The device of slot machines - even ultra-modern ones - is so simple that it is impossible to calculate theoretical probability winning is not a difficult task. But there is one significant “but”: for such calculations, you need to know exactly the number of game symbols on each reel, and in modern slots it can be huge. And emulator developers are not at all eager to reveal main secret, which allows you to calculate your chances of winning in slots.

Probability of winning

Purely theoretically, the probability of winning can be calculated mathematically, but in practice it is impossible to determine which combination of symbols you will get in the next round on the slot machine. The point is that the result of each spin is determined by the generator random numbers- and it is simply impossible to hack its algorithm and see by what principle the generator “produces” the result in the form of matched game symbols.

How to calculate chance

Modern emulators, which can be found in every online casino, work almost the same as the legendary " one-armed bandits" of the past. They have exactly the same a certain amount of reels, and each one is marked with a certain number of different game symbols. The chance of winning is calculated depending on these two numbers by simple exponentiation.

For example, if the slot machine has 3 reels, 20 symbols each, then the number of combinations is 20 to the 3rd power - that is, 20 x 20 x 20 = 8000 combinations. For 3 reels and 32 symbols, the number of combinations will be already 32,768 (32 x 32 x 32). And if the device has 4 reels and only 22 symbols on each, then the number of combinations will be 234,256 (22 x 22 x 22 x 22).

Now that the number of combinations has been calculated, it remains only to determine the chances of a certain sequence of game symbols falling out. For example, the chances of hitting the jackpot (three sevens) on a classic machine with 3 reels and 20 symbols on each of them (provided that only one seven is on each reel) is 1 in 8000 (1/20 x 1/20 x 1/20). If there are two sevens on one reel, and one each on the other two, the probability is calculated as 2/20 x 1/20 x 1/20 - you get a chance of 1 in 4000.

Secrets of big wins

Since to find out the exact number of game symbols in modern slot machines- the task is difficult and clearly not for the average player, secret big wins in slots is not in mathematical calculations. To win, you must first increase your chances of winning by choosing slots. From the above calculations, it is quite clear that less quantity reels, the fewer possible combinations - and the greater the chance of getting a match of game symbols.

Thus, the main secret of winning at slot machines is to choose the simplest slots with a minimum number of reels and paylines. Maybe they look very simple and not interesting, but in the end they will bring the greatest profit.

Gambling on slot machines is characterized by such an indicator as "variance of a random variable". The dispersion index of a random variable is determined from the laws of probability theory and represents the scattering number of a given random variable, in other words, the deviation from mathematical expectation. Most often, this indicator is used when playing poker. I advise you to play 777 slot machines in good quality and test the probability theory for yourself.

To understand how the slot machine works, let's take a closer look at the dispersion indicator.

Dispersion, as noted earlier, is a mathematical deviation from the numerical value of the mathematical expectation of the event under consideration. A simple explanation of this concept can serve as an ordinary children's game "heads or tails?". Throwing a coin twice, provided there are no deformations on it, according to the theory of probability, we get that the coin is one times will fall heads down, the other tails down. But a coin can fall heads down twice. This discrepancy (deviation) of the occurred event from the calculation is called the dispersion.

The dispersion index can be high or low. If the dispersion value is small, then this means that the event that happened is closest to the expected (calculated) event. And, conversely, a high dispersion index indicates a strong spread of the result relative to the calculated value.

Taking into account these mathematical characteristics, it is customary to distinguish three types of dispersion of slot machines, which differ in the game process and the number of winning combinations in it.

The first type is high-dispersion automata. They are characterized by a rare loss of winning combinations for the entire process of the game. However, at the same time winning combinations bring the player a larger amount than in the other two types of machines. If you have a lot of time, a huge supply of patience and a decent initial amount to start, then playing on such a slot will bring you significant income. On a machine of this type, there is a big risk of losing all your funds, "not reaching" one game to winning combination.

The second type is slot machines with medium dispersion. The winnings on them, as a rule, are not large, but exceed the winnings on low-dispersion machines. This type gaming slot perfect for you if you don't have enough large sum money to play "until winning" on high-dispersion slot machines. This type also differs from the machine with high variance by the frequency of occurrence of winnings.

The third type of automata are low dispersion. They very often give out winning combinations, but the amount of winnings on them is very small and rarely exceeds the minimum bid. If you like to feel like a winner during the game, and at the same time have a small amount of money in your pocket, give preference to this particular type of machine.

In addition to these differences, there is another feature of slot machines that arises due to their different variance - the payout ratio. If the final payouts are more than ten thousand times the initial bet amount, this is a high variance machine. Payouts larger than the initial bet by five to ten thousand times are typical for slot machines with medium dispersion. And finally, low variance slots increase the initial bet by less than five thousand times

Showing the amount of dispersion on slot machines is not profitable. For this reason, it is difficult to determine the variance of a single automaton. The question arises: how to place a bet and not make a mistake? How to calculate which machine you are dealing with at the next game. In this problem, the following few elementary tips will be indispensable:

See reviews on the forums to get an idea of ​​the different machines;

Try to play on the machine and evaluate it yourself. At first, you can play in demo mode;

Analyze the rules of the game, the payout table presented on the site.

The level of risk when playing a slot machine is called volatility. The amount of winnings and how often lucky combinations will fall out to the player directly depend on it. Therefore, before playing, clearly determine for yourself what level of risk you can afford, in other words, decide which machine you should play based on the funds you have.

