Roulette Ideas

This thread will be ideas on how to take advantage of the idea that after a certain amount of unique numbers, a number will repeat. The following shows the percentage of time a number will repeat. For example, a number will repeat on the 8th spin 10.24% of the time. If you play all 8 spins, then you will have about a 52% chance to hit (add all percentages up to spin 8)

First Repeat: Percent
2: 2.63%
3: 5.12%
4: 7.27%
5: 8.94%
6: 10.0%
7: 10.43%
8: 10.24%
9: 9.56%
10: 8.48%
11: 7.19%
12: 5.81%
13: 4.52%
14: 3.36%
15: 2.37%
16: 1.6%
17: 1.04%
18: 0.64%
19: 0.37%
20: 0.21%
21: 0.11%
22: 0.05%
23: 0.02%
24: 0.01%
25: 0.01%
26: 0.0%
27: 0.0%
28: 0.0%
29: 0.0%


I have tried flat betting up to spin 8 only, or from spin 5-9 etc, to find some edge. Between outside betting, betting on straight numbers, or the timing of the bets we can figure out some advantage. I would like to hear if you have any ideas to try on this. Thanks!

Has anyone done any calculations on this?

For example, there is a 90% chance of a repeat by spin 13.

So if we wait and bet 11, 12 and 13. 1 unit on 11 number, then 2 units on 12 number and 3 units on 13 numbers.

Would the win out weigh the loss? And what is the effect if the bet is increased? I would rather not write a program to test every permutation if someone already did it and would like to share the results.

2   2.63%   2.63%
3   5.12%   7.75%
4   7.27%   15.02%
5   8.94%   23.96%
6   10.00%   33.96%
7   10.43%   44.39%
8   10.24%   54.63%
9   9.56%   64.19%
10   8.48%   72.67%
11   7.19%   79.86%
12   5.81%   85.67%
13   4.52%   90.19%
14   3.36%   93.55%
15   2.37%   95.92%
16   1.60%   97.52%
17   1.04%   98.56%
18   0.64%   99.20%
19   0.37%   99.57%
20   0.21%   99.78%
21   0.11%   99.89%
22   0.05%   99.94%
23   0.02%   99.96%
24   0.01%   99.97%
25   0.01%   99.98%
26   0.00%   99.98%
27   0.00%   99.98%
28   0.00%   99.98%
29   0.00%   99.98%

in 5684 sequences of 37 spins there have been
e.g. in spin 14: sleepers 24: once hit 13: repeaters 0 : 360 times

7 30 7 0 0 0 0 3055
8 29 8 0 0 0 0 0 2449
9 28 9 0 0 0 0 0 1923
10 27 10 0 0 0 0 0 1465
11 26 11 0 0 0 0 0 1078
12 25 12 0 0 0 0 0 777
13 24 13 0 0 0 0 0 543
14 23 14 0 0 0 0 0 360
15 22 15 0 0 0 0 0 231
16 21 16 0 0 0 0 0 151
17 20 17 0 0 0 0 0 92
18 19 18 0 0 0 0 0 49
19 18 19 0 0 0 0 0 25
20 17 20 0 0 0 0 0 11
21 16 21 0 0 0 0 0 3

I hope you can calculate on this informations

Sorry, somethings wrong with the table

7   30 7 0 0 0 0   3055 8   29 8 0 0 0 0 0   2449 9   28 9 0 0 0 0 0   1923 10   27 10 0 0 0 0 0   1465 11   26 11 0 0 0 0 0   1078 12   25 12 0 0 0 0 0   777 13   24 13 0 0 0 0 0   543 14   23 14 0 0 0 0 0   360 15   22 15 0 0 0 0 0   231 16   21 16 0 0 0 0 0   151 17   20 17 0 0 0 0 0   92 18   19 18 0 0 0 0 0   49 19   18 19 0 0 0 0 0   25 20   17 20 0 0 0 0 0   11 21   16 21 0 0 0 0 0   3

Two posts above, I found this calc tool to be very handy,
to determine the best hit probability vs numbers/spins played;

