37-Spin-Cycle Roulette Betting Options
This is a question to the Wizard and anyone else who has a firm grasp of probability theory and who understands odds.
Based on probability and odds, which of the following betting options on a zero number wheel, is the best?
1) After 31 spins of the wheel, flat bet the 19 numbers that don't show. Bet these numbers 6 times.
2) After 32 spins of the wheel, flat bet the 18 numbers that don't show. Bet these numbers 5 times.
3) After 33 spins of the wheel, flat bet the 17 numbers that don't show. Bet these numbers 4 times.
Thanks in advance for your mathematical analysis.
- Notnab
Quote: NotnabHello,
This is a question to the Wizard and anyone else who has a firm grasp of probability theory and who understands odds.
Based on probability and odds, which of the following betting options on a zero number wheel, is the best?
1) After 31 spins of the wheel, flat bet the 19 numbers that don't show. Bet these numbers 6 times.
2) After 32 spins of the wheel, flat bet the 18 numbers that don't show. Bet these numbers 5 times.
3) After 33 spins of the wheel, flat bet the 17 numbers that don't show. Bet these numbers 4 times.
Thanks in advance for your mathematical analysis.
- Notnab
I am assuming by 'zero number wheel' you really mean a European wheel with a single zero.
By far number 3 is the best of the options.
Number 1 has you making 19 x 6 = 104 bets
Number 2 has you making 18 x 5 = 90 bets
Number 3 has you making 17 x 4 = 68 bets
Since each bet has you losing around 2.6% of your bet, the fewer bets you make means the less money you will tend to lose. So making 68 bad bets is preferable to 90 or 104 bad bets.
Thanks for your response.
Yes, I'm referring to the European wheel and I'm basing the method of play on the following stats:
After 37 spins only 28 numbers will hit (01%) 9 won't hit
After 37 spins only 27 numbers will hit (04%) 10 won't hit
After 37 spins only 26 numbers will hit (09%) 11 won't hit
After 37 spins only 25 numbers will hit (16%) 12 won't hit
After 37 spins only 24 numbers will hit (20%) 13 won't hit
After 37 spins only 23 numbers will hit (20%) 14 won't hit
After 37 spins only 22 numbers will hit (15%) 15 won't hit
After 37 spins only 21 numbers will hit (08%) 16 won't hit
After 37 spins only 20 numbers will hit (04%) 17 won't hit
After 37 spins only 19 numbers will hit(1.2%) 18 won't hit
How does the above affect your perspective?
You are not the first person, and won't be the last, that has asked pretty much this exact question on this site. No numbers are due, the past performance of the wheel has no effect on the future. The expected distribution of #s that hit in 37 spins cannot help you predict which ones will hit next. There is no such thing as a roulette 'cycle'.
Your best strategy at roulette is to make the LEAST amount of bets possible while still having fun. Hence SOOPOO's answer.
Quote: SOOPOOI am assuming by 'zero number wheel' you really mean a European wheel with a single zero.
By far number 3 is the best of the options.
Number 1 has you making 19 x 6 = 104 bets
Number 2 has you making 18 x 5 = 90 bets
Number 3 has you making 17 x 4 = 68 bets
Since each bet has you losing around 2.6% of your bet, the fewer bets you make means the less money you will tend to lose. So making 68 bad bets is preferable to 90 or 104 bad bets.
Doesn't fewer bets result in greater deviation?
Quote: dwheatleyIt doesn't.
You are not the first person, and won't be the last, that has asked pretty much this exact question on this site. No numbers are due, the past performance of the wheel has no effect on the future. The expected distribution of #s that hit in 37 spins cannot help you predict which ones will hit next. There is no such thing as a roulette 'cycle'.
Your best strategy at roulette is to make the LEAST amount of bets possible while still having fun. Hence SOOPOO's answer.
The method doesn't involve prediction / due numbers.
If you spin a roulette wheel every 37 spins, you will not see all 37 numbers land. It will not happen in your lifetime.
A cycle of 37 spins will always result in less than 37 numbers landing.
