Another martingale variation on roulette - curious what the odds are of losing bank roll

Awesome new site, I've been anxiously waiting for your forums after your announcement a few weeks back.

Anyway, I am a bit odds challenged, and I wanted to see what the result of this system would be, as far as probability of losing the bank roll. This is a system that I made up, but I am sure others have also made up the same system in the past. I fully accept and understand that the system will eventually fail, but I wanted to see the system's limits. It should also be noted that I stay away from roulette and usually stick to BJ, video poker, and craps. Here is how it would work:

Bank Roll: 10,000 Units
Variables Starting Values: Count=0, BetUnit=1

How to Play:

On a single 0 roulette wheel, bet on any selected number with 1 BetUnit. If the number fails to come up, then increment Count. Repeat this process until either Count=34 or the number hits.

If Count=34, then reset Count to 0 and double BetUnit
If the number hits, reset Count to 0 and BetUnit to 1

Using this system, what are the odds of losing the bank roll after:
1,000 spins
10,000 spins
100,000 spins
1,000,000 spins

Keep in mind that the number of units won above the number of units bet will vary depending what Count is set to on each win.

Can someone please show me the above probabilities of losing the bank roll, as well as how you arrived at the numbers?

Are you saying that, if you *never* hit your number, you are betting $1 34 times, and then you're betting $2 34 times, then betting $3 34 times, etc?

Yes, I reslize that it's unlikely to happen, but not impossible either, that your number will not hit 68 or more times in a row.

If my interpretation is correct, then you only need to lose 52 times in a row before a win will not cover your losses.

I.E.: If you have 34 $1 losses, then 18 $2 losses, you're down $70. a win on spin 52, if you've still betting $2, will return $70.

In order to ALWAYS have a positive expectation, you have to shorten the count. I.E. Increase your bet when the net would be zero.

Check out this chart:

