Probability with coffee machines. Slot machines from the point of view of probability theory

The design of slot machines - even ultra-modern ones - is so simple that it is difficult to calculate theoretical probability winning does not seem to be a difficult task. But there is one significant “but”: for such calculations you need to know exactly the quantity game symbols on every 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 what combination of symbols you will get in the next round on the slot machine. The fact 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 on what principle the generator “produces” the result in the form of matching game symbols.

How to calculate chance

Modern emulators, which can be found in every online casino, work almost the same as the legendary ones " one-armed bandits" of the past. They also have a certain number of reels, and each has a certain number of different game symbols. The chance of winning is calculated depending on these two numbers by simply raising to a power.

For example, if a 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 machine 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, all that remains is to determine the chances of a certain sequence of game symbols appearing. For example, the odds of hitting the jackpot (three sevens) on a classic slot machine with 3 reels and 20 symbols on each of them (assuming there is only one seven on each reel) are 1 in 8000 (1/20 x 1/20 x 1/20). If there are two sevens on one reel, and one on each of the other two, the probability is calculated as 2/20 x 1/20 x 1/20 - the chance is 1 in 4000.

Secrets of big wins

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

Thus, the main secret of winning in slot machines- is to choose the simplest slots with a minimum number of reels and paylines. They may look very simple and uninteresting, but in the end they will bring the greatest profit.

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

To understand the principle on which 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 in question. A simple explanation of this concept can be the ordinary children's game “heads or tails?” Having tossed a coin twice, provided there are no deformations on it, according to probability theory, we get that the coin is one will fall once heads down, the other tails down. But the coin can even fall heads down twice. This discrepancy (deviation) of the event that occurred from the calculation is called dispersion.

The dispersion index can be high or low. If the dispersion value is small, this means that the event that occurred is closest to the expected (calculated) event. And, conversely, a high dispersion indicator 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 variance in slot machines, which differ in the process of the game and the number of winning combinations in it.

The first type is high-dispersion automatic machines. They are characterized by a rare occurrence of winning combinations throughout the entire game process. However, at the same time winning combinations bring the player a larger amount than in the other two types of slot 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. Automatically of this type there is a big risk of losing all your funds, “falling short” of one game before winning combination.

The second type is machines with average dispersion. The winnings on them, as a rule, are not great, but they exceed in their magnitude the winnings on low-dispersion slot machines. This type game slot perfect for you if you don't have enough large sum money to play until victory on high-variance slot machines. This type is also distinguished from a high variance slot machine by the frequency of winnings.

The third type of machines are low-dispersion machines. They very often give out winning combinations, but the winning amount is very small and rarely exceeds minimum bid. If you like to feel like a winner while playing and have a small amount of money in your pocket, give preference to this type of machine.

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

Showing the dispersion value on slot machines is not profitable. For this reason, difficulties arise in determining the variance of a single machine. The question arises: how to place a bet and not make a mistake? How to calculate which machine you are dealing with the next time you play. In this problem, the following few basic tips will be indispensable:

Look at reviews on forums to get an idea of ​​different machines;

Try playing the slot machine and evaluate it yourself. At first you can play in demo mode;

Analyze the rules of the game and the payout table presented on the website.

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

P.S. My name is Alexander. This is my personal, independent project. I am very glad if you liked the article. Want to help the site? Just look at the advertisement below for what you were recently looking for.

12.12.2017 Lyudmula Abramochkina

Problems in probability theory (11-13)

Problem 11. Two factories produce the same glass for car headlights. The first factory produces 45% of these glasses, the second – 55%. The first factory produces 3% of defective glass, and the second – 1%. Find the probability that glass accidentally purchased in a store will be defective.

Solution:

The probability that the glass was purchased at the first factory and is defective:

P(A1) = 0.45 0.03 = 0.0135

The probability that the glass was purchased from a second factory and is defective:

P(A2) = 0.55 0.01= 0.0055

According to the total probability formula, the probability that glass accidentally purchased in a store will be defective is equal to

0,0135 + 0,0055 = 0,019

Answer: 0.019

Problem 12. IN mall two identical machines sell coffee. The probability that the machine will run out of coffee by the end of the day is 0.3. The probability that both machines will run out of coffee is 0.12. Find the probability that at the end of the day there will be coffee left in both machines.

Let's consider the events:

A = coffee will run out in the first machine,
B = Coffee will run out in the second machine.

A B = coffee will run out in both machines,
A + B = coffee will run out in at least one machine.

