Roulette strategy thoughts
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June 14th, 2018 at 10:44:26 PM
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Newbie here
I found this strategy seems similar to something I was thinking. Does anyone have thoughts.
It was from another site but I cannot post the link because I'm new
"Hey Fellow Roulette Fans
I would like to share my roulette strategy for all. One I use regularly.
I thought I would share my strategy with everyone. Let me preface it by saying…...
I understand nothing can really beat the house edge
Each spin is not dependent on the previous spins
And the martingale is a recipe for disaster
The strategy I use, can be used on any inside bet, but to illustrate I’ll deal with a straight up bet.
I wait for about 50 spins (sometimes I go to around 75) marking down the numbers as they appear. After the final no bet spin I bet on those numbers remaining. Sometimes (rarely there are none left) often there are about 10ish. I realise these past spins have no bearing on the future…....but it eases my conscious.
I then continue to bet on those numbers until they a) appear b) I’m satisfied I have won enough or c) I’ve lost 100 times (you could go longer if you wish….but that is the limit I set myself).
how I bet is follows…...
1 chip for 35
2 chip for 17
3 chip for 12
etc
So the return is greatest at the start of the step and reduces as that step goes on down to the point it would be minimal profitability. Once you take the step to the next level profitability increases again.
So this is the system that I have been employing over the years. Is it 100% perfect and work every time? No! Sometimes the ball just doesn’t drop. But over the long haul I have come out in front. On one occasion I was using it on a 6 number bet…...whilst I was watching the table those 6 numbers went 49+ times before it came up
I also use it on the various other inside bets, I just adjust the numbers and steps to those bet types.
Another more risky variation that I ever so occasionally use is this…........
I treat each of the 37 numbers as a “business”.
I walk in and bet on all 37 s I know I lose on that bet. But that doesn’t worry me. Because that business has now “closed” and it was profitable.
As the numbers come up you “close” those businesses.
I then use the same stepped progression that I outlined above until a) hopefully all the numbers come up within 100 spins b) I’m happy with my winnings or c) 100 spins happens
So in ideal occasions all 37 numbers come up and all business close profitably. On not so successful days there may be a 1 or 2 or maybe more business still operating. And its just weather the winners outweigh the losers.
Is there a program that you could run it on to determine the profitability over a period of time?"
I found this strategy seems similar to something I was thinking. Does anyone have thoughts.
It was from another site but I cannot post the link because I'm new
"Hey Fellow Roulette Fans
I would like to share my roulette strategy for all. One I use regularly.
I thought I would share my strategy with everyone. Let me preface it by saying…...
I understand nothing can really beat the house edge
Each spin is not dependent on the previous spins
And the martingale is a recipe for disaster
The strategy I use, can be used on any inside bet, but to illustrate I’ll deal with a straight up bet.
I wait for about 50 spins (sometimes I go to around 75) marking down the numbers as they appear. After the final no bet spin I bet on those numbers remaining. Sometimes (rarely there are none left) often there are about 10ish. I realise these past spins have no bearing on the future…....but it eases my conscious.
I then continue to bet on those numbers until they a) appear b) I’m satisfied I have won enough or c) I’ve lost 100 times (you could go longer if you wish….but that is the limit I set myself).
how I bet is follows…...
1 chip for 35
2 chip for 17
3 chip for 12
etc
So the return is greatest at the start of the step and reduces as that step goes on down to the point it would be minimal profitability. Once you take the step to the next level profitability increases again.
So this is the system that I have been employing over the years. Is it 100% perfect and work every time? No! Sometimes the ball just doesn’t drop. But over the long haul I have come out in front. On one occasion I was using it on a 6 number bet…...whilst I was watching the table those 6 numbers went 49+ times before it came up
I also use it on the various other inside bets, I just adjust the numbers and steps to those bet types.
Another more risky variation that I ever so occasionally use is this…........
I treat each of the 37 numbers as a “business”.
I walk in and bet on all 37 s I know I lose on that bet. But that doesn’t worry me. Because that business has now “closed” and it was profitable.
As the numbers come up you “close” those businesses.
I then use the same stepped progression that I outlined above until a) hopefully all the numbers come up within 100 spins b) I’m happy with my winnings or c) 100 spins happens
So in ideal occasions all 37 numbers come up and all business close profitably. On not so successful days there may be a 1 or 2 or maybe more business still operating. And its just weather the winners outweigh the losers.
Is there a program that you could run it on to determine the profitability over a period of time?"
June 15th, 2018 at 12:52:02 AM
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The profitability over time would be:
Wager * (0.027) = Casino's Profit
Thank me later.
Wager * (0.027) = Casino's Profit
Thank me later.
June 15th, 2018 at 1:25:06 AM
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Quote: OzrouletteIs there a program that you could run it on to determine the profitability over a period of time?"
You hang with Bourdain.
I am a robot.
June 15th, 2018 at 1:39:34 AM
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Quote: OzrouletteNewbie here
I found this strategy seems similar to something I was thinking. Does anyone have thoughts.
It was from another site but I cannot post the link because I'm new
"Hey Fellow Roulette Fans
I would like to share my roulette strategy for all. One I use regularly....
Hi OzRoulette, Welcome to the forum.
From your user ID and your reference to 'the 37 numbers' do I rightly deduce that you are discussing playing single zero roulette in Australia?
Well, if so, at least you are only throwing 1/37 = 2.7% of your money away on average every spin. In double zero game you throw away 5.4% Nothing ever changes that, unless they have the weird 'en prison' rule.
I've located the post that you quoted from (on a closed down forum)
For the benefit of other posters, I'll point out that the entire 'system' is from that cut and paste, so apart from asking for any thoughts, the original poster here is not confessing to actually advocating or doing this.
Well. Oz, You won't find many members here advocating this system because everything about it is foolish and subject to the same rate of losing money.
Absolutely the only value in ANY roulette system is if it makes playing a little more amusing or if it slows you down. Everything about that system screams "Gamblers' fallacy"
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
June 15th, 2018 at 4:48:32 AM
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Quote: OnceDearHi OzRoulette, Welcome to the forum.
From your user ID and your reference to 'the 37 numbers' do I rightly deduce that you are discussing playing single zero roulette in Australia?
Well, if so, at least you are only throwing 1/37 = 2.7% of your money away on average every spin. In double zero game you throw away 5.4% Nothing ever changes that, unless they have the weird 'en prison' rule.
I've located the post that you quoted from (on a closed down forum)
For the benefit of other posters, I'll point out that the entire 'system' is from that cut and paste, so apart from asking for any thoughts, the original poster here is not confessing to actually advocating or doing this.