Spin	Bet	Invested	Payoff	Net	Expectation
1	1	1	35	34	0.9722
2	1	2	35	33	0.9452
3	1	3	35	32	0.9190
4	1	4	35	31	0.8934
5	1	5	35	30	0.8686
6	1	6	35	29	0.8445
7	1	7	35	28	0.8210
8	1	8	35	27	0.7982
9	1	9	35	26	0.7761
10	1	10	35	25	0.7545
11	1	11	35	24	0.7335
12	1	12	35	23	0.7132
13	1	13	35	22	0.6933
14	1	14	35	21	0.6741
15	1	15	35	20	0.6554
16	1	16	35	19	0.6372
17	1	17	35	18	0.6195
18	1	18	35	17	0.6023
19	1	19	35	16	0.5855
20	1	20	35	15	0.5693
21	1	21	35	14	0.5534
22	1	22	35	13	0.5381
23	1	23	35	12	0.5231
24	1	24	35	11	0.5086
25	1	25	35	10	0.4945
26	1	26	35	9	0.4807
27	1	27	35	8	0.4674
28	1	28	35	7	0.4544
29	1	29	35	6	0.4418
30	1	30	35	5	0.4295
31	1	31	35	4	0.4176
32	1	32	35	3	0.4060
33	1	33	35	2	0.3947
34	1	34	35	1	0.3837
35	2	36	70	34	0.3731
36	2	38	70	32	0.3627
37	2	40	70	30	0.3526
38	2	42	70	28	0.3428
39	2	44	70	26	0.3333
40	2	46	70	24	0.3241
41	2	48	70	22	0.3151
42	2	50	70	20	0.3063
43	2	52	70	18	0.2978
44	2	54	70	16	0.2895
45	2	56	70	14	0.2815
46	2	58	70	12	0.2737
47	2	60	70	10	0.2661
48	2	62	70	8	0.2587
49	2	64	70	6	0.2515
50	2	66	70	4	0.2445
51	2	68	70	2	0.2377
52	3	71	105	34	0.2311
53	3	74	105	31	0.2247
54	3	77	105	28	0.2184
55	3	80	105	25	0.2124
56	3	83	105	22	0.2065
57	3	86	105	19	0.2007
58	3	89	105	16	0.1952
59	3	92	105	13	0.1897
60	3	95	105	10	0.1845
61	3	98	105	7	0.1793
62	3	101	105	4	0.1744
63	3	104	105	1	0.1695
64	4	108	140	32	0.1648
65	4	112	140	28	0.1602
66	4	116	140	24	0.1558
67	4	120	140	20	0.1515
68	4	124	140	16	0.1473
69	4	128	140	12	0.1432
70	4	132	140	8	0.1392
71	4	136	140	4	0.1353
72	5	141	175	34	0.1316
73	5	146	175	29	0.1279
74	5	151	175	24	0.1244
75	5	156	175	19	0.1209
76	5	161	175	14	0.1175
77	5	166	175	9	0.1143
78	5	171	175	4	0.1111
79	6	177	210	33	0.1080
80	6	183	210	27	0.1050
81	6	189	210	21	0.1021
82	6	195	210	15	0.0993
83	6	201	210	9	0.0965
84	6	207	210	3	0.0938
85	7	214	245	31	0.0912
86	7	221	245	24	0.0887
87	7	228	245	17	0.0862
88	7	235	245	10	0.0838
89	7	242	245	3	0.0815
90	8	250	280	30	0.0792
91	8	258	280	22	0.0770
92	8	266	280	14	0.0749
93	8	274	280	6	0.0728
94	9	283	315	32	0.0708
95	9	292	315	23	0.0688
96	9	301	315	14	0.0669
97	9	310	315	5	0.0651
98	10	320	350	30	0.0632
99	10	330	350	20	0.0615
100	10	340	350	10	0.0598
101	11	351	385	34	0.0581
102	11	362	385	23	0.0565
103	11	373	385	12	0.0549
104	11	384	385	1	0.0534
105	12	396	420	24	0.0519
106	12	408	420	12	0.0505
107	13	421	455	34	0.0491
108	13	434	455	21	0.0477
109	13	447	455	8	0.0464
110	14	461	490	29	0.0451
111	14	475	490	15	0.0439
112	14	489	490	1	0.0426
113	15	504	525	21	0.0414
114	15	519	525	6	0.0403
115	16	535	560	25	0.0392
116	16	551	560	9	0.0381
117	17	568	595	27	0.0370
118	17	585	595	10	0.0360
119	18	603	630	27	0.0350
120	18	621	630	9	0.0340
121	19	640	665	25	0.0331
122	19	659	665	6	0.0322
123	20	679	700	21	0.0313
124	20	699	700	1	0.0304
125	21	720	735	15	0.0296
126	22	742	770	28	0.0287
127	22	764	770	6	0.0279
128	23	787	805	18	0.0272
129	24	811	840	29	0.0264
130	24	835	840	5	0.0257
131	25	860	875	15	0.0250
132	26	886	910	24	0.0243
133	27	913	945	32	0.0236
134	27	940	945	5	0.0229
135	28	968	980	12	0.0223
136	29	997	1015	18	0.0217
137	30	1027	1050	23	0.0211
138	31	1058	1085	27	0.0205
139	32	1090	1120	30	0.0199
140	33	1123	1155	32	0.0194
141	34	1157	1190	33	0.0188
142	35	1192	1225	33	0.0183
143	36	1228	1260	32	0.0178
144	37	1265	1295	30	0.0173
145	38	1303	1330	27	0.0168
146	39	1342	1365	23	0.0164
147	40	1382	1400	18	0.0159
148	41	1423	1435	12	0.0155
149	42	1465	1470	5	0.0150
150	43	1508	1505	-3	0.0146
151	44	1552	1540	-12	0.0142
152	45	1597	1575	-22	0.0138
153	46	1643	1610	-33	0.0134
154	47	1690	1645	-45	0.0131
155	48	1738	1680	-58	0.0127
156	49	1787	1715	-72	0.0123
157	50	1837	1750	-87	0.0120
158	51	1888	1785	-103	0.0117
159	52	1940	1820	-120	0.0113
160	53	1993	1855	-138	0.0110

As you can see, 136 consecutive losses results in a $1000 bankroll being insufficient to cover the next bet.

