Roulette and Probability Theory

It is especially important to understand that the probabilities depend on the results of previous draws, and not the following results themselves. This is the main difference between a Player with a capital letter (someone for whom the game is a way to earn money, a profession) and a simple player. The player bets on probabilities, and understands that the result does not depend on either his previous bets or previous results. Only the probability of the next result depends on previous results. In other words, they understand the difference between theory and practice: probability is theory, the result is practice. Highlight 6: A small explanation - n

The subject of probability theory is the theoretical study of such experiments in which, under the same conditions, mutually exclusive events may occur. Under the same conditions. Are the conditions the same in Internet roulette? No, the number that appears is strictly determined by the algorithm and settings of the pseudo-random number generator. There is a generated sequence of n numbers, and 2 spins in a row are just the values of DIFFERENT sequence members with numbers m and m+1. Even if 2 Zeros appear in a row, these are EQUAL values of DIFFERENT sequence members. Are the conditions the same in real roulette? No, the number that appears is strictly determined by the physical properties of the roulette wheel and ball, the psychophysiology, motor skills, skills, etc. of the croupier. So they try to measure the current in kilograms, mix red with sour. The connections between such phenomena are studied by a completely different science - mathematical statistics. A textbook example: in the 19th century in England, a connection was established between the milk yield of cows and the number of old maids in a given area. It turned out that old maids have many cats that eat mice and rats. More wild bees remain in this area (mice and rats destroy their swarms), clover is better pollinated, crops are larger, cows are better fed and give more milk. The formulations of independent events are especially touching. There are no such events in probability theory. These pseudo-specialists confuse the independence of random variables and mutually exclusive events (the occurrence of numbers in each spin are mutually exclusive events, but all events - the occurrence of ANY number from 0 to 36 in one spin relate to the distribution of ONE random variable). Obviously, there is no correspondence between the arrangement of numbers on the table and on the wheel - on the wheel between 0 and 1 there are 23 numbers, if you count clockwise, between 1 and 2 - 20 numbers, between 2 and 3 - 29, between 3 and 4 - 6. And on the table - they are next to each other. Due to the discrepancy in the arrangement on the table, there are only 4 red numbers in the middle column (by the way, there is such a paradox: we bet 10 units on red and 1 unit each on 4 black numbers 6, 15, 24, 33. From the point of view of probability theory, this is equally probable as betting 5 units on the 3rd column and 1 unit each on the 10 remaining red numbers (we have covered all the red numbers and 4 black ones, i.e. the probability of success = 22/37). Why do we bet 14 units in the first case and 15 in the second? Or why are there only 8 even red numbers? Due to the discrepancy in the arrangement, it follows, for example, that a bet on 1, 2, 3 covers a larger arc of the wheel than a bet on 4, 5, 6 (see for yourself what the distance between the numbers is). It is important to understand that we are not betting on numbers, but on certain sectors of the wheel roulette. Based on the fact that the angular and linear velocities of the ball are many times greater than the speed of the roulette wheel, we can VERY roughly assume that the number that appears in each spin allegedly "has no memory" and does not depend on the previous spin. Before the new spin, the wheel stopped in a certain position relative to the dealer. What conclusion can we draw from this? The result of the next spin depends on the position in which the wheel stopped in the previous spin and the number of revolutions of the ball. We can assume that the ball made N full revolutions on a stationary wheel plus some part of a revolution. Then the wheel made M full revolutions plus some part of a revolution and stopped in a new position relative to the dealer. The same is true for an Internet wheel - can there really be an RNG algorithm that would correspond to the placement of numbers on a real roulette wheel? Look - 10 blacks in a row have appeared. The player has his last bet, he bet on red, doubling. Should he continue betting on red, or bet on even? There is an obvious bias not only towards black, but also towards the lower half of the roulette wheel (the top number on the indicator is the last result, the second from the top is the penultimate... The red arrows show the placement of the results of previous spins. It is clear that the casino server in this game was guided not by the sequence generated by the RNG, but by the balance of this player. Or the RNG generates a sequence in which the distribution of numbers DIFFERENT from the distribution of numbers in real roulette.