Well. Oz, You won't find many members here advocating this system because everything about it is foolish and subject to the same rate of losing money.
Absolutely the only value in ANY roulette system is if it makes playing a little more amusing or if it slows you down. Everything about that system screams "Gamblers' fallacy"
I found that post too, before posting. Odd there are no responses. O_O
Btw, double zero roulette is 5.26% HE, not 5.4%. :)
June 15th, 2018 at 5:17:23 AM
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Quote: OzrouletteEach spin is not dependent on the previous spins
And the martingale is a recipe for disaster
The strategy I use, can be used on any inside bet, but to illustrate I’ll deal with a straight up bet.
I wait for about 50 spins (sometimes I go to around 75) marking down the numbers as they appear.
these are contradicting statements .... and this coming from me should be a big hint that you should stay away from roulette and play blackjack.... im the one who knows how roulette is ususally rigged
June 15th, 2018 at 7:21:00 AM
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100% percent WRONG!! as you are most always are in a betting system forum.Quote: RSThe profitability over time would be:
Wager * (0.027) = Casino's Profit
Thank me later.
you NOW say that over 100 attempts one can NEVER show a profit??
or are you NOW taking back your words??
(one knows you use ev and variance for your risk of ruin always, even at 1000 hands played at video poker. you want those posts of yours???. you lead many down the wrong paths, in my opinion, others beware of this poster!)
this is in the betting system forum where it is safe to 'due the math' and dream of systems on betting systems
yours is useless, imo, and you should go some place else ( a different forum) to spread your manure, in my opinion
the OP has a right to ask safe question(s) and get safe answers instead of 'NO ONE EVER CAN SHOW A PROFIT EVEN AFTER X NUMBER OF BETS'
SUCH A WASTE OF EFFORT
OP, I will return with data you can use as well as others (Roulette lovers) instead of useless words that are meaningless in a betting system forum
I say moderators should delete those posts that are meant to belittle those asking honest question(s) in a safe place (the proper forum)
Sally
thank you for sharing
I Heart Vi Hart
June 15th, 2018 at 7:37:22 AM
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Seeing the name of this thread makes me chuckle
Whats next?
Coin flipping strategy?
Whats next?
Coin flipping strategy?
When somebody doesn't believe me, I could care less. Some get totally bent out of shape when not believed. Weird. I believe very little on all forums
June 15th, 2018 at 7:43:26 AM
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NastyQuote: mustangsally100% percent WRONG!! as you are most always are in a betting system forum.
you NOW say that over 100 attempts one can NEVER show a profit??
or are you NOW taking back your words??
(one knows you use ev and variance for your risk of ruin always, even at 1000 hands played at video poker. you want those posts of yours???. you lead many down the wrong paths, in my opinion, others beware of this poster!)
this is in the betting system forum where it is safe to 'due the math' and dream of systems on betting systems
yours is useless, imo, and you should go some place else ( a different forum) to spread your manure, in my opinion
the OP has a right to ask safe question(s) and get safe answers instead of 'NO ONE EVER CAN SHOW A PROFIT EVEN AFTER X NUMBER OF BETS'
SUCH A WASTE OF EFFORT
OP, I will return with data you can use as well as others (Roulette lovers) instead of useless words that are meaningless in a betting system forum
I say moderators should delete those posts that are meant to belittle those asking honest question(s) in a safe place (the proper forum)
Sally
thank you for sharing
I am a robot.
June 15th, 2018 at 8:52:59 AM
permalink
well yes.Quote: Ozroulette
I found this strategy seems similar to something I was thinking. Does anyone have thoughts.
ok. tell what U know then and now. Oh, you did. ok thanks!Quote: OzrouletteIt was from another site but I cannot post the link because I'm new
many do not like to wait and use the history board as a start. That can also be a disaster too. just saying.Quote: Ozroulette"Hey Fellow Roulette Fans
I would like to share my roulette strategy for all.
One I use regularly.
I thought I would share my strategy with everyone.
Let me preface it by saying…...
I understand nothing can really beat the house edge
Each spin is not dependent on the previous spins
And the martingale is a recipe for disaster
The strategy I use, can be used on any inside bet, but to illustrate I’ll deal with a straight up bet.
XXX
I wait for about 50 spins (sometimes I go to around 75) marking down the numbers as they appear.
For 37 or 38 numbers:Quote: OzrouletteAfter the final no bet spin I bet on those numbers remaining.
Sometimes (rarely there are none left) often there are about 10ish.
[1] "Number of coupons:37, draws:50, mean number drawn:27.5976, mean number NOT drawn:9.40244"
[1] "Number of coupons:37, draws:75, mean number drawn:32.2602, mean number NOT drawn:4.7398"
[1] "Number of coupons:38, draws:50, mean number drawn:27.9841, mean number NOT drawn:10.0159"
[1] "Number of coupons:38, draws:75, mean number drawn:32.8579, mean number NOT drawn:5.14213"
complete distribution results:
> cc.draws(37,50)#0 Roulette, 37 numbers and 50 spins
[1] "Number of coupons:37, draws:50, mean number drawn:27.5976, mean number NOT drawn:9.40244"
not drawn Probability cumulative
u=1 36 1.439181344232344e-77 1.439181344232344e-77
u=2 35 2.916673454521503e-61 2.916673454521503e-61