[search, input box[
adjust the max steps -- latter value -- last bet placed
adjust the first/current spin value -- waiting time, first bet placed

then add probabilities of all spins in the betting sequence or block, or see the biggest gaps/distanx=ces in the graph to gauge where (which section) to bet, & the highest probabilities + where those start to diminish

https://www.wolframalpha.com/input?i2d=true&I=birthday+paradox%5C%2844%29+5+people%5C%2844%29+8+possible+birthdays

Ascertaining by your @FreeRoulette array above & with (11+12+13)=36 pointing to the 36u bank per cycle/attempt --

.

best probabilities of a hit & returns with that amount invested:

(5+6+7+8+9)= 35u  --  (8.94+10.00+10.43+10.24+9.56)= 49.17%
49.17/35= 1.40485714

 (6+7+8+9) = 30u  --     (10.00+10.43+10.24+9.56)  = 40.23%
40.23/30= 1.341

  (6+7+8)  = 21u  --       (10.00+10.43+10.24)     = 30.65%
30.65/21= 1.45952380952

.

To draw the correlation:
defining the normalized number of cycles played & precise distinction in probability coefficient

210u, as the common factor
210/35=  6  →  6* 1.40485714      =8.42914284
210/30=  7  →  7* 1.341           =9.387
210/21= 10  → 10* 1.45952380952   =14.5952380952


8.42914284/14.5952380952  = 0.57752691563
9.387/14.5952380952       = 0.643154976
14.5952380952/8.42914284  = 1.73152103034
14.5952380952/9.387       = 1.55483520776

Or simply, (6+7+8) gives you 57% & 64% better probability at the same 210u cycled through.

.

Thereof:
As the best probability for/constituted by the single buck & single number played is the latter one far outweighs.

To put things into perspective:
in addition, it constitutes as ≈DS bet, played over 3x spins.

So, you may ..
•  get a hit in the 2nd cycle too & still be in plus with 1hit only
•  strategically incorporate the recovery as .. two or more hits to finish .. delaying the vertical (vp, vertical progression0 increase further, lowering the overall exposition increase rate & thus volatility (as moving through the cycles = betting blocks)
•  increase vp !after the hit, once the exposition amount us lowered & reinvested those gains as a press (≈one betting block)
•  the latter obviously prevents the unnecessary stacking of units vertically through the out-of-favor variance intervals → thus in turn, as a design-decisions ripple effect, prevents absolutely unnecessary vertical & exposition escalation  = stacking the units, if necessary only! (to close the individual game in nominal profit, new high) →
•  → leveraging the gains only when indicated with the hit by the wheels itself, indicating potential turn of variance in-favor

Here are a couple of formulas related to the topic which you might find useful.

First is the probability, p, that when betting a location for which there are n equally likely outcomes, for s spins, all of the outcomes will be unique.



Where the (!) is the factorial. It means multiply all whole numbers from the chosen number down to 1. eg,
5! = 5 × 4 × 3 × 2 × 1 = 120. All scientific calculators have a factorial function as does Excel.

To find the probability that you will get at least one repeat in the s spins, subtract the result from 1.

Example: what's the probability of 12 spins all being unique - no repeated numbers?

Here, s = 12 and n = 37 (we're considering single numbers, so there are 37 equally likely outcomes)

Just plug the numbers into the formula -

p = 37! ÷ (37¹² × (37 ─ 12)! = 0.1348

That means the probability of at least one repeat is 1 ─ 0.1348 = 0.865

Here's the second formula, which tells you the expected total number of times an outcome will repeat. The symbols n and s stand for the number of equally likely outcomes and the number of spins, respectively.



eg, in 37 spins, how many repeats would you expect to see (considering single numbers again)? To be clear about what the formula tells you, start counting spins and every time you get a repeat, add 1 to the count. After 37 spins your final count will be N, on average.

Again, n = 37, and s also is 37, so the number of repeats, N is

37 ─ 37 + 37 × (36/37)³⁷ = 13.425

Any questions, let me know. I know math isn't everyone's favourite subject.  

