A little about lotteries

In numerical lotteries, a single simple combination is equally probable and is “a single indivisible entity.” In other words, in the space of a complete array, all elements (mentally imagine “cubes”) have the same size, therefore, there are no priority individual combinations. It is impossible to single out “universal combinations” in the entire array that will “always” play better than others, since the lottery machine or circulation generator is equally likely! What is most striking is that even many experienced players do not understand this.

Equally probable distribution of played combinations –
simple proof #1

Let's move on to the most natural statistics in numerical lotteries - combinatorial. To do this, you need to translate all the winning combinations, for example, in the lottery 5 out of 36, into their serial number (index) in the full array. A scatter plot of the distribution of these combinations in the space of the full array can then be plotted, while respecting the interval and location in the circulation history. Each point on this graph represents a combination that actually played in the space of the full array. Since each individual combination is distributed equally likely throughout the entire array, we can divide this space into equal parts (sectors).

Let's divide the full array of 376992 combinations,
let's say - into 12 equal parts - sectors
- 31416 combinations.

All combinations actually played at the moment in the lottery 5 out of 36
(equiprobable distribution), selected sector - any

Let's count the number of matches of each sector over the last 500 draws.
On average, there will be approximately the same number of hits of a combination in any sector - 41 times.
The chance of any sector to match is 376,992/31416 = 1 time in 12 draws (average)
For 500 draws, any sector will play 500/ 12 = 41 times (average) or 4 times for 50 draws or 2 times for 25
If the combination plays in the selected sector, then the chance of the jackpot increases 12 times for one simple combination from this sector, and will be equal to 1 in 31416. If we have 10 combinations in the game, then 1 in 3141.

What is a single combination?

Let's see what a single combination is using the example of lottery 5 out of 36. There are 376,992 such combinations in this lottery. Each combination has its own serial number in the full array (index - cell).

First combination (000001) = 01-02-03-04-05 ...
Last combination (376992) = 32-33-34-35-36 = 376992 pieces

000001 _ 01-02-03-04-05
000002 _ 01-02-03-04-06
000003 _ 01-02-03-04-07
000004 _ 01-02-03-04-08
…….
…….
…….
002024 _ 01-02-07-11-30
002025 _ 01-02-07-11-31
002026 _ 01-02-07-11-32
…….
…….
174078 _ 04-21-25-32-34
174079 _ 04-21-25-32-35
…….
376992 _ 32-33-34-35-36

Absolutely any combination in the full array is no different from others in terms of the probability of a match.
To better understand this, you need to imagine 376,992 individual lottery balls, with all 376,992 combinations labeled.
It is difficult to imagine such a quantity, much less fit it into a picture; I will show only a few balls out of 376,992 pieces.

Let's do a thought experiment- let's place these balls in a huge lottery machine, which throws out only one ball with the combination indicated on this ball for each draw. We should not forget that after each draw, the dropped ball with the combination indicated on it is thrown back into the same lottery drum. Thus, for the next draw, all combinations will be in place again, and when the lottery machine starts, they will be mixed equally with everyone else.

If it is difficult to imagine the option with balls, then let’s try to imagine a huge roulette wheel, where each ball cell represents a combination. There are 376,992 such cells, since such a lined wheel also cannot fit into the picture, so for a general understanding we will draw only a tiny part with combinations - we have highlighted the initial and final ones.

Take a closer look at the picture - the “wheel” is divided into equal cells(equally probable combinations), and the ball (draw generator) can fall into any hole (cell - index), no matter how we designated these cells (even with pictures). After the draw (spin), the wheel does not decrease - all cells remain in place.

• Note: I would like to draw your attention once again - I am writing about a whole simple single combination. For each individual combination (cell), the meaning of any even, odd, sums, intervals between numbers, repetitions, consecutive numbers, etc. is completely lost - since the combination is a single whole and denotes a cell (index) in the full array, and their huge quantity.

We can trace only individual areas of the array (sectors, ranges, groups of numbers) for the upcoming games, therefore, we will increase our chances of winning the main prize (in individual draws) tens and even hundreds of times. Depends on which sector (array, range) we guess.

Equal distribution
combinations played - simple proof No. 2

Let's take an example of 24 numbers (lottery 6 out of 45), chosen at random.