If you have a bigger bankroll, and suffer 149 consecutive losses, you have to increase your betting unit to keep up.

The Expectation column shows the odds of losing that many times in a row - (35/36)^spin

Yeah, it's unlikely that you'll suffer that many losses, but not impossible either. If you doubt me, think of all the times you passed a roulette table, looed at the board which shows the last dozen spins, and remarked at how many doubles there are. The odds are against that too.

The other obvious broblems are:
- It's time consuming to the point of being boring.
- Look at spin # 63. How'd you like to lose 62 times and then finally hit on # 63? Do I sense a change of strategy coming up? Woo boy. Change the whole chart!

All betting systems in roulette lose at the rate


Expected loss = (Total Wagers)*(House Edge).


This system is no exception.


OK, I wrote a computer program to simulate this system. I didn't answer the question you asked exactly as you asked it, but I hope this helps.

This program allows you to make a FULL SIZE bet for your last bet, even if you don't have the funds to cover it, so it is not giving exact results. You didn't specify what to do in that situation, so I took the liberty, since it made the programming easier. The effect of this change was to make the bust times possibly LONGER than they would be if you were limited by your last wager to available funds.


I ran 20 simulations starting with a 1000 unit bankroll and got these number of rounds to bust:


Number of rounds to bust: 4371
Number of rounds to bust: 36080
Number of rounds to bust: 19883
Number of rounds to bust: 422
Number of rounds to bust: 1346
Number of rounds to bust: 196353
Number of rounds to bust: 8103
Number of rounds to bust: 1558
Number of rounds to bust: 2316
Number of rounds to bust: 56821
Number of rounds to bust: 1159
Number of rounds to bust: 874
Number of rounds to bust: 286
Number of rounds to bust: 10016
Number of rounds to bust: 1593
Number of rounds to bust: 15160
Number of rounds to bust: 23162
Number of rounds to bust: 642
Number of rounds to bust: 28209
Number of rounds to bust: 1759

Here are 20 simulations of the same system starting with a 10,000 unit bankroll.


Number of rounds to bust: 118029
Number of rounds to bust: 20969
Number of rounds to bust: 23796
Number of rounds to bust: 36480
Number of rounds to bust: 213409
Number of rounds to bust: 26667
Number of rounds to bust: 15573
Number of rounds to bust: 62811
Number of rounds to bust: 131183
Number of rounds to bust: 11448
Number of rounds to bust: 2762753
Number of rounds to bust: 852206
Number of rounds to bust: 166302
Number of rounds to bust: 141391
Number of rounds to bust: 41295
Number of rounds to bust: 6922
Number of rounds to bust: 144477
Number of rounds to bust: 215548
Number of rounds to bust: 18867
Number of rounds to bust: 201690


Here is the source code (please double check this code -- I didn't bother checking, I just wrote it and ran it).

Quote:

#include
#include
#include

main() {

int bankroll=10000;
int count = 0;
int ball;
int betUnit = 1;
int numberOfRounds=0;
srand(time(NULL));

while (bankroll > 0) {

numberOfRounds++;
ball = rand()%37;

if (ball == 0) {
bankroll += 35*betUnit;
betUnit = 1;
count = 0;
}

else {
count++;
bankroll -= betUnit;

if (count == 34) {
betUnit *= 2;
count = 0;
}

}

}

printf("Number of rounds to bust: %d\n", numberOfRounds);
}

Quote: DJTeddyBear

Are you saying that, if you *never* hit your number, you are betting $1 34 times, and then you're betting $2 34 times, then betting $3 34 times, etc?

Yes, I reslize that it's unlikely to happen, but not impossible either, that your number will not hit 68 or more times in a row.

If my interpretation is correct, then you only need to lose 52 times in a row before a win will not cover your losses.

I.E.: If you have 34 $1 losses, then 18 $2 losses, you're down $70. a win on spin 52, if you've still betting $2, will return $70.

