On average how many times will I see multiple sets of 7 roulette numbers come up 3 times in a row.

Sorry the title is confusing. I didnt know how to word it so let me try and clarify.

If i had multiple groups of 7 assigned numbers, how would I calculate the average times any 3 numbers of a group of 7 would come up 3 times in a row within 35 spins?

So imagine this chart below is a single zero roulette board. There are 12 streets on the board. I assign each street A through L.

A = street 1,2,3
B = street 4,5,6
C = street 7,8,9
and so on.

	A	B	C	D	E	F	G	H	I	J	K	L
	3	6	9	12	15	18	21	24	27	30	33	36
0	2	5	8	11	14	17	20	23	26	29	32	35
	1	4	7	10	13	16	19	22	25	28	31	34

I then assign the groups of 7 by assigning "a street" with "a second street" plus the 0 for a total of 7 numbers.

So the first group of 7 numbers would be A0B or (Street A plus 0 plus Street B)
The second group of 7 would be A0C or (Street A plus 0 plus Street C)
The thrid group of 7 would be A0D or (Street A plus 0 plus Street D)
and so on.

Total I would have 66 assigned groups of 7 numbers.

A0B, A0C, A0D, A0E, A0F, A0G, A0H, A0I, A0J, A0K, A0L, B0C, B0D, B0E, B0F, B0G, B0H, B0I, B0J, B0K, B0L, C0D, C0E, C0F, C0G, C0H, C0I, C0J, C0K, C0L, D0E, D0F, D0G, D0H, D0I, D0J, D0K, D0L, E0F, E0G, E0H, E0I, E0J, E0K, E0L, F0G, F0H, F0I, F0J, F0K, F0L, G0H, G0I, G0J, G0K, G0L, H0I, H0J, H0K, H0L, I0J, I0K, I0L, J0K, J0L, K0L


What I know is that the odds of any 3 numbers within a SINGLE group of 7 numbers coming up 3 times in a row on a single zero wheel is,

(7/37)^3 = 0.0067715634

then i take that and multiply that by the 35 spins,

0.0067715634 * 35 = 0.2370047184

So that tells me I should average any 3 numbers of a SINGLE group of 7 coming up 3 times in a row roughly 1 in 4 rounds of 35 spins.

But now with 66 different ways to make 3 numbers of multiple groups of 7 numbers come up 3 times in a row do I multiply 66 by the 0.2370047184?????

0.2370047184 * 66 = 15.64231141

or in other words I should average 15.64231141 per 35 spins????
or if not how would i calculate it???

Basically how would I calculate the average number of occurrences that any 3 numbers of a group of 7 numbers coming up 3 times in a row within 35 spins when there are 66 groups of 7? haha

I think im doing something wrong cause ive been averaging 10.5 per 35 spins. Obviously my sample sizes are way too small for an accurate average but i wanted to check if i was calculating it right.

Thanks!

Quote: danepeterson

Sorry the title is confusing.
Total I would have 66 assigned groups of 7 numbers.

OK
12 choose 2 = 12*11/2 = 66 got it

Quote: danepeterson

What I know is that the odds of any 3 numbers within a SINGLE group of 7 numbers coming up 3 times in a row on a single zero wheel is,

(7/37)^3 = 0.0067715634

this is just for the very next 3 spins
got it

Quote: danepeterson

then i take that and multiply that by the 35 spins,

0.0067715634 * 35 = 0.2370047184

I do not agree
Because your formula for 3 in a row spins counts overlapping sequences where # of spins is greater than 3
in other words
S=success
F=Fail
SSSS
your way counts as 2, 3 in a row
spins 1-3 and 2-4
The count should only be one 3 in a row

I get by simulation and a different formula
0.183597 as the average # of runs of length 3 in 35 spins
This is correct.
(where SSS,SSS counts as 2 runs of 3 and not just 1 run of 6.
There are many ways to count sequences but each has it's own formula as to what really counts)

Quote: danepeterson

So that tells me I should average any 3 numbers of a
SINGLE group of 7 coming up 3 times in a row
roughly 1 in 4 rounds of 35 spins.

I missed this one.
This is an error.
0.0067715634 * 35 = 0.2370047184
0.2370047184 = average number of times (EV) this event happens in 35 spins.
P*N = EV (basic binomial probability stuff)

sometimes the average and the probability look the same, say .1850
but an average # of times can be >=0
where the probability is expressed between 0 and 1 inclusive

Quote: danepeterson

But now with 66 different ways to make 3 numbers of
multiple groups of 7 numbers come up 3 times in a row do I multiply 66 by the 0.2370047184?????

0.2370047184 * 66 = 15.64231141

or in other words I should average 15.64231141 per 35 spins????
or if not how would i calculate it???

Basically how would I calculate the average number of occurrences that any 3 numbers of a group of 7 numbers
coming up 3 times in a row within 35 spins when there are 66 groups of 7? haha

I think im doing something wrong cause ive been averaging 10.5 per 35 spins.
Obviously my sample sizes are way too small for an accurate average but i wanted to check if i was calculating it right.

Thanks!

edit: 9/23
0.183597 * 66 = about 12.1

my first sim results showed 11.3
a few more larger sims (and doing just 2 in a row also) shows that to be close.
The distribution looks interesting and I get the feeling
we are still over counting the number of successes by just *66
(we only add the many probabilities when the events are mutually excelsive - when more than one event can not
happen at the same time)

added:run distributions

say the first 3 numbers are 3,1,2
we just had 11 groups that won (all the A0...)
inclusion-exclusion time (I need to look into it more maybe)
This is the exact reason I run simulations as to see how close any calculated results are

see you in a few days to show my formulas

I would much appreciate your formulas whenever you have a chance.

Thanks!