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1. In a batch of 10 items, 2 are defective. 3 items are chosen at random. Determine the probability that among these products there will be at least one defective.

Solution: We find the required probability using the classical probability distribution formula. First we find n - total number possible outcomes in this trial. because the order of the items is irrelevant. 3 items out of 10 can choose

ways.

Now let's find the number of favorable outcomes m - the number of outcomes in which there will be at least 1 defective item out of 3 selected ones. Since the number of defective items in the batch is 2, the outcomes will be favorable when out of 3 selected items there will be 1 or 2 defective ones. Let's find the number of favorable outcomes m 1 , when among 3 selected products there is 1 defective one.

Let's find the number of favorable outcomes m 2 when among 3 selected products there are 2 defective ones. . The total number of favorable outcomes. Finally:

2. From a deck of 36 cards, 3 cards are drawn at random. What is the probability that there will be 2 aces among them. Solution: We find the required probability using the classical probability distribution formula. First we find n - the total number of possible outcomes in this trial. Since the order of the cards is indifferent, 3 items out of 36 can be selected

ways.

Now let's find the number of favorable outcomes m - the number of outcomes in which there will be 2 aces from 3 selected cards. 2 aces out of 4 can be drawn in ways. Since each combination of aces can be combined with any combination of the rest of the cards, everything will turn out to be a variant. Finally we get:

3. 12 workers received vouchers to 4 rest houses: 3 - to the first, 3 - to the second, 2 - to the third and 4 - to the fourth. Find the probability P(A) that these three workers go to the same holiday home.

Solution: The probability that these three workers will end up together and end up in any of the 4 holiday homes is equal. Since only 2 vouchers were allocated to the third holiday home, they need to get into the remaining 3 out of 4 holiday homes. The probability of this event is Р(3/4) = 0.75.

Finally we get:

• 4. When manufacturing a part, the workpiece must go through 4 operations. Assuming the occurrence of defects in individual operations as independent events, find the probability of manufacturing a standard part if the probability of occurrence of defects in the first operation is 0.05, in the second - 0.01, in the third - 0.02, in the fourth - 0.03. Solution: The probability of manufacturing a good part P(A) is equal to the product of the probabilities of manufacturing a good part at each operation P(Ai). Hence
• 5. Some mechanism consists of 6 parts, of which 2 are worn out. When the mechanism is running, 2 parts are switched on randomly. Find the probability that the unworn parts are turned on.
  Solution: Using the hypergeometric distribution formula, we determine the desired probability that unworn parts will be included.

6. A volley of two guns was fired at the target. The probability of hitting from the first gun is 0.85, from the second 0.91. Find the probability of hitting the target.

Solution: Denote the probability of hitting from the first gun as P(A), and from the second gun as P(B). When the target is hit, 3 options are possible: when both guns hit the target - the probability of this event is equal to

when only the first gun hit the target, the probability of this event is equal; when only the second gun hit the target, the probability of this event is equal. Then the probability of hitting the target will be equal to the sum of all three probabilities:

7. The worker serves 4 machines. The probability that within an hour the first machine will not require the attention of a worker is 0.7, for the second machine - 0.8, for the third - 0.9, for the fourth - 0.85. Find the probability that during an hour at least one machine does not require the attention of a worker.

Solution: Here the probabilities p 1 = 0.7; p 2 = 0.8; p 3 = 0.9; p 4 \u003d 0.85 there are probabilities that one of the machines will require the attention of a worker within an hour, and q 1 \u003d 0.3; q 2 \u003d 0.2; q 3 \u003d 0.1; q 4 \u003d 0.15 there are probabilities that one of the machines will not require the attention of a worker for an hour. Let us find the probability of the opposite event: the probability that within an hour all the machines will require the attention of the worker

Then the probability that within an hour at least one machine does not require the attention of the worker will be equal to

8. Parts from three machines are included in the assembly. It is known that the first machine gives 0.3% of defects, the second - 0.2% and the third - 0.4%. Find the probability of a defective part getting into the assembly if 1000 parts were received from the first, 2000 from the second and 2500 parts from the third.

Solution: Let's solve the example using the formula full probability. As hypotheses, we accept the following events: H 1 - an arbitrarily selected part, made on the first machine, H 2 - an arbitrarily chosen part, made on the second machine, H 3 - an arbitrarily chosen part, made on the third machine; event A is that the part that got to the collection is defective. According to the total probability formula, we have:

where: - the probability that the selected part is defective, provided that it is from the i-th automaton, respectively; Р(Н i) - probabilities of hypotheses. Let's find probabilities of hypotheses.

Finally we get:

9. There are 10 identical urns, of which 9 contain 2 black and 2 white balls, and one contains 5 white and 1 black ball. What is the probability that a ball is drawn from an urn containing 5 white balls if it is white?

Solution: The probability P(A) that a ball is taken from an urn containing 5 white balls, if it turned out to be white, is equal to the product of the probability that a ball is taken from an urn containing 5 white balls P(H) and the probability that the ball taken from this urn the ball turned out to be white P(B).

P(H) = 1/10; using the hypergeometric distribution formula

Finally we get:

10. 10 customers entered the store. The probability of making a purchase for each entrant is the same and equals 0.2. Find the probability that 6 of them make a purchase.

Solution: Apply the Bernoulli formula. Here n = 10, m = 6, p = 0.2, q = 1 - 0.2 = 0.8. By the Bernoulli formula we get

11. A random variable has a probability distribution represented by the table: