Игровые будни - рулетка

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Евгений CMD

Евгений CMD

Свидетель Заноса
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Окей, буду писать недельки две, в других местах. Там ссылки разрешают ;-)
И не грозят постерам ;-)
Довольны?! :)
До встречи, через две недели :)
не угадал ты ни капли, родной.
Идешь в банду " Еблaннедов" Причем не на 2 недельки - а на ПЖ.
Ты не постер, ты фуфломет.
Вбанен!
 
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astin67

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Всё пропало! Теперь как узнать название проги?
А это система, которая дает "прогноз"

 
Лудобот

Лудобот

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а я только ему репутацию вернул....
 
Shpilevoy

Shpilevoy

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не угадал ты ни капли, родной.
Идешь в банду " Еблaннедов" Причем не на 2 недельки - а на ПЖ.
Ты не постер, ты фуфломет.
Вбанен!
фу как грубо....... под такую формулировку попадает каждый второй

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на первый-второй расчитайсь! чур я первый буду :vdv:

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не закрывать же теперь тему про рулетку ))))

мы ломаем голову над пониманием вот этой портянки:

1 Introduction The game is defined by a list of payouts u1, u2,. . . , u` , and a list of probabilities p1, p2, . . . , p` , P ` i=1 pi = 1. We allow ui to be rational numbers, not just integers, to include games like blackjack, n-play video poker or the banker bet in baccarat. We assume that the casino has the advantage, so P ` i=1 piui < 0. The player can bet any positive integer k up to his current bankroll or the maximum bet b, whichever is smaller, and his bankroll increases by uik with probability pi . To cater for blackjack or poker type games, only the player’s initial bet is limited, and we allow the player to borrow money for splits, doubles or raises if necessary, but he has to stop if he loses and ends up with a negative bankroll. Each game is independent of all others. The player has a phantom bonus m without wagering requirements. At any point he can decide to cash in, if his bankroll is greater than m, then m is deducted from his balance in and the bonus is lost, if his bankroll is m or less, he forfeits his whole balance. 2 The question If the player’s current bankroll is n, what strategy should he follow to maximize the expectation of the real money he can cash in after deducting the bonus, and what is this expectation an? n can be a rational number whose denominator is the least common multiple of the denominators of the ui , for example, when dealing with blackjack, 1 n can be a half integer, but the stake is always an integer and let us also define an = n for n < 0. Define the value of the phantom bonus to be an − max(0, n − m), this is the amount the player expects to gain by playing optimally instead of cashing in immediately. In games involving an element of skill, we assume that the player is playing a fixed strategy, possible adjustments to playing strategy in view of the changed expected values are not considered, strategy will only mean betting strategy. 3 The solution In theory, there is an easy way to calculate to an. Start with an,0 = n for n < 0, an,0 = 0 for 0 ≤ n ≤ m, and an,0 = n − m for n > m. Then for each n, calculate P ` i=1 pian+uik,0 for all integers k, 1 ≤ k ≤ min(n, b), and let an,1 be the maximum of these values and an,0. We are calculating the optimal strategy using an,0 as approximation to an. Repeat with an,1, and so on. In general, let an,j+1 = max³ an,j , nX ` i=1 pian+uik,j |1 ≤ k ≤ min(n, b) o´. an,j ≤ n for all j, since the casino has the advantage, so the expected value of the player’s cash-in cannot exceed current bankroll, even ignoring the bonus. an,j ≤ an,j+1, and it is a well-known theorem in analysis that a bounded monotonically increasing sequence has a limit, so an = lim j→∞ an,j exists. Unfortunately, this procedure as described involves an infinite amount of calculation. I conjecture that the phantom bonus is worthless, i.e., an = n − m if n ≥ N = m P ui<0 piui P ` i=1 piui , (1) and that if the player plays a game in which all the payouts are integers ≥ −1 and b ≥ N, then the optimal strategy is simply to bet everything if n < N and to cash in if n ≥ N. Under these conditions N is exactly the point at 2 which the expected value of betting everything is the same as that of cashing in. If the game only has two possible outcomes, u1 = −1 with probability 1 − p and u2 = u, a positive integer, with probability p, then an can be calculated by finding the integer r such that N/(u+ 1)r ≤ n < N/(u+ 1)r−1 , and then an = p r ((u + 1)rn − m), so an is a piecewise linear function on n. I cannot prove