In order to ALWAYS have a positive expectation, you have to shorten the count. I.E. Increase your bet when the net would be zero.

Check out this chart:

Spin	Bet	Invested	Payoff	Net	Expectation
1	1	1	35	34	0.9722
2	1	2	35	33	0.9452
3	1	3	35	32	0.9190
4	1	4	35	31	0.8934
5	1	5	35	30	0.8686
6	1	6	35	29	0.8445
7	1	7	35	28	0.8210
8	1	8	35	27	0.7982
9	1	9	35	26	0.7761
10	1	10	35	25	0.7545
11	1	11	35	24	0.7335
12	1	12	35	23	0.7132
13	1	13	35	22	0.6933
14	1	14	35	21	0.6741
15	1	15	35	20	0.6554
16	1	16	35	19	0.6372
17	1	17	35	18	0.6195
18	1	18	35	17	0.6023
19	1	19	35	16	0.5855
20	1	20	35	15	0.5693
21	1	21	35	14	0.5534
22	1	22	35	13	0.5381
23	1	23	35	12	0.5231
24	1	24	35	11	0.5086
25	1	25	35	10	0.4945
26	1	26	35	9	0.4807
27	1	27	35	8	0.4674
28	1	28	35	7	0.4544
29	1	29	35	6	0.4418
30	1	30	35	5	0.4295
31	1	31	35	4	0.4176
32	1	32	35	3	0.4060
33	1	33	35	2	0.3947
34	1	34	35	1	0.3837
35	2	36	70	34	0.3731
36	2	38	70	32	0.3627
37	2	40	70	30	0.3526
38	2	42	70	28	0.3428
39	2	44	70	26	0.3333
40	2	46	70	24	0.3241
41	2	48	70	22	0.3151
42	2	50	70	20	0.3063
43	2	52	70	18	0.2978
44	2	54	70	16	0.2895
45	2	56	70	14	0.2815
46	2	58	70	12	0.2737
47	2	60	70	10	0.2661
48	2	62	70	8	0.2587
49	2	64	70	6	0.2515
50	2	66	70	4	0.2445
51	2	68	70	2	0.2377
52	3	71	105	34	0.2311
53	3	74	105	31	0.2247
54	3	77	105	28	0.2184
55	3	80	105	25	0.2124
56	3	83	105	22	0.2065
57	3	86	105	19	0.2007
58	3	89	105	16	0.1952
59	3	92	105	13	0.1897
60	3	95	105	10	0.1845
61	3	98	105	7	0.1793
62	3	101	105	4	0.1744
63	3	104	105	1	0.1695
64	4	108	140	32	0.1648
65	4	112	140	28	0.1602
66	4	116	140	24	0.1558
67	4	120	140	20	0.1515
68	4	124	140	16	0.1473
69	4	128	140	12	0.1432
70	4	132	140	8	0.1392
71	4	136	140	4	0.1353
72	5	141	175	34	0.1316
73	5	146	175	29	0.1279
74	5	151	175	24	0.1244
75	5	156	175	19	0.1209
76	5	161	175	14	0.1175
77	5	166	175	9	0.1143
78	5	171	175	4	0.1111
79	6	177	210	33	0.1080
80	6	183	210	27	0.1050
81	6	189	210	21	0.1021
82	6	195	210	15	0.0993
83	6	201	210	9	0.0965
84	6	207	210	3	0.0938
85	7	214	245	31	0.0912
86	7	221	245	24	0.0887
87	7	228	245	17	0.0862
88	7	235	245	10	0.0838
89	7	242	245	3	0.0815
90	8	250	280	30	0.0792
91	8	258	280	22	0.0770
92	8	266	280	14	0.0749
93	8	274	280	6	0.0728
94	9	283	315	32	0.0708
95	9	292	315	23	0.0688
96	9	301	315	14	0.0669
97	9	310	315	5	0.0651
98	10	320	350	30	0.0632
99	10	330	350	20	0.0615
100	10	340	350	10	0.0598
101	11	351	385	34	0.0581
102	11	362	385	23	0.0565
103	11	373	385	12	0.0549
104	11	384	385	1	0.0534
105	12	396	420	24	0.0519
106	12	408	420	12	0.0505
107	13	421	455	34	0.0491
108	13	434	455	21	0.0477
109	13	447	455	8	0.0464
110	14	461	490	29	0.0451
111	14	475	490	15	0.0439
112	14	489	490	1	0.0426
113	15	504	525	21	0.0414
114	15	519	525	6	0.0403
115	16	535	560	25	0.0392
116	16	551	560	9	0.0381
117	17	568	595	27	0.0370
118	17	585	595	10	0.0360
119	18	603	630	27	0.0350
120	18	621	630	9	0.0340
121	19	640	665	25	0.0331
122	19	659	665	6	0.0322
123	20	679	700	21	0.0313
124	20	699	700	1	0.0304
125	21	720	735	15	0.0296
126	22	742	770	28	0.0287
127	22	764	770	6	0.0279
128	23	787	805	18	0.0272
129	24	811	840	29	0.0264
130	24	835	840	5	0.0257
131	25	860	875	15	0.0250
132	26	886	910	24	0.0243
133	27	913	945	32	0.0236
134	27	940	945	5	0.0229
135	28	968	980	12	0.0223
136	29	997	1015	18	0.0217
137	30	1027	1050	23	0.0211
138	31	1058	1085	27	0.0205
139	32	1090	1120	30	0.0199
140	33	1123	1155	32	0.0194
141	34	1157	1190	33	0.0188
142	35	1192	1225	33	0.0183
143	36	1228	1260	32	0.0178
144	37	1265	1295	30	0.0173
145	38	1303	1330	27	0.0168
146	39	1342	1365	23	0.0164
147	40	1382	1400	18	0.0159
148	41	1423	1435	12	0.0155
149	42	1465	1470	5	0.0150
150	43	1508	1505	-3	0.0146
151	44	1552	1540	-12	0.0142
152	45	1597	1575	-22	0.0138
153	46	1643	1610	-33	0.0134
154	47	1690	1645	-45	0.0131
155	48	1738	1680	-58	0.0127
156	49	1787	1715	-72	0.0123
157	50	1837	1750	-87	0.0120
158	51	1888	1785	-103	0.0117
159	52	1940	1820	-120	0.0113
160	53	1993	1855	-138	0.0110