Let's calculate the probability of complete and partial coincidence based on the real circulation history in a simplified manner (a simple calculation, and quite accurate for a large number of circulations), then use the special HYPERGEOMET function, which is present in Excel spreadsheets. This is a statistical function that can be used to calculate the probability of a complete or partial match.

(click to enlarge)

2311 lottery draws 6-45 have been loaded.

1. One match showed in 128 draws
2311/128 = 1 to 18.1.
HYPERGEOMET = 1 to 16.6.

2. Two matches showed in 472 circulations
2311/472 = 1 to 4.9
HYPERGEOMET = 1 to 4.9

3. Three matches were shown in 754 circulations.
2311/754 = 1 to 3.1
HYPERGEOMET =1 to 3.02

4. Four matches were shown in 659 circulations.
2311/659 = 1 to 3.5
HYPERGEOMET = 1 to 3.6

5. Five matches were shown in 249 circulations.
2311/249 = 1 to 9.3
HYPERGEOMET = 1 to 9.12

6. Six matches were shown in 37 runs.
2311/37 = 1 to 62.5
HYPERGEOMET = 1 to 60.51

As you can see, the probability of complete and partial coincidence almost completely coincides with the calculated values. This means that the lottery generator produces combinations with equal probability. When generating or manually marking any markers, the values ​​will differ slightly, but they will be close to the theoretical ones. The more circulation history is loaded, the closer the result. Due to the fact that the circulation in the archive is catastrophically small, we use groups of numbers of sufficient length.

From the uniform (equiprobable) distribution, another conclusion follows: it doesn’t matter which numbers are included in the group of numbers - even, odd, the top of the playing field or the bottom, etc. The only thing that matters is the number of numbers in the group, on which the probability directly depends. We look at the screenshot - 18 numbers of markers are marked - random, top, even.

(click to enlarge)

There are no significant differences in the intensity of the coincidence of 5 numbers.
In other words, the circulation generator pays attention to any marked markers evenly, no matter what you “draw” on the playing field. Sometimes they “advise” to play with so-called “pieces” - this will not change anything in terms of the probability of a match - any “piece” will play with the same frequency as a “non-piece”...

Now we know for sure - any marked group of numbers, in equal numbers, has the same probability of coincidence. Why? Because it is made up of equally probable simple combinations. In this case, how can we even understand which group may be more likely to play in the coming games?

Strategic combination generators for numerical lotteries

When you realize that a particular combination is equally likely,
then some people are completely confused about common statistics :)

For example, why “even-odd” is played in the “majority” in a certain proportion, or why “sum” plays in the middle range and more. It turns out that the combinations seem to be not equally probable? This question is easy to answer, precisely after fully realizing that a single combination is equally probable. So why do combinations seem to “love to play” in certain proportions, ranges, amounts - if they are equally probable?

• Because we “select” arrays of equally probable single combinations with this information. It is important to know here how many combinations obtained in dedicated sectors. Arrays of combinations, highlighted with statistical information - contain different quantities equally probable combinations, therefore, these arrays have different probability for a coincidence.

Let's look at the example of statistics
even, odd numbers

• Let's try to understand one of the popular tips when choosing a combination:
  choose combinations that contain an equal number of even and odd numbers

Let's figure out why this happens. In lottery 5 out of 36, the most common odds and evens will look like this: 2 even – 3 odd, or 3 even – 2 odd. We count the number (even - odd) of all possible combinations in the lottery 5 out of 36

To better understand why a lottery machine or a draw random number generator tries to throw out such combinations of numbers in combinations, let’s turn for clarity to the roulette wheel, which is nothing more than an equally probable random number generator, unless, of course, it’s skewed :)

Let's distribute all combinations according to odd-even criteria together, and according to the table,
Let's draw a circular graph - imagine that these are marked sectors on a roulette wheel

Mentally add the largest sectors that contain 124848 combinations together = 124848 pieces (2 even - 3 odd) + 124848 pieces (3 odd - 2 even) = 249696 combinations out of 376992 possible, or 66.23%, or the chance of these two sectors is 376992/ 249696 = 1 to 1.5 for each spin (draw) or approximately 33 numbers out of 36.