(6+7+8)

Given that the 7ⁿ (ⁿ - numbers played) has the highest probability of 10.34 ..

you may determine that
•  (7+7+7) is the best option overall, & that will simplify the real-time play calculations as well
•  on the first betting block (after a restart, new high ..in the chain of blocks of 3-spins bet), you may use (6+7+8) .. due to the possibility of having the hit on the first spin = gaining an extra unit on all such occasions/instances

Yes, how would it be calculated:

ie.

a)
playing 21 numbers (21ⁿ) per betting block of 3-spins,how to calculate the probability of having the first hit, as well as, having another hit in the consecutive block after that (completre break-down including the results)


b)
also, doing that above on a spin basis
attack   (6+7+8)
         (6+7 ..
recovery  .. +8)
         (7+7+7) thereof

so attack is 21+13= 34ⁿ
   thereof, the probability of having the first hit at any spin (based on numbers played), & as well, the probability .. after securing that first hit (+33), of the second hit (4*7= 28, 5* 7= 35) within 3,4 or 5 next spins  =  combo hit  ---

& then .. if that is successful, & required to continue, the same for the 3rd hit .. or 3 hits within a certain amount of spins (related to the first hit)  =  triple combo.

Thank you for the math, I will go through it. I wanted to know how long it would take for a certain amount of unique numbers to appear. This for a trial over million spins on a single zero roulette.

For length 10
10: 9242

Shows that out of million, the first 10 spins were unique 9,242 times.
10 uniques came up 35,061 times in a million.

-----------------

For length 2
2: 473409
4: 12751
6: 352
8: 7
10: 1

For length 3
3: 284979
5: 7750
6: 14909
7: 208
8: 815
9: 806
10: 34
11: 61
12: 41
13: 5
14: 3
19: 1

For length 4
4: 183514
6: 4899
7: 9734
8: 13818
9: 495
10: 1275
11: 1533
12: 1134
13: 142
14: 189
15: 166
16: 87
17: 33
18: 27
19: 21
20: 4
21: 7
23: 3
24: 2
25: 1

For length 5
5: 120563
7: 3161
8: 6298
9: 9104
10: 11277
11: 806
12: 1615
13: 1870
14: 1712
15: 1245
16: 326
17: 351
18: 360
19: 298
20: 151
21: 70
22: 78
23: 74
24: 48
25: 38
26: 19
27: 11
28: 7
29: 3
30: 4
31: 2
32: 2
34: 1
38: 1
39: 1

For length 6
6: 77835
8: 2145
9: 4035
10: 5930
11: 7213
12: 8342
13: 970
14: 1674
15: 1906
16: 2029
17: 1714
18: 1254
19: 560
20: 593
21: 592
22: 484
23: 382
24: 247
25: 186
26: 183
27: 144
28: 119
29: 102
30: 72
31: 56
32: 34
33: 40
34: 32
35: 25
36: 18
37: 20
38: 13
39: 8
40: 10
41: 5
42: 4
43: 1
44: 4
45: 1
46: 4
47: 3
48: 1
49: 1
51: 1
53: 2

For length 7
7: 49061
9: 1333
10: 2512
11: 3627
12: 4622
13: 5321
14: 5758
15: 1035
16: 1522
17: 1820
18: 1828
19: 1691
20: 1489
21: 1139
22: 695
23: 727
24: 706
25: 674
26: 527
27: 438
28: 377
29: 291
30: 292
31: 268
32: 201
33: 194
34: 164
35: 132
36: 110
37: 95
38: 84
39: 65
40: 65
41: 50
42: 43
43: 40
44: 46
45: 32
46: 23
47: 25
48: 21
49: 15
50: 13
51: 17
52: 12
53: 3
54: 8
55: 6
56: 5
57: 5
58: 4
59: 5
60: 2
61: 4
62: 1
63: 3
64: 4
65: 2
66: 1
67: 1
70: 1
73: 1
82: 1