As you can see, 136 consecutive losses results in a $1000 bankroll being insufficient to cover the next bet.

If you have a bigger bankroll, and suffer 149 consecutive losses, you have to increase your betting unit to keep up.

The Expectation column shows the odds of losing that many times in a row - (35/36)^spin

Yeah, it's unlikely that you'll suffer that many losses, but not impossible either. If you doubt me, think of all the times you passed a roulette table, looed at the board which shows the last dozen spins, and remarked at how many doubles there are. The odds are against that too.

The other obvious broblems are:
- It's time consuming to the point of being boring.
- Look at spin # 63. How'd you like to lose 62 times and then finally hit on # 63? Do I sense a change of strategy coming up? Woo boy. Change the whole chart!

You misread, it is a martingale variant, so the bet is doubled, not incremented by 1 at the end of a cycle. The wins will always cover the losses. The only question is how long will it be until the bank roll or table limit is reached.

Quote: teliot

All betting systems in roulette lose at the rate


Expected loss = (Total Wagers)*(House Edge).


This system is no exception.


OK, I wrote a computer program to simulate this system. I didn't answer the question you asked exactly as you asked it, but I hope this helps.

This program allows you to make a FULL SIZE bet for your last bet, even if you don't have the funds to cover it, so it is not giving exact results. You didn't specify what to do in that situation, so I took the liberty, since it made the programming easier. The effect of this change was to make the bust times possibly LONGER than they would be if you were limited by your last wager to available funds.