That is why, with each test (roulette spin) of a lottery machine or draw generator, combinations from this sector will tend, in most cases, to play in an odds ratio of 2-3 or 3-2.

• In this example it plays not a separate combination– here a dedicated “huge sector” with combinations plays, in other words, we have marked approximately 33 numbers out of 36, naturally, almost always this number of numbers will “catch” all the prize money!

Why parity in combinations like 2-3 or 3-2? Everything is explained by the costs of the decimal system, which encodes the whole combination. Each individual whole (complete) combination simply represents a cell of 376,992 pieces. Recall the thought experiment with balls, in which the combination is indicated as a whole, or an example with a roulette wheel, where each combination simply designates a cell and is indivisible. But how we select the array of combinations does not matter. It’s just convenient to follow these signs (even-odd) for part of the array - the sector.

If we generate any random combinations for the same number of combinations (2469696 pieces), regardless of these proportions in general, then nothing will change in terms of the probability of matching the resulting array (sector) (1 to 1.5). Any equally probable random combination generator will seem to follow this advice on its own (without any filters) - What’s interesting is that no one specifically programs it this way, putting instructions (algorithm) into it to produce exactly these combinations of numbers.

Don't believe me? Check it out for yourself!

1. Review the draw history - most odd-even combinations will be 2-3, 3-2 (5 out of 36) and 3-3 (6 out of 45).
2. Take any random number generator, combinations - generate and write down the resulting combinations, then check.

Conclusion:

• Most likely, such advice is addressed to those who manually fill out tickets, without any software; even a simple random combination generator will follow this advice on its own.
• This advice is of little use to us, since the sector contains two-thirds of all combinations - not in roulette, because we play for dozens, where the chance is 1 in 3.
• This advice is suitable for lotteries that take place very rarely, although it will not help much.
• It is more correct to try to guess sectors 1-4, 4-1, and with fairly frequent circulations 5-0, 0-5 (we are waiting for the average period)

last known circulation No. 11366 from 2019-12-17 18:00:00. Numbers: [

, + ]. Sum of numbers = 79.

For each lottery participant the most important issue, when purchasing a ticket, is combination selection. Someone bets on their favorite numbers, someone chooses them using a random number generator (RNG) (and not necessarily using a computer one, each of us has our own RNG on our shoulders;)), .... All these methods make we need to think carefully, since abstract thinking with numbers is not the easiest thing to do. Vision- the strongest, most important and developed feeling in a person. Let's take advantage of this strength and visualize lottery combinations. This will greatly facilitate the understanding of the random process and help us make a choice.

Total for the lottery “Gosloto “5 out of 36” Maybe 376992 combinations. Let's number the possible lottery combinations as follows: principle is an ordered list, where each new combination differs from the previous one by adding 1 to the rightmost number. It turns out this sequence: (1) 1,2,3,4,5 (2) 1,2,3,4,6 (3) 1,2,3,4,7 () . . . (32) 1,2,3,4,36 (33) 1,2,3,5,6 (34) 1,2,3,5,7 (35) . . . (376992) 32,33,34,35,36 .

combination: !}

id: !}

Now let's look at the history of the latter 1000 circulations and for each of the combinations we will match the number we received - "id", it turns out scatter plot. In combinations with "id" >= 52361 not present - "1" -unit, in combinations c "id" >= 98737 There is not - "1" -units and "2" -ek, for combinations with "id" >= 139657 No "1" -units, "2" -ek and "3" -ek, etc... .

". Believe it or not, it's up to you! I’m not forcing you, let everyone have their own opinion. One of them imagines himself to be a supergenius, but in reality he doesn’t know how much 6x9 is. Another... I am writing an article for those who have not yet figured out this wonderful program, but are making every effort to win. But there is a lot to learn from practice. People think that they will press a button and millions will fall from the sky. No. This program is an analyzer. If you process the data ineptly, nothing will happen. But, since there is a reference point, so to speak, the forecast of numbers is based on it. That’s why there is an analysis of previous circulations, where, strictly speaking, . But I'll look at another real win at 5 out of 36:

Remember one statement for now: numbers may transfer to a new circulation from the last five, but the entire circulation does not! If you, of course, paid attention to this (if you didn’t pay attention, then sorry, THIS PROGRAM IS NOT FOR YOU), in particular in “Gosloto 6 out of 45” - this often happens. That the numbers are repeated 2-3 times, the same. And there is an “increase” or “decrease” of +/- 1 sign. There is such a thing. Let's assume that the number "6" has passed. This means that there will probably be “6”, but perhaps there will be either “5” or “7”!? Yes. Thanks to the program and one of the game methods, this option can be calculated, minimizing the headache with calculations. We considered one of the options, the so-called “growth-decline”. The method in the program is called “ in trend". (We'll get the numbers to generate later)

But there are also other numbers that need to be excluded? Right. Let’s put “in trend” on the shelf for now and look at the exceptions. In the program, one of the game techniques is called “ exclude". Everything is simple here. We take (I won’t say exactly how much, you need to look at the numbers, old and not repeated. In old editions, as soon as a “complete replacement” occurs, then up to this point, if there were no such numbers. Usually it is 3,4,5,6, 7 circulations.) We copy them, the numbers of ALL THESE circulations and paste them into the field for calculation. We generate. We get numbers that theoretically cannot exist. And yet Gosloto deliberately rigs them so that (someone as smart as you won’t win, knowing this). Remember what I wrote in bold: numbers can change... but now we have excluded them with you. We got a certain result and copied it “to notepad” - right in the program, before switching to a new method.

Let's return to the method " in trend". We take the last 5 draws. Paste it into the field. We generate. We get the growth and decline of numbers. The so-called "side numbers". Why do we need them? These are the very numbers - exceptions of exceptions. But in total, your result in this case will be equal to 10%! (2.3 numbers in total) Why? Because numbers that are repeated too often or are rare are deleted by the program on purpose. In the method, “exclude” all paired or more numbers, and then “side” numbers are built for them if there is growth. For example: there are numbers 1,2,2,4. 2ka rises, and 4ka falls - that means the number will be 3! Do you know what I mean!? This is the essence of the calculation of this method. After all, here we are looking for growth and decline - and nothing more. Copy to notepad in the program.

Next, we copy from the notepad into the calculation field what we got in the first “exclude” method and the second “in trend” method into a pile. We generate. We get a set for generation: 25 5 7 26 27 10 11 1 3 2 4 9 6 Just in case, we copy them “to notepad” in the program. And we switch to a new method of play. (Taken from my logs in the program).

Now that we have numbers to generate, we can use one or the other method. The first is the “selection +/-”, which allows you to set which sign and how much to “move”. If it is +2 (in the first digit, where the number is 25) then we get: 23, 24, 25, 26, 27! We don't need this. After all our task is to reduce the calculated numbers for combinations. And so the easiest way to choose in this case is: "-1/+1" method. For example: If the first sign contains the number 25, then we get: 24, 25, 26. What we need so as not to be “scattered”, because we excluded some numbers ourselves in the literal sense of the word. On the right there is a self-control field “Super Prize Test” to check matches. And here is the result:

Analyzed numbers from circulation archives: 13
Approximate amount for bets: 19 (There may be more, so you need to minimize the numbers for the combinations - in the right way)
Number of bets spent: 19 (Expenses: 19×300 rubles = 5700 approximately, possibly more)
Numbers for combinations: 2 26 8 3 9 4 24 0 5 12 11 25 28 7 10 27 1 6 (why did “0” appear by +1/-1 1-1=0, this is what we get after “filtering” and “processing” using various methods)
Successful numbers, taking into account the processed numbers, became (in one of the generations):

2 4 6 10 26

We generate. The numbers “5 out of 36” came out in draw N1938 dated July 28, 2014: 2 4 6 10 26 the winnings were: 1 105 140 rubles!!! Congratulations to the winner! Thank you for having this program! You've done the numbers correctly.

Learn! Unfortunately, more and more people began to use the program and note that they began to win 1-2 million in state lotto several times a month. We won the largest super prize so far - 10 million! (We played with extended bets, but it justified itself completely, since all the calculations were made using this program. We spent about 300,000 rubles. There was an “investor” in the project!) I am sure that you yourself will be able to pick up and discover your own secrets of state lotto , use various methods and their combinations for filtering and analysis. It’s not stupid to hit the button and then scream. I wish you good luck!