u=3 34 2.169689310782557e-51 2.169689311074224e-51
u=4 33 3.256509307369646e-44 3.256509524338577e-44
u=5 32 1.505798638567348e-38 1.505801895076873e-38
u=6 31 7.304197461521032e-34 7.304348041710539e-34
u=7 30 7.179497809932151e-30 7.180228244736322e-30
u=8 29 2.121765593797005e-26 2.122483616621478e-26
u=9 28 2.432017408248281e-23 2.434139891864902e-23
u=10 27 1.285931681171704e-20 1.288365821063569e-20
u=11 26 3.547531055279904e-18 3.56041471349054e-18
u=12 25 5.589083606082646e-16 5.624687753217551e-16
u=13 24 5.383356155861824e-14 5.439603033393999e-14
u=14 23 3.34092126479809e-12 3.395317295132031e-12
u=15 22 1.392067352520005e-10 1.426020525471325e-10
u=16 21 4.023901277345215e-09 4.166503329892348e-09
u=17 20 8.283924725256978e-08 8.700575058246212e-08
u=18 19 1.240625906322626e-06 1.327631656905088e-06
u=19 18 1.375045122905984e-05 1.507808288596493e-05
u=20 17 0.0001143607416032031 0.000129438824489168
u=21 16 0.0007216381065122463 0.0008510769310014144
u=22 15 0.003484928230698981 0.004336005161700395
u=23 14 0.01296328619088658 0.01729929135258697
u=24 13 0.03731105292310735 0.05461034427569433
u=25 12 0.08330893197328543 0.1379192762489798
u=26 11 0.1444060772175798 0.2823253534665595
u=27 10 0.1940739990523021 0.4763993525188616
u=28 9 0.2015301713524824 0.677929523871344
u=29 8 0.1607354571537463 0.8386649810250904
u=30 7 0.09757780328449239 0.9362427843095827
u=31 6 0.04450320423805204 0.9807459885476348
u=32 5 0.01496820827470736 0.9957141968223422
u=33 4 0.003614667881269277 0.9993288647036115
u=34 3 0.0006021379495103832 0.9999310026531218
u=35 2 6.48743932454933e-05 0.9999958770463673
u=36 1 4.015970349498736e-06 0.9999998930167168
u=37 0 1.069832832034557e-07 1
[1] "Number of coupons:37, draws:50, mean number drawn:27.5976, mean number NOT drawn:9.40244"
> cc.draws(37,75)#0 Roulette, 37 numbers and 75 spins
[1] "Number of coupons:37, draws:75, mean number drawn:32.2602, mean number NOT drawn:4.7398"
not drawn Probability cumulative
u=1 36 8.97578454133916e-117 8.97578454133916e-117
u=2 35 6.103719946869055e-93 6.103719946869055e-93
u=3 34 1.146531042334978e-78 1.146531042334984e-78
u=4 33 2.286704194269901e-68 2.286704194384554e-68
u=5 32 2.799012146681416e-60 2.799012169548458e-60
u=6 31 1.295967409524145e-53 1.295967689425362e-53
u=7 30 6.023390116725599e-48 6.023403076402492e-48
u=8 29 5.048261189072503e-43 5.048321423103268e-43
u=9 28 1.115121080592597e-38 1.115171563806828e-38
u=10 27 8.418854084307068e-35 8.419969255870874e-35
u=11 26 2.615284137492938e-31 2.616126134418525e-31
u=12 25 3.833087872949563e-28 3.835703999083981e-28
u=13 24 2.939818407467977e-25 2.943654111467061e-25
u=14 23 1.278205851026686e-22 1.281149505138153e-22
u=15 22 3.355322489905352e-20 3.368133984956733e-20
u=16 21 5.591609826558039e-18 5.625291166407607e-18
u=17 20 6.160487352374294e-16 6.21674026403837e-16
u=18 19 4.63772146284193e-14 4.699888865482314e-14
u=19 18 2.451038282992436e-12 2.498037171647259e-12
u=20 17 9.298279229513099e-11 9.548082946677825e-11
u=21 16 2.578623784951427e-09 2.674104614418205e-09
u=22 15 5.306139039421077e-08 5.573549500862898e-08
u=23 14 8.199585389310487e-07 8.756940339396777e-07
u=24 13 9.605731422696719e-06 1.04814254566364e-05
u=25 12 8.591580310120604e-05 9.639722855784243e-05
u=26 11 0.0005895494675250511 0.0006859466960828936
u=27 10 0.003111671966865291 0.003797618662948184
u=28 9 0.01263524241407783 0.01643286107702602
u=29 8 0.039375533774433 0.05580839485145902
u=30 7 0.09364574461168275 0.1494541394631418
u=31 6 0.1683414903417944 0.3177956298049362
u=32 5 0.2253100086654047 0.5431056384703409
u=33 4 0.2193756147616125 0.7624812532319534
u=34 3 0.1498461115191942 0.9123273647511476
u=35 2 0.06759223387508045 0.979919598626228
u=36 1 0.01796370592864979 0.9978833045548778
u=37 0 0.002116695445122071 1
[1] "Number of coupons:37, draws:75, mean number drawn:32.2602, mean number NOT drawn:4.7398"
> cc.draws(38,50)#00 Roulette, 38 numbers and 50 spins
[1] "Number of coupons:38, draws:50, mean number drawn:27.9841, mean number NOT drawn:10.0159"
not drawn Probability cumulative
u=1 37 3.895863802749243e-78 3.895863802749243e-78
u=2 36 8.114752481285795e-62 8.114752481285795e-62
u=3 35 6.20896875196668e-52 6.208968752778155e-52
u=4 34 9.5931972844765e-45 9.593197905373375e-45
u=5 33 4.570281276377452e-39 4.570290869575357e-39
u=6 32 2.286190923452339e-34 2.286236626361035e-34
u=7 31 2.319649365836473e-30 2.319877989499109e-30
u=8 30 7.083797162255486e-27 7.086117040244985e-27
u=9 29 8.399600882236821e-24 8.406686999277067e-24
u=10 28 4.599915288493157e-21 4.608321975492434e-21
u=11 27 1.315989460008252e-18 1.320597781983745e-18
u=12 26 2.153065427287628e-16 2.166271405107465e-16
u=13 25 2.156766223665314e-14 2.178428937716389e-14
u=14 24 1.394263852948971e-12 1.416048142326135e-12
u=15 23 6.062090022704067e-11 6.20369483693668e-11
u=16 22 1.831954173169448e-09 1.893991121538815e-09
u=17 21 3.950998076633495e-08 4.140397188787376e-08
u=18 20 6.212992354330983e-07 6.62703207320972e-07
u=19 19 7.248586372017541e-06 7.911289579338513e-06
u=20 18 6.363476242070619e-05 7.15460520000447e-05
u=21 17 0.0004251679372204709 0.0004967139892205157
u=22 16 0.002181543237810482 0.002678257227030997
u=23 15 0.008655931663796607 0.0113341888908276
u=24 14 0.0266931261253937 0.03802731501622131
u=25 13 0.06418568547470954 0.1022130004909308
u=26 12 0.1205297327608814 0.2227427332518122
u=27 11 0.1767114290150352 0.3994541622668474
u=28 10 0.2018505993464849 0.6013047616133324
u=29 9 0.1788789139554827 0.7801836755688151
u=30 8 0.1221661869765338 0.9023498625453489
u=31 7 0.06367709033345331 0.9660269528788021
u=32 6 0.02498667892683492 0.9910136318056371
u=33 5 0.007240830227813003 0.9982544620334501
