), then the standard probability = 19/37 = 0.5135.
3) For the average 8 this probability will be 0.5135 * 1.155 = 0.593.
Then, for 10 spins, the probability of never hitting the distance of 8 (i.e., the average would be 9.4):
P = (1 - 0.593) ^ 10 = 0.407 ^ 10 = 0.000124 = 0.012%, this is for bets on 19 numbers.
For the standard 9.24 this probability is (18/37)^10 = 0.000743 = 0.074%. Thus, decreasing the average distance from 9.24 to 8 gives a more than 6-fold increase in the probability of guessing on 10 spins...

Something like that. Although I'm not a mathematician, I could be wrong.

I really liked this post by Alta.
I will expand on these considerations, as they are directly related to my topic.
I love depth.

Here we are already talking about 19numbers in the interval of 10 spins.
Alt stopped at comparing two microscopic probabilities of not hitting the target at all, i.e.
ten-loss pieces of one and the other.
(LLLLLLLLLL)
......
Let there be two characters:
Regular Player (Avg 9.24)
and the special Mr. O. (cf. 8.0)

Bet 19 numbers,
and they launched the 10th spinning reel 50,000 times.

You know, the theory is such a delight,
The tournament hasn't even started yet, but you can already see the results.

For a regular player

never 37
1 time 391
2 times 1861
3 times 5240
4 times 9679
5 times 12260
6 times 10784
7 times 6504
8 times 2574
9 times 603
10 times 64

At the special Mr. O.

never 6
1 time 90
2 times 595
3 times 2314
4 times 5901
5 times 10318
6 times 12528
7 times 10430
8 times 5699
9 times 1845
10 times 268

It is quite clear (in absolute values),
that by reducing the average distance to 8.0, a special player reduces the ten-loss by 6 times
(never one 37, the other 6)
But at the same time, the probability that a bet of 19 numbers will play at least once in an interval of 10 spins is quite high for both.
1 - 0.00012 = 0.99988 or 99.988% (for special)
1 - 0, 0007427 = 0.9992573 or 99.9257% (for normal)

If you look at the maximum values (9 and 10 times)
then the usual one has 667
the special one has this result 2113

Now I'll compare the MO

For a regular player
MO = 5 times
σ = √D = √2.498 = 1.5 ≈ 2
F(7) - F(2) = 0.9352 - 0.0457 = 0.8893 or 88.9%
or p(3) + p(4) + p(5) + p(6) + p(7) = 0.8893
in the picture



at the special Mr. O.
MO = 6 times
σ = √D = √2.41351 = 1.5535 ≈ 2
F( - F(3) = 0.9577 - 0.0601 = 0.8976 or 89.8%
or p(4) + p(5) + p(6) + p(7) + p( = 0.8976

in the picture



And now interval comparison from 5 to 10 times for one and the other

For a regular player
F(10) - F(4) = 1 - 0.3442 = 0.655 or 65.5%
differently
p(5) + p(6) + p(7) + p( + p(9) + p(10) = 0.655

in the picture




The irresistible Mr. O.
F(10) - F(4) = 1 - 0.1782 = 0.8218 or 82.18%
differently
p(5) + p(6) + p(7) + p( + p(9) + p(10) = 0.8218

in the picture



These were interval comparisons.
and now specific

If we take the average of a regular player for comparison, it is 5 times
and the average of the extraordinary is 6 times
then 6-5 = 1
sigma equals 2
This means that on average they differ by... half of the first sigma.
And this is clearly visible both in the calculations and in the graphs.
In fairness, I can say that such a shift of the peak to the right (even by half of the first sigma)
and also the weighting of the values of the entire right-hand side of the mathematical expectation, not everyone will demonstrate.
But this is provided that it is reliable, the player is really capable of reaching 8.0
And it’s not clear at all how Olynes got there and what he considers a long distance.
this is me from the quote
Olynes writes:

average distance 8.0 on a long distance.

Long distance is long, like a railroad?)
Olynes, when he talks about stars, mountains, the moon... then at such moments he is very romantic)
...
Therefore, even if we assume that this is really the case, and Olynes has a method for achieving such results, I can say that in the short term he may well lose, and he is only in the black in the long run.
His game is also a case, just with a higher frequency of played bets.
(Moreover, he avoids ideal roulettes).

There is a post by Alt, where he suggests taking not 10 spins, but 370 for clarity
It can be done this way too.
I can show the difference for comparison and, looking ahead, I will say that on average the players differ... also by half of the first sigmashuli.

Imagine a tournament with one number and three hundred and seventy spins are launched 50,000 times
Then
for a regular player the calculated results are:

never 2
1 time 22
2 times 113
3 times 378
4 times 946
5 times 1892
6 times 3153
7 times 4504
8 times 5629
9 times 6255
10 times 6256
11 times 5686
12 times 4739
13 times 3645
14 times 2603
15 times 1736
16 times 1084
17 times 638
18 times 354
19 times 187
20 times 94
21 times 45
22 times 20
23 times 9
MO = 10 times
sigma ≈ 3

in the picture




At the special Mr. O.

never -
1 time 6
2 times 34
3 times 131
4 times 376
5 times 863
6 times 1650
7 times 2704
8 times 3878
9 times 4942
10 times 5669
11 times 5911
12 times 5650
13 times 4985
14 times 4084
15 times 3123
16 times 2239
17 times 1511
18 times 963
19 times 581
20 times 333
21 times 182
22 times 95
23 times 47

MO=11 times
sigma ≈ 3
in the picture



If we take the calculated value of MO (without rounding, it is 11.47)
11.47 - 10 = 1.47≈ 1.5

That is, the same thing: the shift of the vertex to the right for Mr. O. is half the first sigma compared to the vertex of an ordinary player.
This is if you look at the graphs

Next, you can compare intervals from 10 to 20

the usual one has it
F(20) - F(9) = 0.9984117 - 0.4579297 = 0.5404 or 54.04%

in the picture



at the special Mr. O.
F(20) - F(9) = 0.992699 - 0.2917503 = 0.7009 or 70.09%
in the picture



This is an analytical comparison of the performance of two players with an average of 9.24 and 8.0
in the coordinate system of the binomial distribution.




My next post is where I will take distances from 0 to 18 on the right and left.
More in the style of math statistics.
We agreed that first I would show what and how I counted, then Olynes would comment.
Perhaps we are looking at the same thing, but from different angles.

I will comment on one of the most striking posts by Alta #810 in the topic "Roulette Mathematics"
100pravda.com/forum/roulette-game/675-al...rosy?start=795#36479

The highlight here is in this quote:

Now let's imagine that the first 25 distances are distributed evenly, i.e. there are approximately equal numbers of them. Then the picture changes. There will no longer be 1.93 times fewer distances than 1 distance, but the same number. Each distance will account for 0.4959 / 25 = 0.0198 of the total number

I'll make a small adjustment now,
instead of 0.0198 I'll take 0.0196
this won't have much of an impact on the results...
you'll understand why later

you will get a table like this



Let's say I didn't know that Alt "forcibly" straightened the exponent in a separate section,
and he would immediately show me this table and ask:
What do you see?
Then I would say:
I see that a roulette game is going on with 51 numbers and the numbers that come up are eliminated from the game...

there is still a subtle touch of hypergeometric distribution here,
but to explain it we need a separate post,
This type of distribution is of greater interest to card players

Next I will show where the 70% for distances 1-25 came from (post #809 in the topic "roulette mathematics"
100pravda.com/forum/roulette-game/675-al...rosy?start=795#36373 )
as another option

Should we continue to wait for life to become calm in Minsk?