I'm working on a kick a** method. I could do this research on my own but perhaps someone knows already. Then again, I really doubt it. lol My question being >> I need to know WHEN the average would be (how many spins) a complete street hits. Any street, in any order. Such as 17, 16, 18. I do NOT mean the three numbers hit back to back to back. Here is a true example >>
12
35
34
7
0
13
22
10
3
29
19
4
11 <<<<< So, its the 10, 11, 12 street. When does this average occure? My question is for the 0 or 00 wheel, I dont care. Thanks, Ken
Ken,
I ran a simulation over 1 million trials - the AVERAGE was 19 spins (actually 18.80) with the MAXIMUM wait being around 70 spins, and the minimum is (obviously) 3 spins. This was for a single zero wheel.
Thanks Bayes! I am a bit confused though. 70 being the max? That means within 69 spins, NOT ONE complete street hit? No way. Ken
I was thinking that myself but what will happen at times is that you will keep getting the first two of a street and the third one will not come. So you could have 12 qualifiers and you are sitting waiting for one of them to complete. I can imagine at times that could take up to 70 spins to hit one of the missing numbers. Mind you it's a long shot. But we know 12 numbers can sleep for a very long time.
Thats the same as a dozen (or column) not hitting for 70 spins. Ken
Quote from: MrJ on Aug 15, 04:51 AM 2010
Thanks Bayes! I am a bit confused though. 70 being the max? That means within 69 spins, NOT ONE complete street hit? No way. Ken
It's a very rare event, but it can happen. Initially I ran the simulation over 100,000 trials and the maximum was 58-60 spins (average of 19 being the same) , increasing to 1M produced the 70. If I ran it for 100 M spins I might get a sleeper of 80 or more. I think that's what is meant by "random does not have limits". :)
If you like I can generate a breakdown of the frequencies (on how many trials the wait was 3,4,5... spins).
That would be great if you could, I dont want to waste a ton of your time. Ken
Ken,
Here is the breakdown over 1 M trials. Numbers in the left column are how many spins it took for a street to hit, and those in the right are how many times that wait occurred (so the total in the right hand column sums to 1 million).
The most frequent wait was actually 17, but the AVERAGE was higher because of the lack of symmetry (as you can see from the graph, there is a skew to the right).
1 0
2 0
3 1429
4 4056
5 7648
6 11811
7 17056
8 22411
9 27764
10 33428
11 38412
12 42800
13 47123
14 50008
15 52154
16 53603
17 54386
18 53616
19 52229
20 50042
21 46708
22 44217
23 40618
24 36694
25 32883
26 28745
27 25400
28 21966
29 18882
30 15961
31 13166
32 11010
33 9001
34 7241
35 5900
36 4814
37 3692
38 2964
39 2374
40 1869
41 1422
42 1112
43 849
44 645
45 467
46 343
47 276
48 233
49 137
50 113
51 94
52 68
53 38
54 31
55 24
56 18
57 12
58 11
59 9
60 4
61 5
62 1
63 1
64 3
65 1
66 0
67 0
68 0
69 0
70 0
71 1
72 0
73 1
74 0
75 0
76 0
77 0
78 0
79 0
80 0
From this you could go on to figure out the % of time you get a full street between any 2 times. e.g. it could be 80% between 10 and 40 spins (I don't know, just guessing).
Very, very useful sir....THANKS a bunch! Ken
looking at them figures over the million spins, it looks like it would only have hit over 40 spins 1 time in 200 and it will hit between spin 20-40 every 1 out of 2 spins.
So is there a way to take advantage between spin 20 and 40?
"So is there a way to take advantage between spin 20 and 40?" >>> Thats why I asked. My way of looking at it is a little different from the stats but the stats are indirectly involved. Ken
With what software do you make the graphs?
I use Gnuplot (link:://gnuplot.sourceforge.net/).
Maybe not the most user-friendly graphing program, but very powerful and has been around a long time.
There is also the charting component from Microsoft :)
link:://:.microsoft.com/downloads/details.aspx?FamilyId=130F7986-BF49-4FE5-9CA8-910AE6EA442C&displaylang=en (link:://:.microsoft.com/downloads/details.aspx?FamilyId=130F7986-BF49-4FE5-9CA8-910AE6EA442C&displaylang=en)
Good on you for using gnuPlot Bayes, it sure is capable!
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I think the question needs some clarification.
There are two ways to interpret it:
1) when a number of a street hits then we wait and see after how many spins the street is completely hit.
(because in Mr J example the first number was a number of the street)
2) we just start throwing spins and wait to see in how many spins a (any one) street is completely hit.
Which from the two interpretations did you Mr J care about, and which one did you Bayes researched? I guess it's probably the 1) bu it doesn't hurt to clear this up.
I was thinking about that. At what POINT are we starting to look for the three numbers? I think the overall results would be roughly the same. Maybe not exactly but close enough for testing purposes. Ken
Hi Mr J,
Version 2) would need much more spins than 1).
Bayes did it at around 1 million trials. Even if he did 5 million, IMO, I dont think the 'average' would be that different. Ken
I took it from Ken's first post that he was looking for 2) :
Quote from: MrJ on Aug 14, 07:10 PM 2010
I need to know WHEN the average would be (how many spins) a complete street hits. Any street, in any order. Such as 17, 16, 18.
If you meant 1), that's quite a different problem. As I understand it, it means whatever the first number is to hit (in your current session) you then see how long it takes the corresponding street to complete. Say 14 is the first number (which belongs to street 13-15), you then wait to see how long it takes for 13 and 15 to hit, but this is the same as waiting for any two particular (preselected) numbers to hit, which could be hundreds of spins.
The reason 1) takes longer than 2) is because you're waiting on a
particular street to complete (which is determined by the very first spin), whereas in 2) you're not.
"you're waiting on a particular street to complete" >>> Correct Bayes. All I was looking for (for betting purposes) is for the first street (any street) to get all 3 hits. Ken
No roulette simulation, software or even real wheel can exactly tell you as to when a street or its numbers turn sleeper. By all these, the maximum that you can do is sampling of possible events, but you can not get any conclusion. Betting on sleeper can be a blunder without comparison.
This method and question have nothing to do with sleepers. Ken
Dear Ken,
By sleeper I mean a number which has yet to turn up for a while. You are looking for a street to complete, so far I could understand your concepts. In such situation, if 2 & 3 have turned up you would bet on 1. Here 1 may turn sleeper. Correct me if I misunderstood.