u=34 4 0.001507738334938373 0.9997622003683885
u=35 3 0.0002165918064271544 0.9999787921748157
u=36 2 2.011177821573497e-05 0.9999989039530314
u=37 1 1.071533839809168e-06 0.9999999754868711
u=38 0 2.451312857572011e-08 1
[1] "Number of coupons:38, draws:50, mean number drawn:27.9841, mean number NOT drawn:10.0159"
> cc.draws(38,75)#00 Roulette, 38 numbers and 75 spins
[1] "Number of coupons:38, draws:75, mean number drawn:32.8579, mean number NOT drawn:5.14213"
not drawn Probability cumulative
u=1 37 1.247423295962515e-117 1.247423295962515e-117
u=2 36 8.71836914495016e-94 8.71836914495016e-94
u=3 35 1.684460877761416e-79 1.684460877761424e-79
u=4 34 3.458392069877782e-69 3.458392070046228e-69
u=5 33 4.361481138188797e-61 4.361481172772717e-61
u=6 32 2.082510775594094e-54 2.082511211742211e-54
u=7 31 9.991310449682353e-49 9.99133127479447e-49
u=8 30 8.652940495876285e-44 8.653040409189032e-44
u=9 29 1.977275492118061e-39 1.977362022522152e-39
u=10 28 1.546101873495314e-35 1.546299609697566e-35
u=11 27 4.980790453050885e-32 4.982336752660583e-32
u=12 26 7.580861860048949e-29 7.585844196801609e-29
u=13 25 6.046772810463679e-26 6.054358654660481e-26
u=14 24 2.738625923901267e-23 2.744680282555928e-23
u=15 23 7.501525453127065e-21 7.528972255952625e-21
u=16 22 1.30694506736379e-18 1.314474039619742e-18
u=17 21 1.508477827151387e-16 1.521622567547585e-16
u=18 20 1.192388617132723e-14 1.207604842808199e-14
u=19 19 6.633452887571066e-13 6.754213371851886e-13
u=20 18 2.656276219792878e-11 2.723818353511397e-11
u=21 17 7.799777576322341e-10 8.072159411673481e-10
u=22 16 1.705303952119028e-08 1.786025546235763e-08
u=23 15 2.810889603657615e-07 2.989492158281191e-07
u=24 14 3.528137962768036e-06 3.827087178596155e-06
u=25 13 3.398387182833988e-05 3.781095900693603e-05
u=26 12 0.0002526283937582426 0.0002904393527651786
u=27 11 0.001454602325445919 0.001745041678211097
u=28 10 0.006497207454121986 0.008242249132333084
u=29 9 0.02249712907624844 0.03073937820858152
u=30 8 0.06019233842121695 0.09093171662979847
u=31 7 0.1236620048277582 0.2145937214575567
u=32 6 0.1930956051092642 0.4076893265668209
u=33 5 0.2256116396933839 0.6333009662602047
u=34 4 0.192632092212368 0.8259330584725727
u=35 3 0.1158560454380705 0.9417891039106433
u=36 2 0.04618586657792974 0.987974970488573
u=37 1 0.01088432574022525 0.9988592962287982
u=38 0 0.001140703771201369 1
[1] "Number of coupons:38, draws:75, mean number drawn:32.8579, mean number NOT drawn:5.14213"
>
as to the math of how many numbers were NOT hit after 50 (75) spins, I used a simple Markov chain to return the values. Others could use the binomial probability distribution (and have) or other methods, like simulation.
the R code I used can be found here for those so interested
https://sites.google.com/view/krapstuff/roulette
and might be more fun playing that way instead of another way more boring.Quote: OzrouletteI realise these past spins have no bearing on the future…....but it eases my conscious.
nothing at all wrong with playing a way to have fun.
all of them? 37?Quote: OzrouletteI then continue to bet on those numbers until they a) appear
the chance that all do happen in 100 spins or less is less than 10%
(about 0.0640027995281959)
I calculated it
> cc.draws(37,100)#0 Roulette, 37 numbers and 100 spins
[1] "Number of coupons:37, draws:100, mean number drawn:34.6107, mean number NOT drawn:2.38935"
not drawn Probability cumulative
u=1 36 5.597953896179504e-156 5.597953896179504e-156
u=2 35 1.277324930977542e-124 1.277324930977542e-124
u=3 34 6.058625160388693e-106 6.058625160388693e-106
u=4 33 1.605708367809435e-92 1.605708367809496e-92
u=5 32 5.202497815395365e-82 5.202497815555936e-82
u=6 31 2.297921604000543e-73 2.297921609203041e-73
u=7 30 5.038221333144894e-66 5.038221562937055e-66
u=8 29 1.189866686068989e-59 1.189867189891145e-59
u=9 28 4.998982061590965e-54 4.998993960262864e-54
u=10 27 5.268697647465129e-49 5.268747637404732e-49
u=11 26 1.781195057294679e-44 1.781247744771053e-44
u=12 25 2.31653426118251e-40 2.316712385956987e-40
u=13 24 1.330532850492093e-36 1.330764521730689e-36
u=14 23 3.756479892889646e-33 3.757810657411377e-33
u=15 22 5.67325555130601e-30 5.677013361963421e-30
u=16 21 4.904597146204352e-27 4.910274159566315e-27
u=17 20 2.564163322450434e-24 2.569073596610001e-24
u=18 19 8.479615906410877e-22 8.505306642376976e-22
u=19 18 1.840679633796656e-19 1.849184940439033e-19
u=20 17 2.704381355267079e-17 2.72287320467147e-17
u=21 16 2.758618555004584e-15 2.785847287051298e-15
u=22 15 1.995255701961448e-13 2.023114174831961e-13
u=23 14 1.041163442355477e-11 1.061394584103796e-11
u=24 13 3.975278762478615e-10 4.081418220888995e-10
u=25 12 1.123015740446178e-08 1.163829922655068e-08
u=26 11 2.367162384597664e-07 2.48354537686317e-07
u=27 10 3.744508462602827e-06 3.992863000289144e-06
u=28 9 4.458690879700829e-05 4.857977179729744e-05
u=29 8 0.0003996771183387274 0.0004482568901360248
u=30 7 0.002688234912218009 0.003136491802354034
u=31 6 0.01346527500378389 0.01660176680613792
u=32 5 0.04957159833245486 0.06617336513859279
u=33 4 0.1312872950702168 0.1974606602088096
u=34 3 0.2416185252853372 0.4390791854941468
u=35 2 0.2913189792734887 0.7303981647676355
u=36 1 0.2055990357041687 0.9359972004718041
u=37 0 0.0640027995281959 1
[1] "Number of coupons:37, draws:100, mean number drawn:34.6107, mean number NOT drawn:2.38935"
>
Oh there is more
that is fine, but very hard for a computer to duplicate as it does not understand 'won enough'Quote: Ozrouletteb) I’m satisfied I have won enough
got it.Quote: Ozrouletteor c) I’ve lost 100 times (you could go longer if you wish….but that is the limit I set myself).
okQuote: Ozroulette<snip>
So this is the system that I have been employing over the years.