For length 8
8: 29466
10: 771
11: 1489
12: 2221
13: 2854
14: 3209
15: 3475
16: 3623
17: 910
18: 1255
19: 1479
20: 1559
21: 1479
22: 1399
23: 1140
24: 937
25: 686
26: 740
27: 738
28: 724
29: 618
30: 616
31: 503
32: 429
33: 344
34: 343
35: 311
36: 284
37: 255
38: 263
39: 195
40: 196
41: 202
42: 158
43: 139
44: 133
45: 104
46: 107
47: 107
48: 93
49: 73
50: 75
51: 49
52: 55
53: 70
54: 39
55: 43
56: 39
57: 32
58: 38
59: 34
60: 19
61: 25
62: 12
63: 17
64: 15
65: 16
66: 23
67: 13
68: 9
69: 13
70: 14
71: 5
72: 8
73: 5
74: 4
75: 6
76: 6
77: 2
78: 8
79: 2
80: 1
81: 2
82: 3
83: 2
84: 1
86: 1
87: 5
88: 1
89: 2
91: 1
96: 1
97: 1
99: 1
100: 1
105: 1
112: 2
139: 1

For length 9
9: 17022
11: 500
12: 883
13: 1228
14: 1670
15: 1796
16: 1962
17: 2138
18: 2226
19: 732
20: 925
21: 1061
22: 1085
23: 1101
24: 1029
25: 936
26: 837
27: 747
28: 642
29: 635
30: 695
31: 631
32: 528
33: 550
34: 455
35: 449
36: 370
37: 384
38: 359
39: 349
40: 314
41: 321
42: 260
43: 240
44: 241
45: 209
46: 231
47: 171
48: 182
49: 167
50: 148
51: 146
52: 159
53: 116
54: 130
55: 113
56: 124
57: 94
58: 88
59: 87
60: 80
61: 75
62: 66
63: 67
64: 54
65: 59
66: 60
67: 56
68: 48
69: 64
70: 43
71: 36
72: 51
73: 37
74: 40
75: 44
76: 30
77: 19
78: 26
79: 23
80: 21
81: 14
82: 15
83: 20
84: 18
85: 18
86: 19
87: 6
88: 17
89: 9
90: 13
91: 8
92: 8
93: 6
94: 13
95: 5
96: 7
97: 9
98: 3
99: 3
100: 7
101: 7
102: 5
103: 6
104: 5
105: 4
106: 3
107: 2
108: 4
109: 2
110: 2
111: 5
112: 8
115: 2
116: 2
117: 2
118: 3
119: 3
120: 4
121: 2
122: 1
123: 2
125: 1
126: 1
128: 1
132: 1
133: 1
135: 1
136: 2
137: 1
139: 1
142: 1
145: 1
150: 1
153: 1
159: 1

For length 10
10: 9242
12: 253
13: 494
14: 683
15: 873
16: 1021
17: 1089
18: 1185
19: 1160
20: 1163
21: 529
22: 596
23: 655
24: 712
25: 711
26: 688
27: 635
28: 647
29: 558
30: 501
31: 489
32: 462
33: 471
34: 440
35: 456
36: 431
37: 354
38: 353
39: 371
40: 344
41: 347
42: 331
43: 301
44: 272
45: 288
46: 267
47: 277
48: 234
49: 237
50: 213
51: 198
52: 212
53: 198
54: 206
55: 170
56: 160
57: 159
58: 148
59: 168
60: 137
61: 108
62: 150
63: 130
64: 105
65: 94
66: 87
67: 107
68: 107
69: 91
70: 79
71: 88
72: 92
73: 73
74: 54
75: 68
76: 74
77: 68
78: 63
79: 55
80: 43
81: 51
82: 59
83: 47
84: 46
85: 47
86: 44
87: 35
88: 43
89: 44
90: 34
91: 31
92: 33
93: 32
94: 30
95: 35
96: 25
97: 28
98: 24
99: 20
100: 24
101: 15
102: 21
103: 19
104: 21
105: 19
106: 24
107: 13
108: 17
109: 17
110: 22
111: 10
112: 18
113: 20
114: 14
115: 11
116: 12
117: 12
118: 7
119: 7
120: 5
121: 13
122: 8
123: 4
124: 7
125: 3
126: 5
127: 6
128: 4
129: 4
130: 8
131: 6
132: 8
133: 4
134: 9
135: 2
136: 4
137: 4
138: 4
139: 7
140: 1
141: 10
142: 3
143: 1
144: 4
145: 1
146: 3
148: 2
149: 2
150: 3
151: 1
152: 3
153: 1
154: 3
155: 2
156: 1
157: 1
159: 1
160: 2
161: 1
162: 1
163: 1
166: 1
167: 2
168: 3
169: 3
170: 2
171: 1
172: 1
173: 3
174: 1
176: 1
177: 2
178: 1
180: 2
182: 2
183: 1
185: 1
187: 2
191: 1
193: 1
196: 2
198: 1
202: 1
206: 1
214: 1
220: 1
221: 2
225: 1
248: 1