I ran 20 simulations starting with a 1000 unit bankroll and got these number of rounds to bust:


Number of rounds to bust: 4371
Number of rounds to bust: 36080
Number of rounds to bust: 19883
Number of rounds to bust: 422
Number of rounds to bust: 1346
Number of rounds to bust: 196353
Number of rounds to bust: 8103
Number of rounds to bust: 1558
Number of rounds to bust: 2316
Number of rounds to bust: 56821
Number of rounds to bust: 1159
Number of rounds to bust: 874
Number of rounds to bust: 286
Number of rounds to bust: 10016
Number of rounds to bust: 1593
Number of rounds to bust: 15160
Number of rounds to bust: 23162
Number of rounds to bust: 642
Number of rounds to bust: 28209
Number of rounds to bust: 1759

Here are 20 simulations of the same system starting with a 10,000 unit bankroll.


Number of rounds to bust: 118029
Number of rounds to bust: 20969
Number of rounds to bust: 23796
Number of rounds to bust: 36480
Number of rounds to bust: 213409
Number of rounds to bust: 26667
Number of rounds to bust: 15573
Number of rounds to bust: 62811
Number of rounds to bust: 131183
Number of rounds to bust: 11448
Number of rounds to bust: 2762753
Number of rounds to bust: 852206
Number of rounds to bust: 166302
Number of rounds to bust: 141391
Number of rounds to bust: 41295
Number of rounds to bust: 6922
Number of rounds to bust: 144477
Number of rounds to bust: 215548
Number of rounds to bust: 18867
Number of rounds to bust: 201690


Here is the source code (please double check this code -- I didn't bother checking, I just wrote it and ran it).

Quote:

#include
#include
#include

main() {

int bankroll=10000;
int count = 0;
int ball;
int betUnit = 1;
int numberOfRounds=0;
srand(time(NULL));

while (bankroll > 0) {

numberOfRounds++;
ball = rand()%37;

if (ball == 0) {
bankroll += 35*betUnit;
betUnit = 1;
count = 0;
}

else {
count++;
bankroll -= betUnit;

if (count == 34) {
betUnit *= 2;
count = 0;
}

}

}

printf("Number of rounds to bust: %d\n", numberOfRounds);
}

Thanks, that's really cool. How many BetUnits does it take to hit a bankroll of 10000?

Perhaps this simpler question may help you understand the Martingale fallacy. Typically a table will let you lose 6 times in a row before you hit the table maximum. Assume no house edge (i.e. coin toss).
For you to lose 6 times in a row out of your first 6 tosses is 1/64 (1/2^6).
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The common fallacy is that losing 6 times in a row is a relatively rare circumstance. But the reality is that it takes only 89 tosses of a coin until you have a 50/50 probability of losing 6 tosses in a row. If you are playing Martingale, by the time you get to 89 tosses, there is a very low probability that you will have doubled your bankroll. The more tosses you do the higher the probability is that you will lose 6 in a row.
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If you find a game that lets you lose 7 8 or 9 times in a row before you hit the table maximum, doesn't change the odds at all. Because your bankroll must be larger you will still never get very far before your odds are over 50% that you will lose the required number of times in a row.
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The house edge only hurts you by decreasing the number of tosses before you hit 50%. Keep in mind that many games pay "bonus amounts" (blackjack, field bet in craps). This has the effect of allowing the game to have a much lower probability of a player loss, while keeping the house edge tolerable. It greatly increases the probability of losing 6,7,8 or 9 in a row.
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Define the term "minimum bankroll" required to be the minimum you must have to cover the losses of 6,7,8,or 9 losses in a row. For example if you are playing a game that will have 6 losses in a row you need 63*minimum bet.
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In summary (1) there is a proven psychological block in all human beings that make people think that long losing streaks are unlikely. It has been verified over and over. (2) The mathematics of streak calculation are very high level graduate math, usually involving the Markov transition matrix. (3) Even without a house edge, the odds of you "losing x times in a row" before you double your "minimum bankroll" (see definition above) are 2:1. It is far from a sure thing, and it doesn't matter if you find a table that large maximum/minimum ratio. The highest I've ever seen in vegas is 9 losses in a row.
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Credentials: Master's degree in mathematics.

There is a game you can play that will show you about streaks (and by extension the Martingale theory). Print some blank pieces of paper with a 15 by 15 grid of empty boxes.

