Roulette Machine Number Sequence

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As what we have said over and over again, roulette, whether you’re talking about live or RNG, is a random game. The wheels being used to play don’t have any capacity to remember the last outcome.

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• 1, 8, 27, 64, 125. 9, 73, 241, 561, 1081, 1849. Divergent sequences: 1, 2, 4, 8, 16, 32. 1, 2, 0, 3, -1, 4, -2. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55. 2, 3, 5, 7, 11, 13, 17, 19, 23. (click on sequence to compute difference table).
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• William Hill roulette machines and Bet Fred roulette machines use the same software, with some land bookies using a separate supplier for their roulette machines. This gives two distinct versions of FOBT (Fixed Odds Betting Terminal) software. As such, roulette machine cheats may work on some betting terminals but not others.

That is, for every event that the ball is spun, the outcome has no direct connection with what came out in the past. The silver edge review. Even more so, it won’t affect what will appear next time. https://loeternalpoker2wildjungle.peatix.com.

Roulette is a casino game named after the French word meaning little wheel.In the game, players may choose to place bets on either a single number, various groupings of numbers, the colors red or black, whether the number is odd or even, or if the numbers are high (19–36) or low (1–18). Rainbow riches online.

In the American roulette wheel, for example, number 17 has the probability of 1/38 (1 chance out of the 38 numbers) coming up for every spin. Suppose 17 has emerged as a winner for 5 consecutive times already. At the sixth round, that number will once again have a 1/38 probability of winning.

With that fact already established, let us look how it can be connected to the roulette number sequences and how you, as a player, can take advantage of this.

Repeating Numbers

Since all numbers have 1/38 odds of appearing each turn, then we can say that for a number to come up again and again won’t be a rare occurrence. Statistically though, a repeat is likely to happen after every 38 spins.

Number Sequences

Expert players usually do their bet placements on groups of adjacent numbers in the roulette wheelhead. The minds behind this game know this for sure. Which is why, they arranged or sequenced the number in a way that if the player is to place wagers on contiguous sectors (based on the roulette wheelhead), he might need to disperse his chips across the roulette betting layout.

Number Spans

Taking the American roulette wheel as basis again, most popular contiguous bet combinations work around the center column. If you take a closer look at it, this betting section features the span from numbers 23 to 5 with 0, 7, 9, 28, 30 in the gaps. Looking at the roulette wheelhead, this bet was able to cover a span of 15 numbers. This is actually 2/5 of the entire wheel perimeter.

Betting Types Based on This Number Sequence

To make use of this number sequence, a player may make use of the following betting combinations:

1. Column + Straight-Up

Player puts a straight-up bet on the numbers 0, 7, 9, 28, and 30, and then on the center column.

1. Column + Street

Player puts a street bet on the row of 7, 8, 9 and on 28, 29, 30. He would then mix it up with a center column.

1. Column Only

A player can definitely go right on the center column. He might have to hope that the ball won’t hit the gaps in between.

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Again, we will reiterate the fact that this there’s no possible way to influence the outcome of a spin, unless of course, it is being acted upon. Each result drawn out from a spin is pure random. These are only some of the best betting placements done by roulette professionals. While they don’t guarantee a sure win, they do, however, provide good coverage of the numbers found in the roulette betting layout. It’s always luck that we can depend upon.

Thread Rating:

OnceDear
Administrator
Hi,
In 1500 spins, there are 1499 possible pairs of spins
For any pair of spins, the chances of both coming up red is (18/38) * (18/38) or 0.2243767313 assuming double zero wheel.
So you can expect to see 2 consecutive reds 1499 * 0.2243767313 times or 336 times.
Of course a sequence of 3 reds is treated as two sequences of 2
If you want to work it out for sequences of 3, then.
In 1500 spins, there are 1498 possible triplets of spins
For any triplet of spins, the chances of all three coming up red is (18/38) * (18/38) * (18/38) or 0.10628371482 assuming double zero wheel.
So you can expect to see 3 consecutive reds 1498 * 0.10628371482 times or 159 times.
Of course a sequence of 5 reds is treated as three sequences of 3.
Meanwhile. Remember, you don't have a winning system. You never will. :o)