Is it 100% perfect and work every time? No!
Sometimes the ball just doesn’t drop.
But over the long haul I have come out in front.
<snip 2nd part now>
yes, of course. there are many actually and ways to calculate too.Quote: OzrouletteIs there a program that you could run it on to determine the profitability over a period of time?"
Some say doing a simulation or doing the math yourself is rewarding to you and has more meaning than others saying things like "you will always lose X% of all the bets you make when you play forever and forever or close to it".
Thank you for sharing!
Sally
I Heart Vi Hart
June 15th, 2018 at 8:56:07 AM
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'NO ONE EVER CAN SHOW A PROFIT EVEN AFTER X NUMBER OF BETS'Quote: onenickelmiracleNasty
I agree
"ALL Blackjack card counters will show a lifetime profit." "Buy my book and learn how to do it!"
"No, buy my book and software. the game of 21 CAN be beaten"
It goes on and on
I agree "NASTY"
Sally
I Heart Vi Hart
June 15th, 2018 at 10:16:25 AM
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Thank you for your reply. Your math skills are far beyond mine!!!
I read the strategy a little different. I read it as though after waiting the 50 (75) spins you then start betting on only remaining numbers for up to 100 more spins using. So I guess the maths would essentially be, what would the odds of all 37 numbers coming up within 150 or 175 spins.
At my local casino the longest losing streak I've seen on the board was about 270 spins.
Ps i didn't mean to cause conflict in the forum
I read the strategy a little different. I read it as though after waiting the 50 (75) spins you then start betting on only remaining numbers for up to 100 more spins using. So I guess the maths would essentially be, what would the odds of all 37 numbers coming up within 150 or 175 spins.
At my local casino the longest losing streak I've seen on the board was about 270 spins.
Ps i didn't mean to cause conflict in the forum
June 15th, 2018 at 10:32:55 AM
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thank you for the nice words. my math skills are really poor, really. I just now how to do some math. not allQuote: OzrouletteThank you for your reply. Your math skills are far beyond mine!!!
conflict comes and goes.Quote: OzrouletteSo I guess the maths would essentially be, what would the odds of all 37 numbers coming up within 150 or 175 spins.
At my local casino the longest losing streak I've seen on the board was about 270 spins.
Ps i didn't mean to cause conflict in the forum
I might have misunderstood what was requested
so for 150 or 175 spins
the calculator I linked too returned this
0.4693558523937973 for NOT getting all 37 numbers in 150 spins
(getting them all at least 1 time: 0 0.5306441476062024)
about the same as a fair coin flip
[1] "Number of numbers:37, spins:150, mean numbers show:36.3928, mean numbers NOT shown:0.607181"
0.2690448249887037 for NOT getting all 37 numbers in 175 spins
(getting them all at least 1 time: 0 0.7309551750112959 ... close to 3 out of 4)
[1] "Number of numbers:37, spins:175, mean numbers show:36.6939, mean numbers NOT shown:0.306082"
> cc.draws(37,150)#0 Roulette, 37 numbers and 150 spins
[1] "Number of numbers:37, spins:150, mean numbers show:36.3928, mean numbers NOT shown:0.607181"
not drawn Probability cumulative
u=1 36 2.177424544122785e-234 2.177424544122785e-234
u=2 35 5.593903481953029e-188 5.593903481953029e-188
u=3 34 1.691806217552409e-160 1.691806217552409e-160
u=4 33 7.917353455855316e-141 7.917353455855316e-141
u=5 32 1.79732212463243e-125 1.797322124632431e-125
u=6 31 7.224562549300243e-113 7.22456254930204e-113
u=7 30 3.524468664093874e-102 3.524468664166119e-102
u=8 29 6.60567133118469e-93 6.605671334709158e-93
u=9 28 1.002191559641911e-84 1.002191566247582e-84
u=10 27 2.049896079225596e-77 2.049896179444752e-77
u=11 26 8.139605169720873e-71 8.139607219617053e-71
u=12 25 8.216206407780575e-65 8.216214547387795e-65
u=13 24 2.587978440302679e-59 2.587986656517227e-59
u=14 23 2.983255293484949e-54 2.983281173351514e-54
u=15 22 1.427881780683184e-49 1.427911613494917e-49
u=16 21 3.141072332524691e-45 3.14121512368604e-45
u=17 20 3.44979520742426e-41 3.450109328936629e-41
u=18 19 2.025068936374158e-37 2.025413947307051e-37
u=19 18 6.723576407237209e-34 6.725601821184516e-34
u=20 17 1.323795287964448e-30 1.324467848146567e-30
u=21 16 1.608223449016317e-27 1.609547916864464e-27
u=22 15 1.246460072185899e-24 1.248069620102764e-24
u=23 14 6.338652290947639e-22 6.351132987148666e-22
u=24 13 2.165107140962139e-19 2.171458273949288e-19
u=25 12 5.064474486323305e-17 5.086189069062798e-17
u=26 11 8.240693411766362e-15 8.29155530245699e-15
u=27 10 9.441856006721884e-13 9.524771559746454e-13
u=28 9 7.684553070333018e-11 7.779800785930483e-11
u=29 8 4.465897058034082e-09 4.543695065893387e-09
u=30 7 1.855594798028902e-07 1.901031748687836e-07
u=31 6 5.493716400334692e-06 5.683819575203476e-06
u=32 5 0.0001148034098288437 0.0001204872294040472
u=33 4 0.001663095455592935 0.001783582684996982
u=34 3 0.01618201085930923 0.01796559354430621
u=35 2 0.09998409037823851 0.1179496839225447
u=36 1 0.3514061684712526 0.4693558523937973
u=37 0 0.5306441476062024 1
[1] "Number of numbers:37, spins:150, mean numbers show:36.3928, mean numbers NOT shown:0.607181"
> cc.draws(37,175)#0 Roulette, 37 numbers and 175 spins
[1] "Number of numbers:37, spins:175, mean numbers show:36.6939, mean numbers NOT shown:0.306082"
not drawn Probability cumulative
u=1 36 1.358000757958306e-273 1.358000757958306e-273
u=2 35 1.170635684660775e-219 1.170635684660775e-219
u=3 34 8.940028082700988e-188 8.940028082700988e-188
u=4 33 5.559512553120152e-165 5.559512553120152e-165
u=5 32 3.340665000923602e-147 3.340665000923602e-147
u=6 31 1.281001721577693e-132 1.281001721577696e-132
u=7 30 2.947824014892597e-120 2.947824014893878e-120
u=8 29 1.556408310825757e-109 1.556408310855235e-109
u=9 28 4.487148616277543e-100 4.487148617833951e-100
u=10 27 1.278466330326644e-91 1.278466334813792e-91
u=11 26 5.500219820489394e-84 5.500219948336027e-84
u=12 25 4.888428938811648e-77 4.888429488833642e-77
u=13 24 1.139019525217914e-70 1.139020014060863e-70