all this, but I have quite good numerical evidence. The idea for the proof I have depends on some kind of convexity of the sequence an. (Convexity means an ≤ (an−1 + an+1)/2, in other words, that the player should always take a bet with no house edge. I would only need something weaker, that an/n increases with n, and an+k ≤ an + k, the latter meaning that value of the phantom bonus decreases as n increases.) If the conjecture is true, then the calculations can be restricted to n < N, which makes each iterative step finite. Another approach to the problem is that if the optimal strategy is known for each n, then we can write down the linear equations relating the an and solve them, this gives exact answers, unfortunately the optimal strategy is not known in advance. The method described below is a combination of iterations and linear equations. Step 1. Calculate N by (1). Set an = n for n < 0, an = 0 for 0 ≤ n ≤ m, and an = n − m for m < n ≤ N. Step 2. Check for each n, 0 < n ≤ N, whether betting the maximum possible amount improves the current value of an, and if so, update the value of an. Do this 10, 20 or 50 times, the exact number does not seem to make much difference in the running time. For some reason I do not understand, going in decreasing order of n gives faster convergence than in increasing order. Step 3. Calculate the optimal strategy using the current values for an, and solve the resulting linear equations. This is equivalent to using the currently optimal strategy and letting it converge. This is the slowest step because it involves solving a large, but sparse system of linear equations, usually consisting of several hundred equations. Step 4. For each n, calculate the expected value of betting any of possible bet sizes using the current values of an. If this does not give an improvement for any n, then the current numbers are the exact values of an, and we can also deduce the optimal strategy. If improvement is possible, then go back to Step 2. The calculations were carried out using Mathematica, which can handle 3 rational numbers with arbitrary numerators and denominators, so it can give exact answers. These exact answers can be used to justify the use of the unproven conjecture retrospectively. This method produced results reasonably quickly for all the games are I looked at. 4 The results The tables in the Appendix show the results for various games. The size of the phantom bonus was always taken to be m = 10. The first column shows the maximum bet, the second column the largest bankroll at which it is still in the player’s interest to play, rather than to cash in. The remaining columns show the value of the phantom bonus, an − max(0, n − m), from n = 5 to 40 in steps of 5. These numbers were rounded to 3 decimal places. The probabilities for blackjack were taken from a simulation by Michael Shackleford, the “Wizard of Odds”, available on the web at . I believe that the problem scales linearly as long as the payouts are integers, so choosing m = 10 is not a real restriction. The tables confirm some of the expected phenomena. The value of the phantom bonus, and also the size of the bankroll at which the player should stop playing both increase with b. The value of the phantom bonus is maximal at n = m. The house edge is not the primary factor in determining the value of the phantom bonus, a large probability of losing and a small probability of winning a large amount is the good for the player. Among all the roulette bets, which all have the same house edge, betting on a single number is the best. Similarly, Jacks or Better video poker with doubling is better than without doubling, while the house edge is the same. The line bet (6 numbers) in roulette with house edge 2.7% is better for the player than coin flip with probability of winning 0.495 and house edge 1%. The numbers for coin flip with probability of winning 0.499 and house edge 0.2% are similar to those for Jacks or Better video poker with house edge 0.46%. 5 Details of the strategy Assume that all the payouts are integers ≥ −1. Empirical evidence suggests and I may be able to prove it rigorously, too, that if player’s bankroll is a 4 multiple of b, he should always bet the maximum, b. In this case the numbers ajb (j = 0, 1, 2, . . .) satisfy the recurrence relation ajb = P ` i=1 pia(j+ui)b . Let u = max{ui |1 ≤ i ≤ `}. The solution is of the form ajb = P ` i=0 ciλ j i , where the λi , i = 0, 1, . . . , u, are the roots of the characteristic equation 1 = P ` i=1 piλ ui , assuming that the roots are distinct. Let kb be the highest multiple of b at which the player should still play. The coefficients ci , i = 0, 1, . . . , u, can be determined from the equations a0 = 0, a(k+1)b = (k+1)b−m, a(k+2)b = (k+2)b−m,. . . , a(k+u)b = (k+u)b−m. Define