Discrete Distributions (Alatissa expert in probability theory)

There are 2 fundamentally different options:
a) when the probability of the entire sequence of events or elements of the system influences the outcome;
b) when what happened before the next event is not important at all.

A)
What is the probability of getting "heads" 2 times in a row? That's right, 0.5*0.5=0.25. That is, 2 times less than the probability of getting "heads" in a single attempt.

b)
You flip a coin 10 times, and all ten times it came up "tails." Let's flip it for the 11th time. Question: What is the probability that it will come up "tails" again?
The correct answer is 0.5! Although I really want to say (1/2)^11 = 0.00049.

"Gambler's fallacy" (roulette alignment)

This is a misconception about randomness. If repeated independent outcomes of a random process have shown deviations from the expected outcome, then future deviations in the opposite direction will become more likely.

Gambler's fallacy: "...it's obvious: if a number doesn't come up for so long, then it must come up any minute now!"

Classical probability theory considers each event separately, statistically independent from the previous ones, and not in a chain of events. Although in TV there is a law of large numbers - the average value of the sample will be close to the mathematical expectation of this distribution.

When 9 heads in a row come up, it seems obvious to many people that the next toss will have a higher probability of heads. It is hard to believe that heads will continue to come up. However, this conclusion is wrong. The probability of the next heads or tails is still 1/2.
However, it is necessary to distinguish between the concepts: the probability of getting heads or tails in each specific case and the probability of getting tails N times in a row. These are different problems.

10 consecutive tails have P=(1/2)^10.
5 eagles in a row P=(1/2)^5 = 1/32 = 0.03125

If we flip a coin 21 times, then the probability of getting 21 heads is 1 in 2,097,152. However, the probability of getting heads after 20 previous heads remains 1/2.
This option is an application of Bayes' theorem, which allows one to determine the probability of an event given that another statistically interdependent event has occurred.

And further on there is a dark forest.
Bayes' theorem says that the outcome of each trial reduces to the base probability.
That is, it is equally probable to throw 21 heads in a row or 20 heads + the next tail.

+1
Mira gets an appreciation for her comment in this thread from a reader of the blog Casino Mining Pool (CMP) about online casinos, programs and winning strategies 100pravda.com/127-roulette/teoriya-veroyatnosti

mous writes:

Okay. I have $20 on my balance.
1. If I want to bet 4 moves in a row on the neighboring "13" numbers for $1, what is the probability of winning?
2. If I want to bet 4 moves in a row on the neighbors of number "13" for $1, but before that its neighbors did not play 10 moves in a row, what is the probability of winning?
3. If I want to bet 4 moves in a row in the neighbors of "13" number for 1$, but before that its neighbors won 3 times in the previous 10 spins, then what is the probability of winning?

Yes, the same answer
Each spin p=5/37

Here I would like to answer in detail the question of whether the history of previous occurrences influences the mathematical expectation.
mous suggests considering three cases, one could say three strategies.
For each case I will find the mathematical expectation.

I'll start with the first one.

p= 5/37 each spin,
the opposite event is usually written with "q"
q= 1-p = 1- 5/37 = 32/37

There are two ways to find the mathematical expectation:
through the binomial distribution and through the geometric one.
I'll start with the binomial.
The player distributes $20 into 4 spins of $5 each.
Combinations he may encounter:

The neighbors of number "13" will never play:
qqqq
Just once:
pqqq
qpqq
qqpq
qqqp
Twice:
ppqq
qppq
qqpp
qpqp
pqpq
pqqp
Three times:
qppp
pppq
pqpp
ppqp
Four times:
pppp

If all these combinations are summed up, we get a complete group of events, which is equal to one.
To make these entries look more condensed, the binomial coefficient is used.
N!/ k! (Nk)!, where N is the number of spins on the required interval, k is the number of successful drops.
then it turns out like this
never 4! / 0! (4-0)! =1
one time 4! / 1! (4-1)! =4
two times 4! / 2! (4-2)! =6
three times 4! / 3! (4-3)! =4
four times 4! / 4! (4-4)! =1

The binomial coefficient denotes, for example,
two times 4! / 2! (4-2)! =6
ppqq
qppq
qqpp
qpqp
pqpq
pqqp

Six options, when a sector of 5 numbers will play only twice in the interval of 4 spins.

Then the whole group will look like this:
q⁴ + 4*q³p +6* q²p² + 4*qp³ + p⁴

Once the entire group is assembled, you can begin to calculate the mathematical expectation.

MO = financial result/turnover (sum of bets)

Final result on one spin: (-5 loss, +31 win)

then for the group
financial res. = q⁴* (-20) + 4* q³p* 16 + 6*q²p²* 52 + 4* qp³ * 88 + p⁴ * 124 =
=(32/37)⁴ *(-20) +4* (32/37)³ * 5/37 * 16 + 6* (32/37)²* (5/37)² *52 + 4* 32/37* (5/37)³* 88 + (5/37)⁴ * 124 = -0.54054..

turnover = q⁴* 20 + 4* q³p* 20 + 6*q²p²* 20 + 4* qp³ * 20 + p⁴ * 20 =
=(32/37)⁴ * 20 +4* (32/37)³ * 5/37 * 20 + 6* (32/37)²* (5/37)² * 20 + 4* 32/37* (5/37)³* 20 + (5/37)⁴ * 20 = 20

Then MO = financial res./turnover = -0.54054 / 20 = - 0.027027...

The second option for calculating mo is through geometric distribution.

The player distributes $20 into 4 spins of $5 each.
Then the group of events through geometric distribution will look like this:
p
qp
qqp
qqqp + qqqq

That is, the player can win immediately when he plays the desired sector,
on the second spin if the first one is a loss,
on the third, if both the first and second are losing.
On the fourth, after three losses, he will either win or lose, but the $20 allocated for the session will run out.

Full group of events:
p + qp + q²p + q³p + q⁴

Then the mathematical expectation is
MO = financial results / turnover

Finnish result = p * 31+ qp* 26 + q²p *21 + q³p * 16+q⁴ *(-20) =
= 5/37 * 31 + 32/37 * 5/37 * 26 + (32/37)² *5/37 * 21 + (32/37)³ *5/37 * 16 + (32/37)⁴ * (-20) = -0.440509..

turnover = p * 5+ qp* 10 + q²p * 15 + q³p * 20+q⁴ * 20 =
= 5/37 * 5 + 32/37 * 5/37 * 10 + (32/37)² *5/37 * 15 + (32/37)³ *5/37 * 20 + (32/37)⁴ * 20 = 16.298837..

MO = financial results / turnover = - 0.440509 / 16.298837 = - 0.027027...

That is, you can calculate it in any way, but the result is still the same.

Next, I will calculate whether such an action on the part of the player as doubling the bet after a loss will affect the mathematical expectation.

And I will also find the expected value when the player starts waiting for the entry point into the game.

God bless Alatiss!