Discrete Distributions (Alatissa expert in probability theory)

Alatissa wrote: Everyone has their own individual story.
If the distance is 50-100 thousand (and more), then the practical value is very close to the calculated one.

Why?
Player A allocated $1,000 to players B and C. Player A sits at home and manages both players' bets via messenger. They don't bet anything themselves. So whose game is it?

Player A.
Players A+B+C.
Players B+C separately.

How does the roulette know that B and C are the "hands" of the same player A?
Any logical answer will lead to further inferential connections that will show the absurdity of the concept of "individual history".

LUCKY-13 writes:

3) Does each table and each player have their own history or are all players, casinos and roulettes of the world united into one common egregor, in which all indicators are averaged?

Everyone has their own individual history.
If the distance is 50-100 thousand (and more), then the practical value is very close to the calculated one.

With a probability of 18/37 the player will get a final result of 1st place (if he plays number 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18)
with a probability of 12/37 will receive a financial result of 0 USD (if the number is 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36)
with a probability of 7/37 will receive a financial result - 3 USD (if the number is 0, 19, 20, 21, 22, 23, 24)

MO = 18/37 * 1 + 12/27 * 0 + 7/37 * (-3) = -0.081081

-0.081081/ 3 = - 0.027027..

If we take this example of Edwerk on rates.
Let the player accumulate a history of 50,000 spins
№ 21, 15, 26, 30, 17, 33, 21, 29, 17, 14, 24, 22, 10, 20, 17...... etc.
Then the final result will look like this
-3, 1, 0, 0, 1, 0, -3, 0, 1, 1, -3, -3, 1, -3, 1....
Rates
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3...

In the array where the final result is, you can sort 1, 0 and -3
Then
+1 spins 24325,
0 spins 16216,
-3 spins 9460
out of 50000.
(these shares are calculated from the theory)

24325* 1 +16216* 0 + 9460* (-3) = - 4055

Turnover (sum of bets) = 3 * 50000 = 150000

then fin res /turnover = -4055/ 150000 = -0.027...

In practice, it is very rare for a player to get exactly -0.027 due to variance.
But according to the law of large numbers, the more history is accumulated, the closer the practical result is to the theoretical one.

Edwerk writes:

And what do you think, Alatissa, will change if we imagine that we decided to only put 2 chips on 1/18 and one chip on the third dozen on one spin. Or we will put 3 chips only on the third dozen once.
What is taken into account as a specific value in your understanding of the expectation? Does the bet itself not affect the outcome of the round? Or is it time that plays a decisive role?)

With a probability of 18/37 the player will get a final result of 1st place (if he plays number 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18)
with a probability of 12/37 will receive a financial result of 0 USD (if the number is 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36)
with a probability of 7/37 will receive a financial result - 3 USD (if the number is 0, 19, 20, 21, 22, 23, 24)

MO = 18/37 * 1 + 12/27 * 0 + 7/37 * (-3) = -0.081081

-0.081081/ 3 = - 0.027027..

If on the third dozen
With a probability of 12/37 the player will receive a final result of 6uе
With a probability of 25/37 the financial result is 3rd place

MO = 12/37 * 6 + 25/37 * (-3) = - 0.081081

-0.081081 / 3= - 0.027027...

Coin wrote: 4) The player has a pile of mixed chips of different denominations 1-25-50-100$, 20 pieces of each (stack). On each spin, the player does not look and places 5 chips on the field in 5 random numbers. In one case, all 5 chips can be $1, in the other extreme case, all 5 chips are $100. On average, you get a mixed bet.

The expected value will remain -2.7% both for the amount of money and in terms of the number of chips.

a) How to correctly calculate the average win for the next spin?
b) Does knowing the history of the results of the previous 3 spins (what denomination was included in the bet and whether they were Win or Loss) affect the chances of guessing in the current spin?

LUCKY-13 wrote: In light of the presentation made by Alatissa, these are not entirely correct questions.

to Coin:
(1+25+50+100) / 4 = 44 average bet per room.
Average win 44*36 = 1584.
Knowing history doesn't affect the probability.

44 is the average bet for five numbers
fin result 272.8
5/37 * 272.8 = 36.86486.. (winnings)
32/ 37 * (-44) = 38.054054.. (loss)
36.86486- 38.054054 = -1.189189...
-1.189189/ 44 = - 0.027027..

LUCKY-13 writes:

Knowledge of history does not affect the probability.

Yes, that is true.
In the theory of belief there is a concept of dependent and independent random events.
In roulette, it is a process of independent events.

You can replace the roulette with another image (type)
Let there be 37 numbered balls from 0 to 36 in a basket. (Or a deck of numbered cards)
They took out the ball, put it back, mixed it up, took it out again....
If a player bets, for example, on #7, then there are always (after returning) 37 balls in the basket, and the probability of getting #7 each time is 1/37.

If the balls are not returned to the basket, then in this case knowledge of the history of the drops helps the player,
If #7 hasn't appeared for a long time, then the probability increases.
Although at the very beginning p=1/37
But this is a different game, it has nothing to do with roulette.