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Math help

Started by FreeRoulette, Jun 13, 08:25 PM 2019

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FreeRoulette

Hello,

I'm trying to create a tool to go with a gap system. But first I need to fine tune the system.

How the system works:
Wait for there to be a maximum gap in the wheel of x numbers, then play a progression on those numbers.

For the test below, I waited for a maximum gap of 4 numbers.

There is clearly a bell curve to this, but need a math person to tell me the best gap number and amount of spins to use before giving up the progression.

Thank for any help

Trials: 1,000,000

This is a count of how many spins it took for the gap to appear. 0 spin means there were no gaps of 4.



Spin) count
0 293998
8 31
9 258
10 1156
11 3156
12 6474
13 11303
14 16947
15 22817
16 28704
17 33662
18 37400
19 39638
20 40697
21 40317
22 40246
23 37972
24 36207
25 33134
26 30918
27 27882
28 25498
29 22860
30 20249
31 18096
32 16138
33 14231
34 12464
35 11111
36 9580
37 8470
38 7346
39 6483
40 5604
41 4992
42 4447
43 3823
44 3249
45 2869
46 2509
47 2266
48 1931
49 1642
50 1464
51 1260
52 1061
53 948
54 832
55 696
56 627
57 554
58 526
59 405
60 364
61 288
62 284
63 238
64 220
65 178
66 174
67 148
68 136
69 88
70 98
71 94
72 85
73 62
74 49
75 36
76 41
77 28
78 31
79 35
80 18
81 16
82 23
83 16
84 15
85 11
86 11
87 12
88 10
89 5
90 6
91 3
92 5
93 1
94 4
95 3
96 1
97 1
98 1
100 3
102 1
103 1
104 3
107 1
114 1
126 1
134 1
135 1

This is the count of how many spins it took to hit the numbers in the gap.

spins it took) count
1 43789
2 43908
3 43174
4 42451
5 40888
6 39181
7 37112
8 35280
9 33413
10 31074
11 28985
12 26644
13 24543
14 22472
15 20733
16 18813
17 17215
18 15595
19 13965
20 13009
21 11530
22 10472
23 9379
24 8581
25 7569
26 6737
27 6179
28 5487
29 5083
30 4450
31 3940
32 3642
33 3211
34 2862
35 2617
36 2303
37 1981
38 1851
39 1676
40 1482
41 1290
42 1180
43 1024
44 985
45 830
46 778
47 726
48 657
49 591
50 478
51 439
52 387
53 363
54 304
55 325
56 261
57 224
58 194
59 173
60 163
61 137
62 113
63 104
64 119
65 85
66 84
67 89
68 64
69 51
70 47
71 61
72 48
73 43
74 30
75 28
76 21
77 31
78 24
79 16
80 21
81 16
82 17
83 7
84 12
85 3
86 9
87 9
88 7
89 6
90 5
91 9
92 1
93 3
94 3
95 5
96 2
97 2
98 3
99 2
100 1
101 2
102 1
103 1
104 1
106 1
107 1
109 1
110 1
111 2
112 2
121 1
125 2

This is a percentage that 4 appeared at a certain spin.

spin) percent
0 29.3998
8 0.0031
9 0.0258
10 0.1156
11 0.3156
12 0.6474
13 1.1303
14 1.6947
15 2.2817
16 2.8704
17 3.3662
18 3.74
19 3.9638
20 4.0697
21 4.0317
22 4.0246
23 3.7972
24 3.6207
25 3.3134
26 3.0918
27 2.7882
28 2.5498
29 2.286
30 2.0249
31 1.8096
32 1.6138
33 1.4231
34 1.2464
35 1.1111
36 0.958
37 0.847
38 0.7346
39 0.6483
40 0.5604
41 0.4992
42 0.4447
43 0.3823
44 0.3249
45 0.2869
46 0.2509
47 0.2266
48 0.1931
49 0.1642
50 0.1464
51 0.126
52 0.1061
53 0.0948
54 0.0832
55 0.0696
56 0.0627
57 0.0554
58 0.0526
59 0.0405
60 0.0364
61 0.0288
62 0.0284
63 0.0238
64 0.022
65 0.0178
66 0.0174
67 0.0148
68 0.0136
69 0.0088
70 0.0098
71 0.0094
72 0.0085
73 0.0062
74 0.0049
75 0.0036
76 0.0041
77 0.0028
78 0.0031
79 0.0035
80 0.0018
81 0.0016
82 0.0023
83 0.0016
84 0.0015
85 0.0011
86 0.0011
87 0.0012
88 0.001
89 0.0005
90 0.0006
91 0.0003
92 0.0005
93 0.0001
94 0.0004
95 0.0003
96 0.0001
97 0.0001
98 0.0001
100 0.0003
102 0.0001
103 0.0001
104 0.0003
107 0.0001
114 0.0001
126 0.0001
134 0.0001
135 0.0001
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Let Me Win

Already done years ago by John Solitude and called 'Raindrops'.


FreeRoulette

Quote from: Let Me Win on Jun 13, 09:15 PM 2019
Already done years ago by John Solitude and called 'Raindrops'.

Do you have a link that I could look at it?
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Let Me Win

link:://:.john-solitude.be/

John has been around for a long time and I found his free ebook very interesting but after around a year of following his method, sorry it just doesnt work.