HardMan and Median Joe,

You guys are math wizards. I don't often get a chance to speak with someone who actually knows the math. So, before I go analyzing this unique numbers idea any more by programming experiments. Do you think the math works out the exact same for both of these situations?

A) Wait for 7 unique numbers to come up, then bet all 7 numbers for 1 spin.

B) Just bet 7 random numbers for 1 spin.

If these are both exactly the same, then the whole unique number idea is a waste of time and I would be better to explore betting on 6, 7, 8 numbers and hope the win multiplier would put me into profit over time.

Probability-wise, yes .. A=B.

Then again, the probability model, as much as it is useful, does not account for all the nuances/aspects of the game.

.

But one thing is a fact, even within the probability model itself: the probability of a hit increases after each 7-unique as a set; & as well, as the unhit-set after an unhit-set pass by .. those as a chain pass reaching a type of extreme (about 300 numbers played) after securing a hit then .. the probability of reaching that extreme again diminishes with each new chain set -- something to rely on, amongst other things.

.

As far as math wizard goes, MJ's math skills & experience are an iceberg meanwhile mine are the tip of the iceberg .. more of case-specific applied math that I learnt along the way & still do .. what is useful, relevant & applicable.

Quote from: HardMan on Oct 25, 2022, 03:40 AMProbability-wise, yes .. A=B.

does not account for all the nuances/aspects of the game.

Betting 'last outcome' eg. DZ ... prevents to bet on the quiscent one by default.

Betting last 7 uniques, similarly, prevents betting on the on average ≈1/3 of the quiescent numbers within the cycle (37-spins) ..

.. after a chain of sets, although it won't constitute ev+ .. as a singular factor being enough to facilitate consistent winning -- it will add one tiny advantage, as a portion of what you need to accomplish that.


So .. NO, A≠B.

Quote from: FreeRoulette on Oct 25, 2022, 01:08 AMDo you think the math works out the exact same for both of these situations?

A) Wait for 7 unique numbers to come up, then bet all 7 numbers for 1 spin.

B) Just bet 7 random numbers for 1 spin.

This is the acid test for bet selection; if picking random numbers is as good as your BS, it means the BS is useless. The problem with waiting for some event to lose a certain number of times is that it flies in the face of the basic probability model for roulette, which is that spins are independent. What players tend to forget when waiting for virtual losses is that every long losing run starts with a shorter losing run, and because the "pool" of spins is essentially infinite, waiting for losses doesn't get you any closer to a win.

This is basic logic, assuming the independent outcomes model is correct (and it's pretty hard to refute). The math also backs up this conclusion.

But what about regression to the mean? Some traders use it and swear by it, so I prefer to keep an open mind on the topic.

The only way to be sure is to compare your bet selection with a random selection. No need to test for a billion spins, just run a few sessions using both and then compare. If there is a difference it's likely to be small, so you really need to use proper statistical test, but that's a topic for another thread.

By the way, math is nice, but coding is the more useful skill because there are many problems and questions where there is no analytical answer, or it's very hard to find, so coding is your only resource. Plus, your math might be wrong.