Quote: NotnabThe method doesn't involve prediction / due numbers.
If you spin a roulette wheel every 37 spins, you will not see all 37 numbers land. It will not happen in your lifetime.
A cycle of 37 spins will always result in less than 37 numbers landing.
Always?
Would you care to demonstrate how 37 spins absolutely cannot result in each of the 37 numbers hitting once? I will tell you the e th probability of 37 spins yielding every number once:
37/37 * 36/37 * 35/37 * 34/37 * 33/37 * 32/37 * 31/37 * 30/37 * 29/37 * 28/37 * 27/37 * 26/37 * 25/37 * 24/37 * 23/37 * 22/37 * 21/37 * 20/37 * 19/37 * 18/37 * 17/37 * 16/37 * 15/37 * 14/37 * 13/37 * 12/37 * 11/37 * 10/37 * 9/37 * 8/37 * 7/37 * 6/37 * 5/37 * 4/37 * 3/37 * 2/37 * 1/37 =
Quick Pause: Interestingly, you are less than 50% to have the first eight numbers all be different!
---Less than 20% to have the first eleven numbers be different!
---Less than 10% on the first thirteen!!!
---Barely over 0.5% on the first eighteen!!!
.0000000000000001303986462 or 0.00000000000001303986462%
I am surprised to find that hitting the same number nine consecutive times is slightly more probable than this result. If it has ever been proven that the same number has hit nine (or more) consecutive times, then I would be willing to be that we've spun for the cycle at least once in the course of human gambling events. It would be an unprovable bet, though.
I would also venture to say it will happen in my lifetime. If you look at all of the casinos, on-line casinos, video games, play-for-fun on-line, video roulette, etc. etc. etc., Every repeated result would be a new series with the repeated number the first of the series, because the 37/37 spin is obviously a given. Whether or not I will be there when it happens, however, is a different matter entirely. ; )
Quote: NotnabDoesn't fewer bets result in greater deviation?
Since you seem like a nice guy making an honest attempt at learning something, let me steer you in the right direction.
Learn these terms.....
Expected value
Variance
Once you have a firm grasp on what those two terms are you can then craft questions that the 'math guys' can then answer in a meaningful way.
Quote: NotnabBased on probability and odds, which of the following betting options on a zero number wheel, is the best?
I can show some values. You can choose what is the best.
You also be doing a lot of waiting. Why?
Where is the value in waiting so long?
Quote: Notnab1) After 31 spins of the wheel, flat bet the 19 numbers that don't show. Bet these numbers 6 times.
4.81% of every 31 spin sets will on average result in 19 numbers not showing
48,112 out of 1 million trials
win prob = 19/37 for 6 spins
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
0 0.0132564 0.0000000 1.0000000 0.0132564 0.9867436
1 0.0839570 0.0132564 0.9867436 0.0972134 0.9027866
2 0.2215533 0.0972134 0.9027866 0.3187667 0.6812333
3 0.3118157 0.3187667 0.6812333 0.6305823 0.3694177
4 0.2468541 0.6305823 0.3694177 0.8774364 0.1225636
5 0.1042273 0.8774364 0.1225636 0.9816637 0.0183363
6 0.0183363 0.9816637 0.0183363 1.0000000 0.0000000
7.97% of every 32 spin sets will on average result in 18 numbers not showingQuote: Notnab2) After 32 spins of the wheel, flat bet the 18 numbers that don't show. Bet these numbers 5 times.
79,657 out of 1 million trials
win prob = 18/37 for 5 spins
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
0 0.0357075 0.0000000 1.0000000 0.0357075 0.9642925
1 0.1691408 0.0357075 0.9642925 0.2048483 0.7951517
2 0.3204772 0.2048483 0.7951517 0.5253255 0.4746745
3 0.3036100 0.5253255 0.4746745 0.8289355 0.1710645
4 0.1438153 0.8289355 0.1710645 0.9727508 0.0272492
5 0.0272492 0.9727508 0.0272492 1.0000000 0.0000000
11.91% of every 33 spin sets will on average result in 17 numbers not showingQuote: Notnab3) After 33 spins of the wheel, flat bet the 17 numbers that don't show. Bet these numbers 4 times.