By condition P(A) = P(B) = 0.3; P(AB)= 0.12

Events A and B are joint, the probability of the sum of the two joint events is equal to the sum of the probabilities of these events, reduced by the probability of their occurrence:

P(A+B) = P(A) + P(B) – P(A B) = 0.3 + 0.3 – 0.12 = 0.48

Then, the probability of the opposite event, that the coffee will remain in both machines, is 1 – 0.48 = 0.52

Answer: 0.52

Problem 13. Before a soccer match begins, the referee tosses a coin to determine which team will have first possession of the ball. The “White” team takes turns playing with the “Red”, “Blue”, “Green” teams. Find the probability that in exactly two out of three matches the White team will have the first possession of the ball.

We make a list of all possible outcomes in these three games with “Reds” (R), “Blues” (C) and “Greens” (G).
P – first in possession of the ball, N – not.

PPP
PPN
PNP
NPP
PIN
NPN
NNP
NNN

and see how many of them contain exactly 2 times P, i.e. in exactly two matches the White team will have the first possession of the ball.
There are 3 such options, and there are 8 options in total. Then the required probability is equal to

Task 1. At the geometry exam, the student gets one question from the list of exam questions. The probability that this is a question on External Angles is 0.35. The probability that this is an inscribed circle question is 0.2. There are no questions that simultaneously relate to these two topics. Find the probability that a student will get a question on one of these two topics in the exam.

Solution:

Events “You will get a question on the topic Inscribed Angles” and “You will get a question on the topic Inscribed Circle” – . This means that the probability that a student will get a question on one of these two topics in the exam is equal to the sum of the probabilities of these events: 0.35 + 0.2 = 0.55.

Answer: 0.55.

Task 2. Two factories produce the same glass for car headlights. The first factory produces 70% of these glasses, the second – 30%. The first factory produces 1% of defective glass, and the second – 3%. Find the probability that glass accidentally purchased in a store will be defective.

Solution:

Situation 1:

The glass comes from the first factory (event probability 0.7) and (multiplication) it is defective (probability of event 0.01).

That is, both events must happen. In the language of probability theory, this means each of the events:

Situation 2:

The glass comes from the second factory (event probability 0.3) And it is defective (probability of event 0.03):

Because when buying glass we find ourselves in situation 1 or (amount) in situation 2, then we get:

Answer: 0.016.

Task 3. In the shopping center, two identical machines sell coffee. The probability that the machine will run out of coffee by the end of the day is 0.3. The probability that both machines will run out of coffee is 0.16. Find the probability that at the end of the day there will be coffee left in both machines.

Solution:

The probability of event A: “the coffee will run out in the first machine” P(A) is equal to 0.3.

The probability of event B: “the coffee will run out in the second machine” P(B) is equal to 0.3.

The probability of event AB: “the coffee will run out in both machines” P(AB) is equal to 0.16.

The probability of the sum of two joint events A+B is the sum of their probabilities without the probability of event AB:

We are interested in the probability of an event opposite to event A+B. Indeed, a total of 4 events are possible, three of them marked yellow, correspond to event A+B:

Answer: 0.56.

Task 4. There are two payment machines in the store. Each of them can be faulty with probability 0.12, regardless of the other machine. Find the probability that at least one machine is working.

Solution:

Both machines are faulty with probability

At least one machine is working (good+faulty, faulty+faulty, good+good) – this is an event opposite to the event “both machines are faulty”, so its probability is

Answer: 0.9856.

Solution:

Biathlete hits the target for the first time and (multiplication) second, And third:

Since the probability of hitting the target is , then the probability of the opposite event, a miss, is

The biathlete missed the fourth shot And at the fifth:

Then the probability that the biathlete hit the target the first 3 times is ( And!) the last two missed are as follows:

Answer: 0.01.

Task 6. The likelihood that a new vacuum cleaner will last more than a year, is equal to 0.92. The probability that it will last more than two years is 0.84. Find the probability that it will last less than two years but more than a year.

Solution:

Consider the following events:

A – “the vacuum cleaner will last more than a year, but less than 2”,

B – “the vacuum cleaner will last more than 2 years”,

C – “the vacuum cleaner will last more than a year.”

Event C is the sum of joint events A and B, that is

But, since both A and B cannot happen at the same time.

Answer: 0.08.

Solution:

Probability of all three light bulbs burning out within a year

Then the probability of the opposite event - at least one lamp will not burn out - is

Answer: 0.999657.

Solution:

Method I

Let the probability that the egg purchased from the agricultural company is from farm I be . Then the probability that the egg purchased from the agricultural company is from farm II is .

1) from farm I And Category I

2) from farm II And I category,

II method

Let be the number of eggs of the first farm, then the number of eggs of the highest category in this farm is .

Let be the number of eggs of the second farm, then the number of eggs of the highest category in this farm is .

Since, according to the condition, 60% of the eggs receive the highest category, and all eggs purchased by the agricultural company, of which the highest category, then

That is, many times more eggs are purchased from the first farm.

Then the probability that the egg purchased from this agricultural company will be from the first farm is

Answer: 0.6.