Having a table of the odds of all the roulette elements falling out in front of you, you can easily calculate the odds of your bet falling out. Roulette element on which you are betting Probability of winning Payouts On six numbers 16.20% 5 to 1 On four numbers 10.80% 8 to 1 On three numbers 8.10% 11 to 1 On two numbers 5.40% 17 to 1 On one number (Straight-Up Bet) 2.70% 35 to 1 On a dozen or a column 32.40% 2 to 1 On Odds 48.6% 1 to 1 For example, if you bet on 10 numbers on European roulette, according to the table above, the odds of winning will be 27%. Knowing all the odds of the roulette elements falling out, you can calculate the win or loss of a roulette strategy or system.

5-5-10-5-5-10 we calculate the probabilities for practitioners and people who answer the question "what is the probability of a number coming up on roulette" with 1/37 "35 guessed numbers did not play even once during 6 spins in a row" probability (2/37)^6 for theorists and people who answer the question "what is the probability of a number coming up on roulette" with 100% "35 any numbers did not play even once during 6 spins in a row" probability 1* 1 *(2/37)^4 (the first is any, the second is any, including the same as the first) ///////////////// notice what difficulties people have who answer the question "what is the probability of a number coming up on roulette" with 100% //////////////////////////////// "will play" = "will win" even a fool can understand that "that they will play 6 spins in a row" there is no hint anywhere that "both at the same time" is absurd for roulette so there is nothing to talk about ////////////// I make decisions on bets on the nearest spins, so I am not interested in what can happen in 100-500 spins, so I calculate the probabilities here and now ////////////////// thanks 26o32 "If you close any 2 numbers on the field, what is the probability that they will play 6 spins in a row?" P=(2/37)^6 = 0.0000000249442 I agree accepted //////////////////////// for people who have "the probability of getting a number on roulette" = 100% let's put things in order in our heads P=(2/37)^6 is the probability for any two numbers that we decided to bet and it is exactly the same as for any ONE combination of two other numbers from all possible combinations and their permutations (in theory) DevilMayCry assumes that we cannot do without combinatorics let's calculate the proportion of suitable outcomes in their total mass and we will consider this the required probability we have 37 numbers a total of options for them to be laid out in 6 spins 37^6 = 2565726409 - the total mass of outcomes for 2 numbers in 6 spins 2^6 = 64 ways to be laid out -2 xxxxxx and yyyyyy in 37 numbers of combinations of 2 numbers = C372 = 37! / (2! * (37-2)!) = 13763753091226300000000000000000000000000000000.00 / (2 * 1033314796638610000000000000000000000000.00) = 666.00 total required outcomes 666 * 62 = 41292 in total mass 2565726409 = 41292 / 2565726409 = 0.0000160 /////////////////////////// the question now remains simple 0.0000000249442 * 666 = 0.0000166128 vs. 0.000016093 1) why are they not equal? Is it because of the excluded xxxxxx yyyyyy? 2) to reconcile practitioners and theorists, you need to calculate in a standard way without any fuss for 1 event and multiply by the number of acceptable combinations. 3) but why do we need this TOTAL coefficient? it is not suitable for assessing the risk of the bet; you can ride a four-wheeled bicycle, but why complicate it so much? I think that on roulette you need to think according to the principle of 1/37 and not 100% ///////////////////// combinatorics answered the question about all possible combinations of two numbers for 6 spins in a row, but it could not answer about the specific XXYXXY and double series, and through P everything is considered an element (there are no numbers with 50 significant before the decimal point) but the truth is for the next 6 spins, which is exactly what practitioners need, and not in general all possible for theorists

if we consider not millions of spins but 100-500 spins, then there is no pre-calculated average? is it changeable and moves chaotically during the game? and the average for the session in the first 100 spins will be completely different from the average for the second 100 spins? and even their average will be far from the average for the 3rd hundred spins? //////////// is the question clear or will we again find fault with every comma? /////////////////////////////////////////////////////// Yura: A pre-calculated average exists and it is called MO. But the practical one over such a short distance can shake quite a bit. //////////////////// I also came to the conclusion that for 100-200 spins the games are average abstractions that cannot be screwed on anywhere in any way. There is a clear individual pattern of the game. This pattern needs to be caught and continued until it runs out of steam.

Basically, any long-term flat roulette game leads to losses. It doesn't matter what strategy is used. However, for any spin statistics there is always(!) a winning way. Good players capture several opportunities: a combination of several strategies, attracting progression, stop-losses in profits and losses, strategies for ensuring profit, changing the game and much more. Can a set of these tools win in the long run or not? The basic empirical data, laws, strategies, combinations of bets and different tools are so complex that they cannot be identified using mathematical formulas. Only consideration in the context of specific data and conditions on a large sample of practical tests can give a convincing answer.