119,080 out of 1 million trials
win prob = 17/37
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
0 0.0853715 0.0000000 1.0000000 0.0853715 0.9146285
1 0.2902632 0.0853715 0.9146285 0.3756348 0.6243652
2 0.3700856 0.3756348 0.6243652 0.7457204 0.2542796
3 0.2097152 0.7457204 0.2542796 0.9554355 0.0445645
4 0.0445645 0.9554355 0.0445645 1.0000000 0.0000000
Thanks in advance for your mathematical analysis.
- Notnab
It should never affect anyone's perspective.Quote: NotnabSoopoo,
Thanks for your response.
Yes, I'm referring to the European wheel and I'm basing the method of play on the following stats:
After 37 spins only 28 numbers will hit (01%) 9 won't hit
After 37 spins only 27 numbers will hit (04%) 10 won't hit
After 37 spins only 26 numbers will hit (09%) 11 won't hit
After 37 spins only 25 numbers will hit (16%) 12 won't hit
After 37 spins only 24 numbers will hit (20%) 13 won't hit
After 37 spins only 23 numbers will hit (20%) 14 won't hit
After 37 spins only 22 numbers will hit (15%) 15 won't hit
After 37 spins only 21 numbers will hit (08%) 16 won't hit
After 37 spins only 20 numbers will hit (04%) 17 won't hit
After 37 spins only 19 numbers will hit(1.2%) 18 won't hit
How does the above affect your perspective?
The past only shows what has happened in a sequence of events
and
does no better than a blind squirrel picking the next numbers to win.
Any computer simulation will prove this to be absolutely true using RNG numbers or internet listed actual spins.
Where are the results of your computer simulations?
If you have none, time for you to learn to do them.
There will be power in your findings that none of us here can ever give you.
BTW, your %s are really rounded
| no show | unique #s | probability |
|---|---|---|
| 20 | 17 | 0.051300% |
| 19 | 18 | 0.282190% |
| 18 | 19 | 1.174500% |
| 17 | 20 | 3.605390% |
| 16 | 21 | 8.384150% |
| 15 | 22 | 14.835930% |
| 14 | 23 | 19.936310% |
| 13 | 24 | 20.475660% |
| 12 | 25 | 15.925890% |
| 11 | 26 | 9.406580% |
| 10 | 27 | 4.159970% |
| 9 | 28 | 1.364840% |
| 8 | 29 | 0.328120% |
| 7 | 30 | 0.054960% |
Quote: Mission146Always?
Would you care to demonstrate how 37 spins absolutely cannot result in each of the 37 numbers hitting once? I will tell you the e th probability of 37 spins yielding every number once:
37/37 * 36/37 * 35/37 * 34/37 * 33/37 * 32/37 * 31/37 * 30/37 * 29/37 * 28/37 * 27/37 * 26/37 * 25/37 * 24/37 * 23/37 * 22/37 * 21/37 * 20/37 * 19/37 * 18/37 * 17/37 * 16/37 * 15/37 * 14/37 * 13/37 * 12/37 * 11/37 * 10/37 * 9/37 * 8/37 * 7/37 * 6/37 * 5/37 * 4/37 * 3/37 * 2/37 * 1/37 =
Quick Pause: Interestingly, you are less than 50% to have the first eight numbers all be different!
---Less than 20% to have the first eleven numbers be different!
---Less than 10% on the first thirteen!!!
---Barely over 0.5% on the first eighteen!!!
.0000000000000001303986462 or 0.00000000000001303986462%
I am surprised to find that hitting the same number nine consecutive times is slightly more probable than this result. If it has ever been proven that the same number has hit nine (or more) consecutive times, then I would be willing to be that we've spun for the cycle at least once in the course of human gambling events. It would be an unprovable bet, though.