Hi,
I'm currently researching some roulette data. I would like to know how I can calculate the expected number of times a sequence of red comes up. So for example, in 1500 roulette spins, how often do I expect a sequence of 2 reds in a row to come up, how often do I expect a sequence of three reds in a row to come up and so on. Hope somebody can help me with that.
Cheers,
Peter

Don't worry about it, and move onto your next project. Your idea won't beat roulette.

I would like to know how I can calculate the expected number of times a sequence of red comes up.

it is easy.
just remember dealing with averages or expected numbers is not the same as the probability.
2 totally different animals
*****
for 3 or more run (in Excel for example)
parameters:
p=(18/38)
length=3
trials=1500
q=(20/38)
THE FORMULA (without proof - that is internet stuff)
=(p^length)*(1+((trials-length)*q))
for exactly length of 3
calculate 4 and 3
subtract 4 from 3
here is my Excel in Google if want to see (easy)
https://goo.gl/98yjKp

So for example, in 1500 roulette spins,
how often do I expect a sequence of 2 reds in a row to come up,
how often do I expect a sequence of three reds in a row to come up and so on.

here is a table of data.
I simulated this 1 million times (1 million sets of 1500 spins)
and calculated it also (rounded to 4 decimals)
super close I do say

.	sim	sim	sim	calc	calc	.
length	freq	exact 3	3 or more	expected #	exact 3	length
1	197014850	197.0149	374.1835	374.1856	197.0579	1
2	93283735	93.2837	177.1686	177.1277	93.2811	2
3	44168570	44.1686	83.8849	83.8467	44.1563	3
4	20912156	20.9122	39.7163	39.6903	20.9022	4
5	9898616	9.8986	18.8042	18.7881	9.8944	5
6	4689428	4.6894	8.9056	8.8937	4.6837	6
7	2217719	2.2177	4.2161	4.2100	2.2171	7
8	1050869	1.0509	1.9984	1.9929	1.0495	8
9	498950	0.4990	0.9475	0.9434	0.4968	9
10	236031	0.2360	0.4486	0.4466	0.2352	10
11	111977	0.1120	0.2126	0.2114	0.1113	11
12	53159	0.0532	0.1006	0.1001	0.0527	12
13	25012	0.0250	0.0474	0.0474	0.0249	13
14	11900	0.0119	0.0224	0.0224	0.0118	14
15	5520	0.0055	0.0105	0.0106	0.0056	15
16	2602	0.0026	0.0050	0.0050	0.0026	16
17	1232	0.0012	0.0024	0.0024	0.0013	17
18	613	0.0006	0.0012	0.0011	0.0006	18
19	272	0.0003	0.0005	0.0005	0.0003	19
20	140	0.0001	0.0003	0.0003	0.0001	20
21	59	0.0001	0.0001	0.0001	0.0001	21
22	30	0.0000	0.0001	0.0001	0.0000	22
23	20	0.0000	0.0000	0.0000	0.0000	23
24	11	0.0000	0.0000	0.0000	0.0000	24
25	5	0.0000	0.0000	0.0000	0.0000	25
26	5	0.0000	0.0000	0.0000	0.0000	26
27	0	0.0000	0.0000	0.0000	0.0000	27
28	0	0.0000	0.0000	0.0000	0.0000	28
29	1	0.0000	0.0000	0.0000	0.0000	29

have fun
hope this helps some(sum)

How Roulette Machines Work

SallyThanks all for the answers. Don't worry, I'm not looking to find a way to beat the wheel, I'm actually trying to show a sample size I have is random and I like to take a look at all kinds of different statistics from it. In this case the distribution of number of red numbers in a row.

