What is the longest number of spins without a repeater?

[RouletteFanatic]

I heard no one has ever had 36 unique spins in 36 spins. So what is the longest anyone heard or have personally encountered before a repeater hits?

[FreeRoulette]

I tried a few tests. The longests one I saw was 21, but mostly a repeat happened within 10-15 spins.

You can test it out here.
nolinks://nolinks.loothog.com/Systems/PHP/coldNumbers.php

[bombus]

I've seen 22.

[VLS]

Perhaps change the question from "what's the longest" to what's the most common.

You may actually do better by focusing on the "belly" of the distribution curve than trying to put limits on the extremes and pumping more and more money on the progressions to counter the ever-increasing deviations.

There have to be a point you have to let random "walk", a "let go" point. (It is better to have it based on knowledge for the distribution of your target selection than having your let go point being simply when all your bankroll's money has been exhausted --In my humble opinion).

I do hope you achieve a suitable strategy for repeaters that works for you.

Kind regards.

[Steve]

It's not uncommon to have sleepers for 200-300 or more spins. The question you should be asking is "Does it help knowing this?"

[Davey-Jones]

seven

[bombus]

seven

Bullshit Davey!

wtf is seven?

[bombus]

Perhaps change the question from "what's the longest" to what's the most common.

You may actually do better by focusing on the "belly" of the distribution curve than trying to put limits on the extremes and pumping more and more money on the progressions to counter the ever-increasing deviations.

There have to be a point you have to let random "walk", a "let go" point. (It is better to have it based on knowledge for the distribution of your target selection than having your let go point being simply when all your bankroll's money has been exhausted --In my humble opinion).

I do hope you achieve a suitable strategy for repeaters that works for you.

Kind regards.

And what would that most common be?

By most common do you mean the average?

What would the "belly" be?

I'd guess the "belly" is about 16.

Cheers.

[Davey-Jones]

Bullshit Davey!

uh... yeah!

[Carpanta]

seven

Average for first repeater is 7, 30542 spin, so something around seven, yes!!!! Take into account you have a first number due to first spin, then 1 + 7,3 spins = 8,3 spin average to have first repetition.

Here you have the equations in spanish if you dont mind (I do not feel like translating, sorry)

The font is from forum grupojoker:

• Contaremos las tiradas a partir del primer número salido.

• Probabilidad de que el primer número salido se repita en la 1a tirada:

      p1 = 1/37 = 0,027027

• Probabilidad de que haya repetición de uno de los dos primeros números en la 2ª tirada:

      p2 = (36/37) x (2/37) = 0,0525931

• Probabilidad de que haya repetición de uno de los tres primeros en la 3ª:

      p3 = (36/37) x (35/37) x (3/37) = 0,0746253

• Probabilidad de repetición en la 4ª:

      p4 = (36/37) x (35/37) x (34/37) x (4/37) = 0,0914329

• Probabilidad repetición en la 5ª:

      p5 = (36/37) x (35/37) x (34/37) x (33/37) x (5/37) = 0,1019353
...
• Y, probabilidad de repetición en la jugada n:

      p(n) = (37!/(37-n)!) x (1/37) x n

Si ahora hacemos para obtener el número medio de tiradas:

      M = p1 x 1 + p2 x 2 + p3 x 3 + p4 x 4 +...+ (37!/37) x 36 x 36

Y el resultado es: M = 7,30542

Lo que quiere decir que: 7,3 es el número medio de tiradas que se darán para que se produzca la primera repetición de un número a partir del primero. Es decir, que después de que se produzca una primera repetición, deberán pasar 1 + 7,3 = 8,3 tiradas de media para que se produzca la siguiente primera repetición. Por tanto es 1ª repetición cada 8,3 tiradas.

Por ejemplo, la permanencia: a1, a2, a3, a4, a5, a3, a6, a7, a2, a9, a4, a8, a10, a11, a9

La descomponemos en dos secuencias, quedando:

      a1, a2, a3, a4, a5, a3

Aquí se produce una repetición, el a3. Ahora empezamos otra nueva.

      a6, a7, a2, a9, a4, a8, a10, a11, a9

Y aquí otra repetición, el a9.

Nótese que en la segunda permanencia tenemos a2 y a4, que son repeticiones, pero de números de la secuencia anterior, que ya está cerrada.

Lo que nos dice el cálculo es que, a partir del primer número, el a6 de la segunda permanencia, aparecerá una repetición en 7,3 tiradas en media, en este caso el a9. O dicho de otra manera, que desde la última repetición, el a3 de la primera permanencia, han de pasar 8,3 tiradas, en media, hasta que se produzca la siguiente repetición, el a9 de la segunda permanencia.

[NS]

Perhaps change the question from "what's the longest" to what's the most common.

You may actually do better by focusing on the "belly" of the distribution curve than trying to put limits on the extremes and pumping more and more money on the progressions to counter the ever-increasing deviations.

This is the best answer to these questions which never stop popping up in roulette forums.

The best because it should lead to investigating and understanding the normal distribution curve in more depth, which, once done, leads to understanding why there is absolutely no way to beat the mathematical model of roulette.

[Carpanta]

And here you have averages for a whole rotation of 37 numbers. It is what could be expected on average terms:

▪ 23,574 show once or more times
▪ 9,776 show twice or more times
▪ 0,632 show 4 or more times

This numbers have been the result of a study by Mark Serra who produced his famous System Roul2 for repeaters.

[Bayes]

Here's some empirical evidence for the theory. Based on a 1M trials. The left column is how long you had to wait for a repeat, and the right column tells you how many times that wait occurred. The average is 8.3 (exactly as theory says), and the most common is 6. In this particular sample the longest before a repeat was 27 spins.

WAIT   FREQUENCY

1         3241
2         6452
3         8923
4         10974
5         12358
6         12671 *
7         12440
8         11614
9         10095
10       8454
11       6914
12       5349
13       3771
14       2659
15       1818
16       1161
17       695
18       413
19       203
20       109
21       44
22       21
23       12
24       7
25       3
26       1
27       1
28       0
29       0
30       0
31       0
32       0
33       0
34       0
35       0

[FreeRoulette]

Lets say on average a repeat happens by 7 spins.  It would cost $28 to cover it and the minimum return, if you won on spin 7 would be $8.  Could you win enough times to make it profitable?

1. bet 1 on 1 numbers
2. bet 1 on 2 numbers
3. bet 1 on 3 numbers
4. bet 1 on 4 numbers
5. bet 1 on 5 numbers
6. bet 1 on 6 numbers
7. bet 1 on 7 numbers

[FreeRoulette]

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