I would also venture to say it will happen in my lifetime. If you look at all of the casinos, on-line casinos, video games, play-for-fun on-line, video roulette, etc. etc. etc., Every repeated result would be a new series with the repeated number the first of the series, because the 37/37 spin is obviously a given. Whether or not I will be there when it happens, however, is a different matter entirely. ; )
Has there been anyone throughout the history of roulette ever seen ALL 37 numbers (0 - 36) land within exactly 37 spins of the wheel?
It will not happen in your lifetime.
Quote: NotnabDoesn't fewer bets result in greater deviation?
Fewer bets do have higher standard deviation than more (smaller) bets for the same total amount. That is a good thing in a negative expectation game, because standard deviation is a measure of likelihood than your result will end up farther from its expected value. In other words, in a negative expectation game, like roulette, the higher the standard deviation, the more likely you are to win (or to lose a lot). If standard deviation were zero, it would be impossible to ever win, you'd just keep losing a little bit on every bet. You can simulate that situation by betting on every number in roulette. Such a bet would have the lowest possible variance, and it would never win.
Quote: NotnabHas there been anyone throughout the history of roulette ever seen ALL 37 numbers (0 - 36) land after 37 spins of the wheel?
I guess, it has not happened. Must mean it's due! :) Any day now!
The 1 in 766,879,127,067,901 chance of it happening could have happened already at least one time over all the years of RouletteQuote: NotnabHas there been anyone throughout the history of roulette ever seen ALL 37 numbers (0 - 36) land within exactly 37 spins of the wheel?
It will not happen in your lifetime.
the game has been around a long time and no one could have been watching for it.
Just like the 154 roll craps hand for 1 in 5.6 billion chance, it would take many years (200 years) to see that, but it happened.
Still no public proof of it, just a story.
Where is the video??
added:
Just knowing no one has seen that event, makes it no more likely that you can pick winning numbers.
I still say a blind squirrel can choose winning Roulette numbers just as well as any human or computer can.
But will that blind squirrel ever be able to sell his system ??
Quote: guido111The 1 in 766,879,127,067,901 chance
I'm predicting that it won't happen in YOUR lifetime, while YOU'RE playing roulette.
Quote: guido111The 1 in 766,879,127,067,901 chance
I'm predicting that it won't happen in YOUR lifetime, while YOU'RE playing roulette.
Quote: NotnabI'm predicting that it won't happen in YOUR lifetime, while YOU'RE playing roulette.
Sure thing, but it doesn't matter to improve your chances of picking the right numbers on the next spin.
Quote: guido111
I still say a blind squirrel can choose winning Roulette numbers just as well as any human or computer can.
Absolute junk.
Hmmm ... I guess squirrels understand probability more than mathematicians.
How does that help you in choosing winning Roulette numbers?Quote: NotnabI'm predicting that it won't happen in YOUR lifetime, while YOU'RE playing roulette.
You want to believe that a Xspin distribution will always turn into a X+Y distribution.
How much are you willing to lose to prove you are correct?
Your betting methods can be easily simulated.
I see also that guido111 shows you to be waiting many,
many spins for just the right combos of numbers to show and not show before you even bet one time.
What kind of bankroll do you have an idea to use?
and
How many times do you expect to make these bets before your bankroll is ruined?
Even I would like to know those numbers.
Sally
Quote: thecesspitSure thing, but it doesn't matter to improve your chances of picking the right numbers on the next spin.
The fact that it's unlikely to happen is a reliable "stat" that can be borne in mind when developing a strategy that's based on past and current reality.
Roulette is like the weather.
And a roulette player should be like a weatherman studying the roulette table conditions.
When you see particular types of clouds, it's more likely to rain or the sun to shine or to snow.
You can't predict a specific number(s), but you can take advantage when certain conditions naturally arise on a regular basis and at random regular intervals.
Sometimes it snows; sometimes it rains; sometimes the sun shines. And just like the real weather, a real roulette wheel provides it's own natural signals.
I've heard what you've said, but never met a weatherman who can describe the nature of the snow or sun.
Quote: guido111I can show some values. You can choose what is the best.
