Distribution of red/black sequences in number of roulette spins

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

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)

Quote: Peter

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.

It depends on whether or not how many times you count a sequence of, say, 5 reds as "a sequence of 2 reds."
Zero, since you are counting how many times red comes up exactly twice in a row?
One?
Two, since it is "two pairs of reds" (1 & 2, 3 & 4)?
Four, since spins 1 & 2 are consecutive, as are 2 & 3, 3 & 4, and 4 & 5?

But as SM777 has pointed out, if you are looking for some mysterious way to beat roulette, don't bother. Remember, after two (or four, or 257, or zero) consecutive reds, the probability that the next spin will be red on a double-zero wheel is still 9/19 (and black is 9/19, and green is 1/19).

Quote: Peter

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

Quote: Peter

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)
Sally

Thanks 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.

Quote: mustangsally

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

Thanks,
Peter

I have to ask.... Why 1500 instead of an "even" number like 1,000, or 10,000.

At least it would be easier to turn into percentages.

Russ