u=14 23 8.373835925572125e-65 8.373847315772266e-65
u=15 22 2.249583466579336e-59 2.249591840426652e-59
u=16 21 2.485319020306598e-54 2.485341516225002e-54
u=17 20 1.243429022617506e-49 1.243453876032668e-49
u=18 19 3.050239234945467e-45 3.05036358033307e-45
u=19 18 3.919563936949707e-41 3.91986897330774e-41
u=20 17 2.788644100561646e-37 2.789036087458977e-37
u=21 16 1.150961611062996e-33 1.151240514671742e-33
u=22 15 2.866495402035986e-30 2.867646642550658e-30
u=23 14 4.453665420764376e-27 4.456533067406927e-27
u=24 13 4.439460918751094e-24 4.443917451818502e-24
u=25 12 2.906398039341283e-21 2.910841956793102e-21
u=26 11 1.273946124615658e-18 1.276856966572451e-18
u=27 10 3.796591089135989e-16 3.809359658801713e-16
u=28 9 7.782606538932018e-14 7.820700135520034e-14
u=29 8 1.10590189515704e-11 1.11372259529256e-11
u=30 7 1.093259392123883e-09 1.104396618076809e-09
u=31 6 7.508811995441778e-08 7.619251657249459e-08
u=32 5 3.556100103331281e-06 3.632292619903776e-06
u=33 4 0.0001142459980573045 0.0001178782906772083
u=34 3 0.002416160947304723 0.002534039237981931
u=35 2 0.03184740726844108 0.03438144650642302
u=36 1 0.2346633784822807 0.2690448249887037
u=37 0 0.7309551750112959 1
[1] "Number of numbers:37, spins:175, mean numbers show:36.6939, mean numbers NOT shown:0.306082"
>
Hope this helps
ask if you have more questions or other observations that are interesting to you
it is your thread
again
thank you for sharing
added:
maybe it would still be fun to play that system but only go for a subset of the numbers remaining.
In other words, not all 37 numbers, maybe 34 of them, for example
I have some R code for that too!
Sally
I Heart Vi Hart
June 15th, 2018 at 11:21:07 AM
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add some more to what looks like your question.Quote: OzrouletteIs there a program that you could run it on to determine the profitability over a period of time?"
Some parameters need to be more specific.
like, exactly how many #s are going to be hit before betting system stops and starts again?
not
I then continue to bet on those numbers until they b) I’m satisfied I have won enough
that can be made into a condition the computer can use like win $100 and start over
or win $56 and start over making these exact wagers (xxxxxxxxxx).
*****
the bottom line that will show with simulation and/or calculations
the longer one makes bets against a casino game that has a house edge,
the less chance one has of showing a profit
or the greater the chance one has of showing a net loss.
NOT, as many other so called experts want all to believe, you will lose this $$$ every time...
How do you think you would play the game of Roulette and what would be your goals?
Sally
I Heart Vi Hart
June 15th, 2018 at 2:47:31 PM
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In relation to this response, please excuse my ignorance of the maths, the mean number of numbers spun would be 36.7 approx. And the odds of them all coming out would be 75%.
As far as I understand the strategy, after waiting the spins 50 (75) the gambler would bet like this on the remaining numbers.
35 games x 1 chip
17 x 2
12 x 3
9 x 4
7 x 5
6 x 6
5 x 7
4 x 8
4 x 9
3 x 10
If I'm correct in interpreting what you said 25% of the time a person would lose 345 units for each number that doesn't come up during the 100 spins. On the flip side 75% of the time you would come out in front as all the numbers would be hit.
As far as I understand the strategy, after waiting the spins 50 (75) the gambler would bet like this on the remaining numbers.
35 games x 1 chip
17 x 2
12 x 3
9 x 4
7 x 5
6 x 6
5 x 7
4 x 8
4 x 9
3 x 10
If I'm correct in interpreting what you said 25% of the time a person would lose 345 units for each number that doesn't come up during the 100 spins. On the flip side 75% of the time you would come out in front as all the numbers would be hit.
June 15th, 2018 at 7:10:56 PM
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26.9% for NOT getting all 37 numbers in 175 spinsQuote: OzrouletteOn the flip side 75% of the time you would come out in front as all the numbers would be hit.
getting them all at least 1 time: 73.1% ... close to 3 out of 4
that about 75% was for 175 spins
not the 150 spins
from there you need the average win given a win (the sumproduct)
and one can then easily see
how well this system would 'work' over any number of times attempted
that could add or take away from the fun
Sally
I Heart Vi Hart
June 15th, 2018 at 9:44:42 PM
permalink
Quote: mustangsally100% percent WRONG!! as you are most always are in a betting system forum.
you NOW say that over 100 attempts one can NEVER show a profit??
or are you NOW taking back your words??
(one knows you use ev and variance for your risk of ruin always, even at 1000 hands played at video poker. you want those posts of yours???. you lead many down the wrong paths, in my opinion, others beware of this poster!)
this is in the betting system forum where it is safe to 'due the math' and dream of systems on betting systems
yours is useless, imo, and you should go some place else ( a different forum) to spread your manure, in my opinion
the OP has a right to ask safe question(s) and get safe answers instead of 'NO ONE EVER CAN SHOW A PROFIT EVEN AFTER X NUMBER OF BETS'
SUCH A WASTE OF EFFORT
OP, I will return with data you can use as well as others (Roulette lovers) instead of useless words that are meaningless in a betting system forum
I say moderators should delete those posts that are meant to belittle those asking honest question(s) in a safe place (the proper forum)
Sally
thank you for sharing
What's wrong?
June 15th, 2018 at 9:46:57 PM
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Just looking at my WinCraps results for the boxcars (12) frequency and out of 393 of them, 25 were 101 rolls or more between them including 3 that were more than 200 rolls. That's out of 14,138 rolls though. So that's nearly 1 out of 16 times it will take more than 100 rolls for a boxcar (12) to show up. Looking at snake eyes (2) I've got 19 over 100 rolls with 1 over 200 rolls, out of 403 snake eyes. So that's closer to 1 out of 21 times waiting over 100 rolls for snake eyes.