You also be doing a lot of waiting. Why?
Where is the value in waiting so long?
4.81% of every 31 spin sets will on average result in 19 numbers not showing
48,112 out of 1 million trials
win prob = 19/37 for 6 spinsx prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
0 0.0132564 0.0000000 1.0000000 0.0132564 0.9867436
1 0.0839570 0.0132564 0.9867436 0.0972134 0.9027866
2 0.2215533 0.0972134 0.9027866 0.3187667 0.6812333
3 0.3118157 0.3187667 0.6812333 0.6305823 0.3694177
4 0.2468541 0.6305823 0.3694177 0.8774364 0.1225636
5 0.1042273 0.8774364 0.1225636 0.9816637 0.0183363
6 0.0183363 0.9816637 0.0183363 1.0000000 0.0000000
7.97% of every 32 spin sets will on average result in 18 numbers not showing
79,657 out of 1 million trials
win prob = 18/37 for 5 spins
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
0 0.0357075 0.0000000 1.0000000 0.0357075 0.9642925
1 0.1691408 0.0357075 0.9642925 0.2048483 0.7951517
2 0.3204772 0.2048483 0.7951517 0.5253255 0.4746745
3 0.3036100 0.5253255 0.4746745 0.8289355 0.1710645
4 0.1438153 0.8289355 0.1710645 0.9727508 0.0272492
5 0.0272492 0.9727508 0.0272492 1.0000000 0.0000000
11.91% of every 33 spin sets will on average result in 17 numbers not showing
119,080 out of 1 million trials
win prob = 17/37x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
0 0.0853715 0.0000000 1.0000000 0.0853715 0.9146285
1 0.2902632 0.0853715 0.9146285 0.3756348 0.6243652
2 0.3700856 0.3756348 0.6243652 0.7457204 0.2542796
3 0.2097152 0.7457204 0.2542796 0.9554355 0.0445645
4 0.0445645 0.9554355 0.0445645 1.0000000 0.0000000
Thanks in advance for your mathematical analysis.
- Notnab
It should never affect anyone's perspective.
The past only shows what has happened in a sequence of events
and
does no better than a blind squirrel picking the next numbers to win.
Any computer simulation will prove this to be absolutely true using RNG numbers or internet listed actual spins.
Where are the results of your computer simulations?
If you have none, time for you to learn to do them.
There will be power in your findings that none of us here can ever give you.
BTW, your %s are really rounded
| no show | unique #s | probability |
|---|---|---|
| 20 | 17 | 0.051300% |
| 19 | 18 | 0.282190% |
| 18 | 19 | 1.174500% |
| 17 | 20 | 3.605390% |
| 16 | 21 | 8.384150% |
| 15 | 22 | 14.835930% |
| 14 | 23 | 19.936310% |
| 13 | 24 | 20.475660% |
| 12 | 25 | 15.925890% |
| 11 | 26 | 9.406580% |
| 10 | 27 | 4.159970% |
| 9 | 28 | 1.364840% |
| 8 | 29 | 0.328120% |
| 7 | 30 | 0.054960% |
Thanks.
Helpful stats. Might prove useful in making a long-term decision.
My main thought has been:
Is it better to bet on 17 numbers with profit (£36 - £17 = £19); with probability 17/37 = 0.4594594, or
Is it better to bet on 18 numbers with profit (£36 - £18 = £18); with probability 18/37 = 0.4864864, or
Is it better to bet on 19 numbers with profit (£36 - £19 = £17); with probability 19/37 = 0.5135135
You remind me of Mauisunset. Are you two friends?
No it is not.Quote: NotnabRoulette is like the weather.
And a roulette player should be like a weatherman studying the roulette table conditions.
weather predicting is about dependent events.
The probabilities are always changing and can never be exactly calculated.
Roulette spins, either one or a sequence of many, are independent events.
The probabilities are known and are always the same.
The past results, as a pointer to what numbers to bet in the next upcoming results,
in no way affects the next spin or a number of next spins.
