Theorem on the presence of memory in random sequences
To test the hypothesis you can use a random number generator in Excel (Roundbetween(1;6)) and check the above rule for say 1000 cases. I do not have a mathematical advantage. Although it is necessary to check with the author, what he proposes to do under the condition X1=X2.
To test the hypothesis you can use a random number generator in Excel (Roundbetween(1;6)) and check the above rule for say 1000 cases. I do not have a mathematical advantage. Although it is necessary to check with the author, what he proposes to do under the condition X1=X2.
To test the hypothesis you can use a random number generator in Excel (Roundbetween(1;6)) and check the above rule for say 1000 cases. I do not have a mathematical advantage. Although, I need to check what the author suggests to do with X1=X2.
What for? Because it can be simpler and more exact.
Let on a roulette drum n numbers, from 0 to n - 1 inclusive.
Suppose that in case if the ball hits the number with bet, then dealer returns the number retTo make it easier to understand, let's make a table. We have three consecutive spins x1, x2, x3 can fall one maximum (max), one minimum (min) and one average (mid).
- If the roll of the last spin coincides with the number of the penultimate spin, we skip a move.
- If x1 > x2, then bet on all numbers greater than x2. We have such numbers: n - 1 - x2
- If x1 < x2, then bet on all numbers lower than x2. We have such numbers: x2
Then we have this result:
| Combination | Penultimate spin - x1 | Last spin - x2 | Future spin - x3 | Betting size | Win size |
|---|---|---|---|---|---|
| 1 | min | mid | max | mid | -mid |
| 2 | min | max | mid | max | ret - max |
| 3 | mid | min | max | n - 1 - min | ret - n + 1 + min |
| 4 | mid | max | min | max | ret - max |
| 5 | max | min | mid | n - 1 - min | ret - n + 1 + min |
| 6 | max | mid | min | n - 1 - mid | n - 1 - mid |
| Total: | 3 * n + 2 * max - 2 * min - 3 | 4 * ret - 3 * n - 2 * max + 2 * min + 3 |
That's all. Now we just need to write a program and check all variants in the nested loop.
For European roulette: n = 37, ret = 35
In Java, such a program would look like this
public class Main { public static void main(String[] args) { // Количество чисел на барабане int n = 37; double dn = n; // Возврат денег дилером в случае если ставка выиграет int ret = 35; double total = 0 d; // Счётчик спинов int score = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (i != j) { int max = Math.max(i, j); int min = Math.min(i, j); double dmax = max; double dmin = min; double result = 4 d * ret - 3 d * dn - 2 d * dmax + 2 d * dmin + 3 d; System.out.println("Max = " + max + ", Min = " + min + ", Result = " + result); total = total + result; } score++; } } double dscore = score * 6; total = total / dscore; // Математическое ожидание выигрыша с одного спина System.out.println("Total = " + total); } }
Let's run it and check it:
... Max = 36, Min = 28, Result = 16.0 Max = 36, Min = 29, Result = 18.0 Max = 36, Min = 30, Result = 20.0 Max = 36, Min = 31, Result = 22.0 Max = 36, Min = 32, Result = 24.0 Max = 36, Min = 33, Result = 26.0 Max = 36, Min = 34, Result = 28.0 Max = 36, Min = 35, Result = 30.0 Total = 1.0810810810810811It turns out the profit is a bit more than a quid per spin
It's easy to send, but it's easy to prove mathematically that the person is right or wrong...
To test the hypothesis you can use a random number generator in Excel (Roundbetween(1;6)) and check the above rule for say 1000 cases. I do not have a mathematical advantage. Although it is necessary to check with the author, what he proposes to do under the condition X1=X2.
Easier to check on the online casino ...
I do not advise you to "test" such strategies in online casinos. Because in virtual casinos, unlike in real casinos, the theory of probability does not rule. There algorithm is set up so that the casino will never get into minus, ie, if the current spin, it does not receive a profit, which is specified in the settings, the algorithm will automatically pick up a "falling out" number that was not bet on - an artificial loss.