I'm a new poster so I have to wait some time before I can post enough to respond. So my postings will be light at first.
I wonder if craps tables are any nicer than WinCraps, because this game has a deathwish on me after my beginner's luck.
I'm a new poster so I have to wait some time before I can post enough to respond. So my postings will be light at first.
I wonder if craps tables are any nicer than WinCraps, because this game has a deathwish on me after my beginner's luck.
June 15th, 2018 at 10:40:28 PM
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Hi and welcome.Quote: ChumpChangeI'm a new poster so I have to wait some time before I can post enough to respond. So my postings will be light at first.
I wonder if craps tables are any nicer than WinCraps, because this game has a deathwish on me after my beginner's luck.
Yes, I agree the 1 in 36 can sleep for many rolls too at Craps.
I have actual dice rolls on my WinCraps I could look at.
with Roulette, there are many sources for actual spin results one can look at too.
here is the data for 200 spins.
0 Roulette prob of 36 or less unique numbers (not all): 0.1447156008857194
00 Roulette prob of 37 or less unique numbers (not all): 0.1698457156512413
> cc.draws(37,200)#0 Roulette, 37 numbers and 200 spins
[1] "Number of numbers:37, spins:200, mean numbers show:36.8457, mean numbers NOT shown:0.154297"
not drawn Probability cumulative
u=1 36 8.469483195610663e-313 8.469483195610663e-313
u=2 35 2.449788257202378e-251 2.449788257202378e-251
u=3 34 4.72418775213577e-215 4.72418775213577e-215
u=4 33 3.90385246795345e-189 3.90385246795345e-189
u=5 32 6.209261264548263e-169 6.209261264548263e-169
u=6 31 2.271369926517092e-152 2.271369926517092e-152
u=7 30 2.465525231083706e-138 2.465525231083729e-138
u=8 29 3.667162158827238e-126 3.667162158829704e-126
u=9 28 2.009046978757217e-115 2.009046978793889e-115
u=10 27 7.973449223141271e-106 7.973449225150318e-106
u=11 26 3.716672732277447e-97 3.716672740250896e-97
u=12 25 2.908429105444836e-89 2.908429142611563e-89
u=13 24 5.012748372409591e-82 5.012748663252505e-82
u=14 23 2.350142474962236e-75 2.350142976237102e-75
u=15 22 3.54299909699412e-69 3.543001447137096e-69
u=16 21 1.965204999691636e-63 1.965208542693084e-63
u=17 20 4.476547188119941e-58 4.476566840205368e-58
u=18 19 4.585351631265854e-53 4.585396396934256e-53
u=19 18 2.277773378340647e-48 2.277819232304616e-48
u=20 17 5.846467615999853e-44 5.846695397923083e-44
u=21 16 8.180125398341328e-40 8.18071006788112e-40
u=22 15 6.528173018811442e-36 6.52899108981823e-36
u=23 14 3.088020085020313e-32 3.088672984129294e-32
u=24 13 8.944520149112384e-29 8.947608822096513e-29
u=25 12 1.630487665788381e-25 1.631382426670591e-25
u=26 11 1.913609328281341e-22 1.915240710708011e-22
u=27 10 1.473009179098158e-19 1.474924419808866e-19
u=28 9 7.544620694912325e-17 7.559369939110414e-17
u=29 8 2.597875342739557e-14 2.605434712678667e-14
u=30 7 6.049119918114898e-12 6.075174265241684e-12
u=31 6 9.531912595853448e-10 9.592664338505865e-10
u=32 5 1.01066394093541e-07 1.020256605273916e-07
u=33 4 7.105916152915381e-06 7.207941813442773e-06
u=34 3 0.0003219932110543061 0.0003292011528677489
u=35 2 0.008915537359908991 0.00924473851277674
u=36 1 0.1354708623729426 0.1447156008857194
u=37 0 0.8552843991142802 1
[1] "Number of numbers:37, spins:200, mean numbers show:36.8457, mean numbers NOT shown:0.154297"
> cc.draws(38,200)#0 Roulette, 38 numbers and 200 spins
[1] "Number of numbers:38, spins:200, mean numbers show:37.8166, mean numbers NOT shown:0.183404"
not drawn Probability cumulative
u=1 37 4.198211107412155e-315 4.198211107412155e-315
u=2 36 1.248059051850993e-253 1.248059051850993e-253
u=3 35 2.475529987972786e-217 2.475529987972786e-217
u=4 34 2.105831250496224e-191 2.105831250496224e-191
u=5 33 3.450921502925237e-171 3.450921502925237e-171
u=6 32 1.301808130611776e-154 1.301808130611776e-154
u=7 31 1.458669304861897e-140 1.45866930486191e-140
u=8 30 2.241908824637661e-128 2.24190882463912e-128
u=9 29 1.270577560328177e-117 1.270577560350596e-117
u=10 28 5.222726555094276e-108 5.222726556364853e-108
u=11 27 2.52464107603269e-99 2.524641081255417e-99
u=12 26 2.051607229532814e-91 2.051607254779224e-91
u=13 25 3.677434809281644e-84 3.677435014442369e-84
u=14 24 1.79594088206979e-77 1.795941249813291e-77
u=15 23 2.825220131919505e-71 2.825221927860755e-71
u=16 22 1.638303149638541e-65 1.638305974860469e-65
u=17 21 3.909605695975998e-60 3.909622079035747e-60
u=18 20 4.204861896326384e-55 4.204900992547174e-55
u=19 19 2.198700111790809e-50 2.198742160800735e-50
u=20 18 5.957034701305324e-46 5.957254575521404e-46
u=21 17 8.825109984668984e-42 8.825705710126536e-42
u=22 16 7.483086409408148e-38 7.48396897997916e-38
u=23 15 3.775704034576675e-34 3.776452431474673e-34
u=24 14 1.171758459808469e-30 1.172136105051616e-30
u=25 13 2.300293499699335e-27 2.301465635804387e-27
u=26 12 2.924698618559932e-24 2.927000084195736e-24
u=27 11 2.455963197673213e-21 2.458890197757408e-21
u=28 10 1.383714516721352e-18 1.386173406919109e-18
u=29 9 5.294011558026385e-16 5.307873292095576e-16
u=30 8 1.386791891626898e-13 1.392099764918994e-13
u=31 7 2.497417197602163e-11 2.511338195251353e-11
u=32 6 3.089332172661289e-09 3.114445554613802e-09
u=33 5 2.606508592810835e-07 2.637653048356973e-07
u=34 4 1.476372035250276e-05 1.502748565733846e-05
u=35 3 0.0005450487602642354 0.0005600762459215739
u=36 2 0.01242295712656632 0.01298303337248789
u=37 1 0.1568626822787534 0.1698457156512413
u=38 0 0.8301542843487586 1
[1] "Number of numbers:38, spins:200, mean numbers show:37.8166, mean numbers NOT shown:0.183404"
00 Roulette has about a 1 in 6 chance (the same as rolling a 7 at Craps)
of sleeping for at least 200 spins (wow!)