If you are choosing to bet 17 numbers that did not show in the last 33 spins
and I choose any 17 numbers at random or by a coin toss.
We both will win and lose the next sequences at the same rate, percentage or average.
You need to understand the difference between dependent events and independent events.
And to make you better than the rest, know the difference between independent events and mutually exclusive events.
These are basic statistical facts.
Then people will listen to you.
a good starting point is here
http://stattrek.com/probability/probability.aspx?Tutorial=Stat
They will not try to sell you anything :)
A computer simulation will easily show the differences between all of these.
Quote: NotnabMy main thought has been:
Is it better to bet on 17 numbers with profit (£36 - £17 = £19); with probability 17/37 = 0.4594594, or
Is it better to bet on 18 numbers with profit (£36 - £18 = £18); with probability 18/37 = 0.4864864, or
Is it better to bet on 19 numbers with profit (£36 - £19 = £17); with probability 19/37 = 0.5135135
from guido111 binomial distribution tables
to bet on 17 numbers you have a 0.0853715 chance of NOT winning at least 1 time
to bet on 18 numbers you have a 0.0357075 chance of NOT winning at least 1 time
to bet on 19 numbers you have a 0.0132564 chance of NOT winning at least 1 time
Looks right to me.
I'll tell you what I am going to do for fun on WOO free Roulette game later, or if I had a lot of money that I didn't really want. I'd wait for seven different consecutive numbers, which could take awhile, and then run the following system:
Eighth Spin: Seven Numbers at $1.00/each, straight bet getting 36:1.
If win: $36 - $7 = $29
If loss: -$7 AND
Eight numbers at $1.00 each
If win: $36 -$8 -$7 = $21
If loss: -$15 AND
Nine Numbers at $1.00 each
If win: $36 -$9 -$8 -$7 = $12
If loss: -$24 AND:
Ten Numbers at $1.00 each
If win: $36 -$10 -$9 -$8 -$7 = $2
If loss -$34 AND:
Eleven Numbers at $2.00 each
If win: $72 -$22, -$10, -$9, -$8 -$7 = $16
If loss: -$56 AND
Twelve Numbers at $3.00 each
If win: $108 -$36 -$22 -$10 -$9 -$8 -$7 = $16
If loss: -$92 AND:
Thirteen Numbers at $5.00 each
If win: $180 -$65 -$92 = $23
If loss: -$157 AND
Fourteen Numbers at $8.00 each
If win $288 -$112 -$157 = $19
If loss: -$269 AND
Fifteen Numbers at $13 each
If win: $468 -$195 -$269 =$4
If Loss: -$464 AND
Sixteen Numbers at $24 each
If win: $864 - $384 -$464 = $16
If loss: -$848 AND
---In my experience, the system would end in failure at this point because most $5.00 Min. tables I have seen have a $500 Maximum. If the maximum were $1,000, however:
Seventeen Numbers at $45 each:
If win: $1,620 -$848 -$765 = $7
If Loss: -$1,613
PROBABILITY OF FAILURE
30/37 * 29/37 * 28/37 * 27/37 * 26/37 * 25/37 * 24/37 * 23/37 * 22/37 * 21/37 ($500 Limit) * 20/37 ($1,000 Limit)
$500 Limit: .022673482 or 2.2673482%
EV on Failure = .022673482 * 848 = -$19.2271126 ($19.23)
$1000 Limit = .012255936 or 1.2255936%
EV on Failure = .012255936 * 1613 = $19.76882497 ($19.77)
Expected Returns
29 * (7/37) = $5.49
21 * (7/37) * (8/37) = $0.86
12 * (7/37) * (8/37) * (9/37) = $0.12
2 * (7/37) * (8/37) * (9/37) * (10/37) = $0.01 (Rounding Up by .0463)
16 * (7/37) * (8/37) * (9/37) * (10/37) * (11/37) = $0.01 (Rounding down .2791886)
16 * Above () and (12/37) = $0.01 (Rounding up impermissibly by .00585128)
23 * Above () and (13/37) = $0.00 (Not even close) In fact, we're going to say the rest of the results total and MAYBE round up to $0.01
Expected Return (Any Success): $6.51
Total Expected Values
$500 Limit
-$19.23 + $ 6.51 = -$12.72
$1,000 Limit
-$19.77 + $6.51 = -$13.26
Hold Percentage: 12.72/848 = 0.015 or 1.5% of total bets for $500 Limit
Hold Percentage: 13.26/1613 = .00822 or .822% of total bets for $1,000 Limit
CAN SOMEONE PLEASE VERIFY MY HOLD PERCENTAGES?