It's a betting game with you and two players. Player A will fill out the matrix with H, or T filled in "randomly" by the player. He must fill out the rows in by himself (row by row). Player B will fill out the matrix using some kind of device (flipping a coin or rolling a dice) using H or T (heads or Tails). A dice takes less time because it is tedious to flip a coin 215 times.

You don't know which method the players used. Bet them $100 each that you can tell them if they made up the data, or if they used a dice (or coin) just by looking at their papers. Of course, the game requires them to be honest.

Most people will take this bet thinking there is no way for you to know how the data was derived. Almost everyone who is making up data will think a run of 6 in a row (either H or T) is very unlikely and will stop streaks that are this long. In reality there is much less than a 3% chance that you will NOT have a streak of 6 or more out of 215. Just look at the paper and look for streaks of 6 or more in a row. That will be the data generated by a dice (or coin).

It is harder than most people think to make up random data.

Is it possible to lose 20 times in a row?How often does this happen?

Quote: markos

Is it possible to lose 20 times in a row?How often does this happen?

Not only is it 'possible', it's the normal result. You'll lose 20 times in a row, about 57% of the time.

To stretch things out a bit, about 10% of the time, you'll get 82 losses in a row!

I'm not as much a mathematician as most of you seem to be, but I was a roulette dealer for a casino party company in Virginia before moving to Las Vegas. Frequently, before guests arrived for a party, I would make a number of spins of the wheel and put a chip on each number that came up, to see how many spins it would take to cover every number on the table. The least number of spins I ever made was 88, the most well over 200 (the guests arrived before the number 34 hit) and most frequently the number was between 120 and 130 spins.

The norm was that most numbers would have appeared after about 90 or 100 spins, but the last two or three numbers just wouldn't drop. Incidentally, we were using standard Vegas wheels, not some "play" type wheels that you might think could influence the outcome.

Another roulette "system" which is "guaranteed" to work is the Steele system, where a bet is made on the dozen (pay at 2:1) which came up furthest from the current spin, then increasing the bet in an incremental series (1,1,2,3,6,9,14,21,31 . . .), each payoff of which covers all previous losses. It's an easy setup to check on Excel. HOWEVER, one soon realizes that a very large bet has to be made after a number of consecutive losses, and the table limit makes it imperative to switch to inside bets to cover the losses for a very small gain. SO, I certainly don't recommend it. I did win using it once in Atlantic City, then abandoned it the second time around, since that time the second 12 didn't hit for 24 consecutive spins. (I quit long before that.)

Quote: DJTeddyBear

Quote: markos

Is it possible to lose 20 times in a row?How often does this happen?

Not only is it 'possible', it's the normal result. You'll lose 20 times in a row, about 57% of the time.

To stretch things out a bit, about 10% of the time, you'll get 82 losses in a row!

I think DJTeddyBear meant is it possible to lose 20 times in a row if you were betting black (or red). The odds of coming up 20 times in a row as black or red (where 0 or 00 doesn't break the streak) out of 1000 spins are 0.11844%.

I believe that the all time record is still 26 set almost a century ago.

Actually, I was talking about the chance of losing a NUMBER bet in roulette. I didn't realize that the thread took a turn and was talking about the even money bets.


But regarding your comment: The odds of seeing Red hit 'x' times in a row, is identical to the odds of correctly betting on Red or Black 'x' times in a row.

Note: You can change your bet, even change it to one of the other even money bets. The key is to simplify it. Seeing Red is an 'event'. Correctly guessing is an 'event'. Since the individual odds of the events are the same, so are the odds of consecutive events.

Quote: markos

Is it possible to lose 20 times in a row?How often does this happen?

In roulette, you are expected to lose 20 times in a row on average after 793,564 spins. Considering there are 467 roulette wheels in the state of Nevada, that would amount to only 1700 even money bets per day per wheel. So there is a high probability that someone will lose that many times somewhere in NV every day.

But that does not mean you will be that person. Get on that next plan to Vegas NOW !!