PS: You can test the program using the Mozilla Firefox browser. Testing time and some functionality are limited. ADVICE: Use first one, then the other, then the first method again - process the numbers “in the right way” and apply your head to everything, of course. If something is not clear, re-read this article again. And until you learn, my advice is not to play for money. We also have such comrades.

Dear Gosloto 5 of 36 players, we offer you quick access to historical data on all past draws. This data is indispensable for collecting statistics, for example, about which numbers appear more often and which ones appear less often. You will receive results for any circulation in a convenient visual form. In addition, you do not need to go through different pages of the site in search of the necessary information - all data will be available from this page. All you have to do is enter the number of the desired circulation. Simply put, if you are developing your own lottery system, then we provide you with a convenient tool for this.

Obtaining the Gosloto 5 out of 36 draw table

All you have to do is enter the circulation number in the form on the page. Next, click on the “Get circulation table” button and see the result. The Fives draw table, consisting of the numbers of the five drawn balls, will be displayed directly below the form.

If you do not know the circulation number, but you have a ticket, then the number will be written on the ticket. You only need to enter the number itself, consisting of numbers, without spaces or other extraneous characters. Please note that the table will be available only after the draw, but not in advance! We wish you good luck in building your own profitable system for playing Gosloto 5 out of 36!

Is it possible to win the lottery? What are the chances of matching the required number of numbers and winning the jackpot or junior category prize? The probability of winning is easy to calculate; anyone can do it themselves.

How is the probability of winning the lottery generally calculated?

Numerical lotteries are conducted according to certain formulas and the chances of each event (winning a particular category) are calculated mathematically. Moreover, this probability is calculated for any desired value, be it “5 out of 36”, “6 out of 45”, or “7 out of 49” and it does not change, since it depends only on the total number of numbers (balls, numbers) and the fact how many of them need to be guessed.

For example, for the “5 out of 36” lottery the probabilities are always as follows

• guess two numbers - 1:8
• guess three numbers - 1:81
• guess four numbers - 1: 2,432
• guess five numbers - 1: 376,992

In other words, if you mark one combination (5 numbers) on a ticket, then the chance of guessing “two” is only 1 in 8. But catching “five” numbers is much more difficult, this is already 1 chance in 376,992. This is exactly the number (376 thousand) There are all possible combinations in the “5 out of 36” lottery and you are guaranteed to win it if you only fill them all. True, the amount of winnings in this case will not justify the investment: if a ticket costs 80 rubles, then marking all the combinations will cost 30,159,360 rubles. The jackpot is usually much smaller.

In general, all probabilities have long been known, all that remains is to find them or calculate them yourself, using the appropriate formulas.

For those who are too lazy to look, we present the winning probabilities for the main Stoloto numerical lotteries - they are presented in this table

Necessary clarifications

The lotto widget allows you to calculate the probabilities of winning for lotteries with one lottery machine (without bonus balls) or two lottery machines. You can also calculate the probabilities of deployed bets

Probability calculation for lotteries with one lottery machine (without bonus balls)

Only the first two fields are used, in which the numerical formula of the lottery is used, for example: - “5 out of 36”, “6 out of 45”, “7 out of 49”. In principle, you can calculate almost any world lottery. There are only two restrictions: the first value should not exceed 30, and the second - 99.

If the lottery does not use additional numbers*, then after selecting a numerical formula, all you have to do is click the calculate button and the result is ready. It doesn’t matter what probability of an event you want to know - winning a jackpot, a second/third category prize, or just finding out whether it is difficult to guess 2-3 numbers out of the required number - the result is calculated almost instantly!