Task 9. Cowboy John has a 0.9 chance of hitting a fly on the wall if he fires a zeroed revolver. If John fires an unfired revolver, he hits the fly with probability 0.3. There are 10 revolvers on the table, only 4 of which have been shot. Cowboy John sees a fly on the wall, randomly grabs the first revolver he comes across and shoots the fly. Find the probability that John misses.

Solution:

John grabs the sighted revolver (possibly) And misses (probability). Probability of this event

John grabs an unfired revolver (possibly) And misses (probability). The probability of this event

John can grab a sighted revolver and miss or grab an unfired revolver and miss, so the required probability is:

Answer: 0.46.

Solution:

Let event A: “the student solves 12 problems correctly,”

event B: “the student will solve more than 12 problems”,

event C: “the student will solve more than 11 problems.”

In this case, the probability of event C is the sum of the probabilities of events A and B:

– this is the desired probability.

Answer: 0.1.

Solution:

(We marked “X” for “good weather”, “O” for “excellent weather”)

Event D: XXXO will occur with probability

Event F: XXOO will occur with probability

Event J: ХООО will occur with probability

Event H: XOXO will occur with probability

Answer: 0.392.

Problem 12. The picture shows a labyrinth. The spider crawls into the maze at the Entrance point. The spider cannot turn around and crawl back, so at each branch the spider chooses one of the paths along which it has not yet crawled. Believing that the choice further path purely random, determine with what probability the spider will come to exit D.

Solution:

On its way, the spider encounters four forks. And at each fork, the spider can choose the path leading to exit D with probability 0.5 (after all, at each fork, two independent equally possible events are possible: “choosing the right path” and “choosing the wrong path”). The spider will reach exit D if he chooses the “right path” at the first fork And On the second, And on third, And on the fourth, that is, the spider will come to exit D with a probability equal to
Answer: 0.0625.

Solution:

Let be the probability that a patient admitted with suspected hepatitis really sick hepatitis.

Then is the probability that a patient admitted with suspected hepatitis not sick hepatitis.

The analysis gives a positive result in cases

the patient is sick And (multiplication) the test is positive

or (addition)

the patient is not sick And the test is false positive

Since, according to the conditions of the task, in 6% of patients with suspected hepatitis, the analysis gives a positive result, then

Round to the nearest thousand: .

Answer: 0.056.

Solution:

Let's reformulate the problem question:

How many shots will it take for the miss probability to be less than 0.02?

With one shot, the probability of miss is 0.6.

With two shots, the probability of a miss is (the first shot is a miss and the second shot is a miss).

With three shots, the probability of a miss is –

With four shots, the probability of a miss is –

With five shots, the probability of a miss is –

We notice that .

So, five shots are enough for the probability of destroying the target to be at least 0.98.

Condition

In a shopping center, two identical machines sell coffee. The machines are serviced in the evenings after the center closes. It is known that the probability of the event “By evening the first machine will run out of coffee” is 0.25. The probability of the event “By evening the second machine will run out of coffee” is the same. The probability that both machines will run out of coffee by evening is 0.15. Find the probability that by evening there will be coffee left in both machines.

Solution

Consider the events

\[\text( : """")\],

\[\text( : """")\].

\[\text(A)\cdot \text(B = """")\],

\[\text(A + B = """")\].

By condition

& \text(P)\left(\text(A) \right)\text( = P)\left(\text(B) \right)\text( = 0)\text(,25; ) \\

& \text(P)\left(\text(A)\cdot \text(B) \right)\text( = 0)\text(,15) \\

Events A and B are joint, since they can occur simultaneously, therefore, the probability of the sum of two joint events is equal to the sum of the probabilities of these events, reduced by the probability of their occurrence:

\[\text(P)\left(\text(A + B) \right)\text( = P)\left(\text(A) \right)\text( + P)\left(\text(B ) \right)\text( - P)\left(\text(A)\cdot \text(B) \right)\text( = 0)\text(,25 + 0)\text(,25 - 0) \text(,15 = 0)\text(,35)\].

Therefore, the probability of the opposite event, that the coffee will remain in both machines, is 1 − 0.35 = 0.65.

Let's give another solution

The probability that the coffee will remain in the first machine is 1 − 0.25 = 0.75. The probability that the coffee will remain in the second machine is 1 − 0.25 = 0.75. The probability that coffee will remain in the first or second machine is 1 − 0.15 = 0.85. Since P(A + B) = P(A) + P(B) − P(AB), we have: 0.85 = 0.75 + 0.75 − X, where the desired probability comes from X = 0,65.

Note.

Note that events A and B are not independent. Indeed, the probability of producing independent events would be equal to the product of the probabilities of these events: \[\text(P)\left(\text(A)\cdot \text(B) \right)=0.25\cdot 0.25=0 ,0625\], however, according to the condition, this probability is equal to 0.15.