... which is probably already done by the author :)
The author prefers stock (not kitchen) trading. The above strategy for trading also rules. Real casinos are forbidden here.
I do not advise you to "test" such strategies in online casinos. Because in virtual casinos, unlike in real casinos, the theory of probability does not rule. There, the algorithm is set up so that the casino will never get into minus, ie, if the current spin, it does not receive a profit, which is specified in the settings, the algorithm will automatically pick up a "falling out" number that was not bet on - an artificial loss.
The author prefers stock (not kitchen) trading. The aforementioned trading strategy also rules. Real casinos are banned here.
Yes, I know that real casinos are banned throughout the former Soviet Union.
Vegas was a free-for-all :)
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The essence of the theorem is that if analysis of prehistory of random sequences on one depth gives zero mathematical expectation, it doesn't mean that the analysis of prehistory on the other depth will give the same expectation.
To simplify it, in order to prove the presence of memory in a random sequence, one should analyze it to its full depth.
Sometimes the presence of memory is confused with an aftereffect. An aftereffect is the presence of an opportunity for a conditional probability that will not equal an unconditional probability. However, the presence of an aftereffect does not at all imply a change in the expectation of the game.
To make it easier to understand how it can be useful for us in practice, even if our knowledge in mathematics is not very good, it would be better to give a concrete example. We won't take roulette from a casino as an example (especially since roulette has two varieties: European and American), but rather a simpler case to make it easier to understand. Let's take a playing cube. Suppose we were to bet $1 each on a number between 1 and 6 (the number of edges of the cube).
It is very easy to calculate winnings or losses, because if we each bet a dollar on different numbers, then if after rolling the die at least one of the numbers turns out under our bet, the dealer will return $6, which corresponds to a win of $6 - n, where n is the number of numbers on which $1 was bet. If none of the numbers under the bet turns out after rolling the die, then the dealer will take all the money we bet.
We skip the first two die rolls that resulted in x1 and x2. And bet on the third roll - x3, but according to the rules of conditional probabilities:
Suppose we have three numbers in three tosses: 2, 3 and 5 (in fact, the theorem proves that it makes no difference which numbers fell out). In what order these three numbers also fell out there is no particular difference, because there are only six options, and they are all equal probability.
Now look at the results (the red colour shows the bets on the numbers less than x2):
It turns out that we got expectation +$4, despite the fact that all combinations of numbers 2, 3 and 5 are equal probability.
Some would probably say no way? Trust but verify. Because for demonstration is chosen playing cube, which has only 6 numbers on its edges and it is difficult to confuse even a schoolboy.
For example, the first combination. We bet on numbers less than x2 = 3, and there are only two: 1 and 2. Correspondingly, the size of our bet was $2. But x3, was equal to 5, i.e., none of the numbers we bet was equal to 5 and we lost all our bets, i.e., $2.
The second combination: a bet on numbers smaller than x2 = 5. There are four: 1, 2, 3, 4, i.e. we gave the dealer $4. x3 = 3 came out. The bet won. The dealer gave us back $6. As a result, our deposit was replenished by a win of +$2.
And so on and so forth.
The theorem proves that if we always bet according to the above conditional probabilities when x1 <> x2, then no matter what values x1, x2 and x3 have and in what order, the mathematical expectation will always be positive.
But someone will object again, that a dealer will hardly want to return us in case of successful betting $6, but rather he will try to decrease our expectation, for example, giving us only $5 in case of winning. Well, then it is easy to calculate that we will have a zero expectation. That is, the game will be fair, despite the fact that the dealer will think that he will earn on it.
OK. Some may start to argue that casinos are illegal in the RF, but stock market speculation is allowed. However, if stock quotes are represented as an equal probability Bernoulli scheme with some missing data (holes in history), the theorem again proves that the expectation under the same conditional probabilities will be positive.
If you are not convinced, the text of the theorem is not secret and can be found in the attached archive. Try to find mistakes in it.