αjb,t by αjb,t = P ` i=1 piα(j+ui)b,t and α0,t = 0, α(t+1)b,t = (t + 1)b − m, α(t+2)b,t = (t + 2)b − m,. . . , α(t+u)b,t = (t + u)b − m. These are the values of the ajb using the assumption that k = t. k can be found as the largest value of t for which αtb,t ≥ tb − m, but unfortunately, there is no exact method of solving for k. Example 1. Coin flip with the probability of winning p The characteristic equation is 1 = (1 − p)λ −1 + pλ, the roots are λ0 = 1, λ1 = (1 − p)/p, and the solution is αjb,t = (((1 − p)/p) j − 1)(tb − m) ((1 − p)/p) t+1 − 1 . Now let m = b = 10, p = 0.495. The table below shows the value of αtb,t calculated using the above formula for various values of t. t 1 2 3 4 5 6 7 8 9 10 tb − m 0 10 20 30 40 50 60 70 80 90 αtb,t 4.95 13.2 22.3 31.7 41.2 50.9 60.6 70.4 80.2 89.97 For t = 9, αtb,t > tb − m, for t = 10, αtb < tb − m, so k = 9. This means that the player should still play with a bankroll of 90, but not with a bankroll of 100, which agrees with the table in the appendix, which says that largest bankroll at which the player should still play is 98. Example 2. Betting on a single number in roulette The characteristic equation is 1 = 36λ −1/37 + λ 35/37, it cannot be solved 5 exactly, but Mathematica is able to provide numerical solutions. Let m = 10, b = 50, the results are in the table below. t 1 2 3 4 5 6 7 tb − m 40 90 140 190 240 290 340 αtb,t 48.4 96.8 145.3 193.8 242.3 290.9 339.5 For t = 6, αtb > tb − m, for t = 7, αtb,t < tb − m, so k = 6. This means that the player should still play with a bankroll of 300, but not with a bankroll of 350, which agrees with the table in the appendix, which says that largest bankroll at which the player should still play is 332. There are some curious phenomena that I do not understand. The table below shows some values of n and the corresponding optimal bet size when betting on a single number in roulette with m = b = 10. If the last digit of n is 7, 8 or 9, then it is never correct to bet 10, but only 7, 8, or 9, respectively. There are many other numbers for which betting the maximum is not correct, in all of these cases the correct bet size is the last digit of n. n 1 2 3 4 5 6 7 8 9 10 bet 1 2 3 4 5 6 7 8 9 10 n 11 12 13 14 15 16 17 18 19 20 bet 10 10 10 10 10 10 7 8 9 10 n 101 102 103 104 105 106 107 108 109 110 bet 10 10 10 10 10 6 7 8 9 10 n 141 142 143 144 145 146 147 148 149 150 bet 10 10 10 10 5 6 7 8 9 10 n 171 172 173 174 175 176 177 178 179 180 bet 10 10 10 4 5 6 7 8 9 10 n 201 202 203 204 205 206 207 208 209 210 bet 10 10 3 4 5 6 7 8 9 10 n 221 222 223 224 225 226 227 228 229 230 bet 10 2 3 4 5 6 7 8 9 10 n 241 242 243 244 245 246 247 248 249 250 bet 1 2 3 4 5 6 7 8 9 10 If we consider the strategy for the split bet with the same m and b, then this phenomenon does not start until much later, the player should bet the maximum possible for all n ≤ 157. 6 I can only guess at the reasons. It seems that multiples are b are somehow preferable, so when the optimal bet is not b, it is the difference between n and the largest multiple of b less than n, so that if the player loses, his bankroll will be a multiple of b and from that point on he will always bet b. This phenomenon also occurs in other games especially near the upper bound. The reason why 7, 8 and 9 are different in the first example seems to be that the maximum bankroll at which the player should still play in the first example is 252, and winning bet with a stake of 7 or more would get him to this bound or above it. In the second example, the maximum bankroll at which the player should still play is 198, this cannot be reached by betting 10 or less, this is why the optimal strategy behaves differently. Other examples also support this, but I have no rigorous explanation for this observation, I can only speculate. 6 The effects of using the wrong strategy The traditional wisdom about phantom bonuses is that the player should bet big, but the previous section shows that betting the maximum is not always correct. I also calculated the expectations if the player only bet min(n, b) and the difference from the an was very small, typically only a few thousandths or even less, so for practical purposes always betting the maximum possible is a good strategy. I also considered what happens if the player chooses a different target, for example he gets a 100% match bonus on his deposit and aims to increase his bankroll tenfold, so if m = 10 as in the previous calculations, he aims for 200. The typical situation is that if the optimal strategy suggests that he should aim higher, say, for 400, he does not gives up much in terms of expectation by stopping at 200. On the other hand, if the optimal strategy suggests stopping earlier, say, at 100, then trying to go for 200 can be expensive.