that could spell disaster to a bankroll using the OP betting system for all the #s to hit.
thank you for sharing
Sally
Last edited by: mustangsally on Jun 16, 2018
I Heart Vi Hart
July 14th, 2018 at 1:50:06 PM
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Can you help me do the math on this American roulette bet, assuming a $5 min table:
On the inside,
bet 3 $dollars on the 10-13 line (same as a $1 dollar split on 10-13, plus $1 dollar split on 11-14 plus $1 split on 12-15) pays $17
$3 dollars on the 22-25 line pays $17
$1 dollar on the 0-00 split pays $17
On the outside,
$7 dollars on the 2nd third pays $14
so, $14 dollars risked on each roll
I figure this covers 20 out of the 38 spots,
any one of those spots hitting pays at least $14 (six spots: 16 thru 21),
some pay $17 (eight spots: 0,00, 10,11,12,25,26,27)
and some pay "both ways" inside and outside bet $14+$17 = $31 (6 spots: 13,14,15,22,23,24)
I lose $14 on 18 spots: 1 thru 9 and 28 thru 36
Over time, shouldn't I come out with a slight edge?
On the inside,
bet 3 $dollars on the 10-13 line (same as a $1 dollar split on 10-13, plus $1 dollar split on 11-14 plus $1 split on 12-15) pays $17
$3 dollars on the 22-25 line pays $17
$1 dollar on the 0-00 split pays $17
On the outside,
$7 dollars on the 2nd third pays $14
so, $14 dollars risked on each roll
I figure this covers 20 out of the 38 spots,
any one of those spots hitting pays at least $14 (six spots: 16 thru 21),
some pay $17 (eight spots: 0,00, 10,11,12,25,26,27)
and some pay "both ways" inside and outside bet $14+$17 = $31 (6 spots: 13,14,15,22,23,24)
I lose $14 on 18 spots: 1 thru 9 and 28 thru 36
Over time, shouldn't I come out with a slight edge?
July 14th, 2018 at 2:18:21 PM
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Quote: zippy11777Can you help me do the math on this American roulette bet, assuming a $5 min table:
On the inside,
bet 3 $dollars on the 10-13 line (same as a $1 dollar split on 10-13, plus $1 dollar split on 11-14 plus $1 split on 12-15) pays $17
$3 dollars on the 22-25 line pays $17
$1 dollar on the 0-00 split pays $17
On the outside,
$7 dollars on the 2nd third pays $14
so, $14 dollars risked on each roll
I figure this covers 20 out of the 38 spots,
any one of those spots hitting pays at least $14 (six spots: 16 thru 21),
some pay $17 (eight spots: 0,00, 10,11,12,25,26,27)
and some pay "both ways" inside and outside bet $14+$17 = $31 (6 spots: 13,14,15,22,23,24)
I lose $14 on 18 spots: 1 thru 9 and 28 thru 36
Over time, shouldn't I come out with a slight edge?
Hi, welcome to the forum.
Let's see now...
Easy maths double 0 wheel.
On average, you pick up 36/38 or 94.7368% of what you put down in chips. That's immutable rules of roulette unless you are daft enough to play the top-line bet which has 7.89% house edge.
I started to do the long calculation, but lost interest. Maybe someone else will oblige. But no.... You have no edge. No way, no how.
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
July 14th, 2018 at 2:52:58 PM
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My thoughts:
It is hard to overcome a 5 % house advantage.
It is easier to have fun and play against a 0.5 % house advantage and throw in a tiny bit of 8 % house advantage house advantage slots to "keep 'em guessing".
- Get some free ( $ 1 ) drinks while you are at it.
Yet I digress.....
It is hard to overcome a 5 % house advantage.
It is easier to have fun and play against a 0.5 % house advantage and throw in a tiny bit of 8 % house advantage house advantage slots to "keep 'em guessing".
- Get some free ( $ 1 ) drinks while you are at it.
Yet I digress.....
There's emptiness behind their eyes
There's dust in all their hearts
They just want to steal us all and take us all apart
July 14th, 2018 at 3:19:22 PM
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Quote: zippy11777Can you help me do the math on this American roulette bet, assuming a $5 min table:
On the inside,
bet 3 $dollars on the 10-13 line (same as a $1 dollar split on 10-13, plus $1 dollar split on 11-14 plus $1 split on 12-15) pays $17
$3 dollars on the 22-25 line pays $17
$1 dollar on the 0-00 split pays $17
On the outside,
$7 dollars on the 2nd third pays $14
so, $14 dollars risked on each roll
I figure this covers 20 out of the 38 spots,
any one of those spots hitting pays at least $14 (six spots: 16 thru 21),
some pay $17 (eight spots: 0,00, 10,11,12,25,26,27)
and some pay "both ways" inside and outside bet $14+$17 = $31 (6 spots: 13,14,15,22,23,24)
I lose $14 on 18 spots: 1 thru 9 and 28 thru 36
Over time, shouldn't I come out with a slight edge?
Let's look at the numbers:
Assuming the "10-13 line" is 1 on 10/13, 1 on 11/14, and 1 on 12/15, and the 22-27 line is similar:
0, 00: win 4 (17 for 0/00 - 6 for the other six lines - 7 for the dozen bet)
1 through 9: lose 14
10 through 12: win 4 (same as 0-00)
13 through 15: win 25 (14 for the dozen + 17 for the 10/13 - 6 for the other six lines)
16 through 21: win 7 (14 for the dozen - 7 for the seven lines)
22 through 24: win 25 (same as 13-15)
25 through 27: win 4 (same as 10-12)
28 through 36: lose 14
Total = 8 - 126 + 12 + 75 + 42 + 75 + 12 - 126 = -28
The per-spin loss is -28/38 = -7/19
This is 1/19 of the total bet...and that is not a coincidence in double-zero roulette; except for the 0-00-1-2-3 bet, expect to lose 1/19 of your total bet. (On a single-zero wheel, this becomes 1/37 of your total bet.)
The problem in your calculation: you forgot to subtract your losing bets when one of your winning numbers comes up.
