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 wait is a little over 8 spins, and the most common is 6.
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
Excellent contribution.
It is amazing how results from different numerical samples can match together in tests made by independent researchers.
Similar figures has been managed by the several studious of roulette as the common average wait. If I recall right, Pierre Basieaux places the average at around ~7. And now we have yours also.
Thank you dear Bayes! :thumbsup:
Dear Bayes,
since we have started checking samples, would you please check for me in your samples how long does it take for a dosen to repeat itself consecutively on average
It is my impression that, for example, in a 10 spin sample we are almost certain to observe a dosen which has repeated consecutively at least once.
I hope you understood my request
Thanks Vic. :)
@ sekuritati,
You are correct!, the average is actually once every 10 spins (precisely, 9.5 spins). But I will do some tests to see what the dispersion is.
Actually, it's only 1 in 10 if you're talking about a particular dozen, if you mean ANY of the dozens repeating then this happens once every 3 spins, on average.
This means the distribution for repeating dozens is the same as that for any particular dozen, so for example, the longest you have to wait for any dozen to repeat is the same length as a particular dozen might sleep, which is anything up to 35 spins.
Hi Bayes, talking about repeaters , i ask this because i never play
this way , just curious:
Let s say i wait 10 spins without any repeat,
than start to bet that last 10 numbers, than comes new number
bet last 11 numbers, than new number and bet the last 12 nmbrs.,
than add new number and so on until i bet 18 numbers, so the
last 18 numbers . Progresion first on 10 numbers, second on 12 nmbrs,
third steep on 13 numbers...so on until 18 numbers.
Stop lose at around -280 ( 9 progression steeps).
What chances ,or avarage have this ? Probably none, just curious.
Or better wait 7 spins without repeat, than bet like above,
on the last 7 numbers, than 8 numbers etc... around the last 15 numbers.
Thanks.
Bayes, I don't know if you are still active re this post but I need a little help understanding your results. Is this saying, for example, that in a 1M spin trial there are no occasions where a number repeated at 29 spins? or 30? etc. The numbers seem very low. I tried to replicate these results and get a very different picture.
wait number of occasions for this wait
1 27284
2 26380
3 26188
4 24636
5 23672
6 23584
7 22580
8 22300
9 21700
10 21423
11 20496
12 20136
13 19516
14 18739
15 18308
16 17739
17 17339
18 17024
19 16058
20 16256
21 15227
22 15004
23 14596
24 14680
25 14000
26 13468
27 13336
28 13292
29 12419
30 12276
31 11848
32 11603
33 11284
34 11136
35 10508
36 10288
37 9960
38+ 363680
Roger
@Roger
It shows the longest period you had to wait before a drawn number repeated.
Woods
Can any of you calculate something else I would like to know?
I wanted to know.... how long between occurences of straigth repeats (the same number appearing twice repeatedly).
e.g: 0-0- 1-5-7-10-20 -20 -2- 10 -34-34
The numbers marked o Bolt are the repeaters.
In this example we have an interval of 4 numbers, then of 2 numbers.
Chees and thanks.
37 spins. :thumbsup:
Quote from: Bayes on Oct 25, 03:24 PM 2011
37 spins. :thumbsup:
Thats on average? It's not the maximum, I think maximum it goes way more then 100 spins...