I would like to know how I can calculate the expected number of times a sequence of red comes up.

it is easy.
just remember dealing with averages or expected numbers is not the same as the probability.
2 totally different animals
*****
for 3 or more run (in Excel for example)
parameters:
p=(18/38)
length=3
trials=1500
q=(20/38)
THE FORMULA (without proof - that is internet stuff)
=(p^length)*(1+((trials-length)*q))
for exactly length of 3
calculate 4 and 3
subtract 4 from 3
here is my Excel in Google if want to see (easy)
https://goo.gl/98yjKp

So for example, in 1500 roulette spins,
how often do I expect a sequence of 2 reds in a row to come up,
how often do I expect a sequence of three reds in a row to come up and so on.

here is a table of data.
I simulated this 1 million times (1 million sets of 1500 spins)
and calculated it also (rounded to 4 decimals)
super close I do say

.	sim	sim	sim	calc	calc	.
length	freq	exact 3	3 or more	expected #	exact 3	length
1	197014850	197.0149	374.1835	374.1856	197.0579	1
2	93283735	93.2837	177.1686	177.1277	93.2811	2
3	44168570	44.1686	83.8849	83.8467	44.1563	3
4	20912156	20.9122	39.7163	39.6903	20.9022	4
5	9898616	9.8986	18.8042	18.7881	9.8944	5
6	4689428	4.6894	8.9056	8.8937	4.6837	6
7	2217719	2.2177	4.2161	4.2100	2.2171	7
8	1050869	1.0509	1.9984	1.9929	1.0495	8
9	498950	0.4990	0.9475	0.9434	0.4968	9
10	236031	0.2360	0.4486	0.4466	0.2352	10
11	111977	0.1120	0.2126	0.2114	0.1113	11
12	53159	0.0532	0.1006	0.1001	0.0527	12
13	25012	0.0250	0.0474	0.0474	0.0249	13
14	11900	0.0119	0.0224	0.0224	0.0118	14
15	5520	0.0055	0.0105	0.0106	0.0056	15
16	2602	0.0026	0.0050	0.0050	0.0026	16
17	1232	0.0012	0.0024	0.0024	0.0013	17
18	613	0.0006	0.0012	0.0011	0.0006	18
19	272	0.0003	0.0005	0.0005	0.0003	19
20	140	0.0001	0.0003	0.0003	0.0001	20
21	59	0.0001	0.0001	0.0001	0.0001	21
22	30	0.0000	0.0001	0.0001	0.0000	22
23	20	0.0000	0.0000	0.0000	0.0000	23
24	11	0.0000	0.0000	0.0000	0.0000	24
25	5	0.0000	0.0000	0.0000	0.0000	25
26	5	0.0000	0.0000	0.0000	0.0000	26
27	0	0.0000	0.0000	0.0000	0.0000	27
28	0	0.0000	0.0000	0.0000	0.0000	28
29	1	0.0000	0.0000	0.0000	0.0000	29

have fun
hope this helps some(sum)

How Roulette Machines Work

SallyThanks all for the answers. Don't worry, I'm not looking to find a way to beat the wheel, I'm actually trying to show a sample size I have is random and I like to take a look at all kinds of different statistics from it. In this case the distribution of number of red numbers in a row.
I guess I wasn't totally clear: I'm looking for 'exactly two in a row', 'exactly three in a row', etc. So a series of four reds in a row will be counted as just that, four in a row. Looks like the best way is to start with 1500 in a row (in a sample size of 1500) and calculate 1499 in a row as suggested with substracting 1500 in a row and work my way down to one.
Or apparently using the below formula. Could you point me in the right direction for the proof? I haven't been able to find anything about this subject on the web, hence I got here.

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THE FORMULA (without proof - that is internet stuff)
=(p^length)*(1+((trials-length)*q))

Thanks,
Peter

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