Roulette Odds 4 Reds In A Row

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1. This is a question regarding fixed odds betting. If you say that the odds are 4 to 1 on something happening, then is that equivalent to saying the probability is 1 in 4 i.e. If you consistently bet on 4 to 1 shots would you simply break even over time?
2. The chance of red four times in a row is 5.6% but if you come to the table immediately after a red number has appeared, the probability that you will witness a further three reds (making a total of four times in a row) is 11.5% – purely because you will be observing only a series of three rounds.
3. The following chart highlights the probabilities of the same colour appearing over a certain number of spins of the roulette wheel. Example: The probability of the same colour showing up 4 times in a row is 5.6%. A graph to show the probability of seeing the same colour of red/black (or any 'evens' bet result for that matter) over multiple spins.

Hi,
In 1500 spins, there are 1499 possible pairs of spins

The odds that the colour red will hit three times in a row is 48.6% x 48.6% x 48.6% = 11.4%.

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)

Poker tournaments medicine hat. 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.

Acrylic poker chip case 500. 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

Roulette Odds 4 Reds In A Row 4

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.

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

Roulette Odds 4 Reds In A Row 4

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)
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.

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

Thanks,

Best Odds For Roulette

Peter