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формулы поломаны, но хотя бы суть изложенного кто-то подскажет? :neo::neo::neo:
 
astin67

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Перевёл - ни чё не понятно. При чём тут блекджек и покер?
 
Владимир

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не в обиду, но вы ребята, помешанные на рулетке немного не от мира сего:fool: в хорошем смысле слова:D
 
astin67

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Я вот счас играю в турнире по слотам, и у меня нет ни какого желания играть на деньги в слоты . Хотя вижу , что я в плюсе . Есть лудомания , а я АНТИЛУДОМАН.
 
Shpilevoy

Shpilevoy

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у нас тута тайные учителя завелись ))))))

2016-09-01_201240.jpg
 
Shpilevoy

Shpilevoy

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а че...... работает )))))
правда не мартин

2016-07-22_183710.jpg


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2016-07-23_010252.jpg
2016-07-23_012014.jpg
 
Franky

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Хто здесь? Так и барабан сломать можно!
Играл "по системе" и выиграл авто-мобиль! Но сикретам ни падилюс!
 
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Александра

Александра

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Хто здесь? Так и барабан сломать можно!
Играл "по системе" и выиграл авто-мобиль! Но сикретам ни падилюс!
Дима , недавно зашла в блэк-джек (amatic) , пыталась играть по тем табличкам которые ты мне давал:)
Итог : -800 руб :(
Ты мне ещё чего-то не рассказал???
 
Александра

Александра

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Аматик дно. Микро и нетент пободрее БЖ держат
Буду знать!!!
Я вообще сначала хотела зайти в лайф , но столы почему-то были грубые:( ( мин.ставка от 500руб.) :(
 
Александра

Александра

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Аматик дно. Микро и нетент пободрее БЖ держат
Даня ты кстати видел что у нас в городе открыли слоты оффлайн? Кстати официально:) целый клуб с огромной рекламой:)
 
Shpilevoy

Shpilevoy

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если это слотоматы то забудь
30-40% пейаут заложенный в билеты

бывают чудеса
с 1000 на 7000 выпрыгивают
но это слишком редко (в психозе чел резко подымает линии и ставки и угадывает момент)
 
Xoomax

Xoomax

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Играю каждый день. Об этом и буду писать.
Играю. Новый подход, пока без названия (Плазма, Капли дождя, Лото - такие варианты). Идея, внутри - частью из Тараканьих бегов, другой частью - из системы Хорватия. Сыграно, и выведено - 4 захода, плюс. Микрогейминг играет отлично, рулетка Сильвер - Исофтбет - через раз, поэтому пока пауза, для данного софта.
 
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