Calculation example. The chance of guessing 5 out of 36 is 1 in 376,992

Examples. Probabilities of winning the main prize for lotteries:
“5 out of 36” (Gosloto, Russia) – 1:376 922
“6 out of 45” (Gosloto, Russia; Saturday Lotto, Australia; Lotto, Austria) - 1:8 145 060
“6 out of 49” (Sportloto, Russia; La Primitiva, Spain; Lotto 6/49, Canada) - 1:13 983 816
“6 out of 52” (Super Loto, Ukraine; Illinois Lotto, USA; Mega TOTO, Malaysia) - 1:20 358 520
“7 out of 49” (Gosloto, Russia; Lotto Max, Canada) - 1:85 900 584

Lotteries with two lottery machines (+ bonus ball)

If the lottery uses two lottery machines, then all 4 fields must be filled in for calculation. In the first two - the numerical formula of the lottery (5 out of 36, 6 out of 45, etc.), in the third and fourth fields the number of bonus balls is indicated (x out of n). Important: this calculation can only be used for lotteries with two lottery machines. If the bonus ball is taken from the main lottery machine, then the probability of winning in this particular category is calculated differently.

* Since when using two lottery machines, the chance of winning is calculated by multiplying the probabilities by each other, then for the correct calculation of lotteries with one lottery machine, the choice of an additional number by default is 1 out of 1, that is, it is not taken into account.

Examples. Probabilities of winning the main prize for lotteries:
“5 out of 36 + 1 out of 4” (Gosloto, Russia) – 1:1 507 978
“4 out of 20 + 4 out of 20” (Gosloto, Russia) – 1:23 474 025
“6 out of 42 + 1 out of 10” (Megalot, Ukraine) – 1:52 457 860
“5 out of 50 + 2 out of 10” (EuroJackpot) – 1:95 344 200
“5 out of 69 + 1 out of 26” (Powerball, USA) - 1: 292,201,338

Example calculation. The chance of guessing 4 out of 20 twice (in two fields) is 1 in 23,474,025

A good illustration of the complexity of playing with two lottery machines is the Gosloto 4 out of 20 lottery. The probability of guessing 4 numbers out of 20 in one field is quite gentle, the chance of this is 1 in 4,845. But when you need to guess correctly and win both fields... then the probability is calculated by multiplying them. That is, in this case, we multiply 4,845 by 4,845, which gives 23,474,025. So, the simplicity of this lottery is deceptive; winning the main prize in it is more difficult than in “6 out of 45” or “6 out of 49”

Probability calculation (expanded bets)

In this case, the probability of winning when using expanded bets is calculated. For example, if there are 6 out of 45 in the lottery, mark 8 numbers, then the probability of winning the main prize (6 out of 45) will be 1 chance in 290,895. Whether to use expanded bets is up to you. Taking into account the fact that their cost is very high (in this case, 8 marked numbers are 28 options), it is worth knowing how this increases the chances of winning. Moreover, it is now very easy to do this!

Calculation of the probability of winning (6 out of 45) using the example of an expanded bet (8 numbers marked)

And other possibilities

Using our widget, you can calculate the probability of winning in bingo lotteries, for example, Russian Lotto. The main thing that needs to be taken into account is the number of moves allocated for the onset of winning. To make it clearer: for a long time in the Russian Lotto lottery, the jackpot could be won if 15 numbers ( in one field) closed in 15 moves. The probability of such an event is absolutely fantastic, 1 chance in 45,795,673,964,460,800 (you can check and get this value yourself). This is why, by the way, for many years in the Russian Lotto lottery no one could hit the jackpot, and it was distributed forcibly.

On March 20, 2016, the rules of the Russian Lotto lottery were changed. The jackpot can now be won if 15 numbers (out of 30) were closed in 15 moves. It turns out to be an analogue of an expanded bet - after all, 15 numbers are guessed out of 30 available! And this is a completely different possibility:

Chance to win the jackpot (according to new rules) in the Russian Lotto lottery

And in conclusion, we present the probability of winning in lotteries using a bonus ball from the main lottery drum (our widget does not count such values). Of the most famous

Sportsloto “6 out of 49”(Gosloto, Russia), La Primitiva “6 out of 49” (Spain)
Category "5 + bonus ball": probability 1:2 330 636

SuperEnalotto "6 out of 90"(Italy)
Category "5 + bonus ball": probability 1:103,769,105

Oz Lotto "7 out of 45"(Australia)
Category "6 + bonus ball": probability 1:3 241 401
“5 + 1” – probability 1:29,602
“3 +1” – probability 1:87

Lotto "6 out of 59"(United Kingdom)
Category "5 + 1 bonus ball": probability 1:7 509 579