I dont have anything personally against him but he is basically saying that eventually your winning number will hit so spread your bets to the sleepers and eventually you will win.

Now that I know better I dont think John has a clue what he is talking about.

Firefox

Quote from: Let Me Win on Jun 14, 08:01 AM 2019
link:://:.john-solitude.be/

John has been around for a long time and I found his free ebook very interesting but after around a year of following his method, sorry it just doesnt work.

I dont have anything personally against him but he is basically saying that eventually your winning number will hit so spread your bets to the sleepers and eventually you will win.

Now that I know better I dont think John has a clue what he is talking about.

A good website.  He is right about a lot of things though poor on advantage play. He doesn't fully understand the concepts. And if he advocates a sleeper method  (didn't read that bit), that's clearly wrong too. A lot of other good advice though.

FreeRoulette

Quote from: Let Me Win on Jun 14, 08:01 AM 2019
link:://:.john-solitude.be/

John has been around for a long time and I found his free ebook very interesting but after around a year of following his method, sorry it just doesnt work.

I dont have anything personally against him but he is basically saying that eventually your winning number will hit so spread your bets to the sleepers and eventually you will win.

Now that I know better I dont think John has a clue what he is talking about.

I wouldn't count on the numbers to eventually hit and keep playing them. Instead, I want to win more trials than I lose to come out ahead over time. That might mean placing flat bets on 4 numbers for 6 spins when the ocurrance of the 4 gap happen at spin y. I just want to play the premium spot on the bell curve.
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Firefox

Quote from: FreeRoulette on Jun 14, 06:40 PM 2019
I wouldn't count on the numbers to eventually hit and keep playing them. Instead, I want to win more trials than I lose to come out ahead over time. That might mean placing flat bets on 4 numbers for 6 spins when the ocurrance of the 4 gap happen at spin y. I just want to play the premium spot on the bell curve.

That won't work either. By relying on immediately past results you won't improve prediction accuracy. Every spin resets the clock.

Your idea will lose to the house edge.

FreeRoulette

Quote from: Firefox on Jun 14, 06:54 PM 2019
That won't work either. By relying on immediately past results you won't improve prediction accuracy. Every spin resets the clock.

Your idea will lose to the house edge.

If every spin resets the clock, then why is there a bell curve? The spin percent peaks between 20 and 22.  The amount of spins it takes to hit also decreases over spins, with the most hits happening in the first spin.
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Firefox

Quote from: FreeRoulette on Jun 14, 08:05 PM 2019
If every spin resets the clock, then why is there a bell curve? The spin percent peaks between 20 and 22.  The amount of spins it takes to hit also decreases over spins, with the most hits happening in the first spin.

It's just a property of the maths of the game. It doesn't mean you can exploit it.

The wheel can't remember what numbers have been hit.

The  next spin is a fresh trial. Any number you bet on will have a 1/37 chance for which you only get 36 units back if it hits.

There's no way to win with a maths based staking system.

FreeRoulette

Quote from: Firefox on Jun 15, 03:13 AM 2019
It's just a property of the maths of the game. It doesn't mean you can exploit it.

The wheel can't remember what numbers have been hit.

The  next spin is a fresh trial. Any number you bet on will have a 1/37 chance for which you only get 36 units back if it hits.

There's no way to win with a maths based staking system.

Do you agree that over time a number will come up at least 1 in 37 spins? 20 spins are gone and at least 4 numbers have not hit. A number can be asleep for hundreds of spins, but what are the chances that 4 numbers are asleep that long and that you picked those numbers? And in the unlikely event that you did, there is a stop loss in place. I'm working on the tool, just to see what happens. No systems win all the time, but maybe I don't need it to win all the time, or even recover losses on a win. The flat bet might do the trick.
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Firefox

Quote from: FreeRoulette on Jun 15, 04:45 AM 2019
Do you agree that over time a number will come up at least 1 in 37 spins? 20 spins are gone and at least 4 numbers have not hit. A number can be asleep for hundreds of spins, but what are the chances that 4 numbers are asleep that long and that you picked those numbers? And in the unlikely event that you did, there is a stop loss in place. I'm working on the tool, just to see what happens. No systems win all the time, but maybe I don't need it to win all the time, or even recover losses on a win. The flat bet might do the trick.

No there's no guarantee a number will come up once in 37 spins.

Take a long period of time say one million spins, but you can make it larger if you wish.

The mean number of appearances of one number is 1,000,000/37 =27,027

The standard deviation is SQRT (npq) =(1/37 x 36/37 x 1,000,000) = 162

link:s://en.m.wikipedia.org/wiki/Standard_deviation

According to the normal distribution curve (see above link) there is a 15.8% chance that the given number will be at least 162 (or many more) occurences short of the 1/37 mean of 27,027. The divergence from the mean in absolute terms will be even larger as the trials get larger.

The gambler thinking his number *must* catch up will be disappointed. It may never do. Statistics dictate this. Even if it does catch up to 20,027 he will still lose due to unequal payout. Due to the payout his number needs to significantly ahead to show a profit

FreeRoulette

I made the Raindrop tool for anyone who wants to play around with it.

link:://loothog.com/Systems/raindrop/raindrop.html
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FreeRoulette

Made a chart to show percentages of when certain segment sizes would appear as the last segment of that size to not hit. It also shows the percent chance the segment will win by a certain spin once found.

link:://loothog.com/Systems/raindrop/raindropchart.html
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