In the event that my Math is correct at the end (I'm good on everything before that) this system actually, somehow, reduces the HE. The problem is that no win after the second play exceeds the total amount that you have out in bets, so system failure is obviously devastating. In fact, the third play only gives you a profit of HALF as much as you have out in current/previous bets at that point.
Coincidentally, it is also the third bet where you are more than 50% to succeed (or have already succeeded) by then:
30/37 * 29/37 * 28/37 = .480919195 or 48.1% chance of three consecutive failures, inversely, 51.9% chance of success.
By the time you reach the point that probability says you should succeed, the total risk is no longer worth the reward.
By comparison, laying the 4 for $24 in Craps would yield a profit of $12, if successful, but you have a 66.666-% chance of that happening!
You don't even have a fixed profit total for this system! It will be anywhere between $2-$29 depending upon what stage of the system at which you win.
With the $500 Limit, you fail 1:44.1 attempts.
With the $1000 Limit, you fail 1:81.59 attempts.
Martingaling the Red starting at $5, however, gives you seven shots at a $500 Limit starting at $5 and eight shots at a $1,000 Limit:
The probability of Martingale failure is:
(19/37) * (19/37) * (19/37) * (19/37) *(19/37) * (19/37) * (19/37)-$500 or * (19/37) -$1,000
$500 = .009415928 or .94%
$1000 = .004835206 OR .48%
$500 Limit Loss EV ($635 Lost) 635 * .009415928 = 5.97911428 -$5.98
$1000 Limit Loss EV ($1275 Lost) 1275 * .004835206 = 6.16488765 or -$6.16
$5.00 Win (Any Point) = 5 * .990584072 = 4.95292036 or $4.95 on $500 Limit
or
$5.00 Win (Any Point) = 5 * .995164794 = 4.97582397 or $4.98 on $1,000 Limit
EV on $500 Limit -$1.03
EV on $1000 Limit -$1.18
1.03/635 = .001622 or .1622% House Hold on $635 Bet
1.18/1275 = .00092549 or .0925% House Hold on $1275 Bet
COMPARISON
THE GOOD:
1.) My goof-off system up top has a better return percentage in relation to the amount bet (assuming a win) than the Martingale with exception only to the Ten Number Pick vs. the Fourth Martingale Spin:
The fourth Martingale spin has you winning $5 against $75 bet or 6.666% of total bets
The Goof-Off system has you winning $2 on a bet of $34 or 5.8824% of total bets
Remember, this is a return only in relation to the bets. It does not reflect the EV of actually winning at this time, nor does it reflect the EV of winning any individual trial. Comparing the two systems, the Martingale is ALWAYS more likely to win any individual trial. If we assume the $500 Limit, the goof-off system is 16/37 to succeed on the final bet before failure and the Martingale is always 18/37 to succeed on any individual turn.
2.) You will usuallly have less same spin money riding (incuding current and previous-lost-bets) with the goof-off system. The only exceptions are the 1st Spin $7 v. $5 and the second spin, which is $15 in both cases.
THE BAD:
Everything Else, specifically:
1.) The goof-off system has a higher chance of failure. It fails slightly less than 2.5 times more than the Martingale at a $500 limit and slightly more than 2.5 times the Martingale at the $1,000 limit. 2.56, to be specific.
The goof-off system also has a higher risk of failure on any individual spin.
The average spin at the $500 limit has a 11.5/37 single spin chance of winning, or 31.08%
The average spin at the $1000 imit has a 12/37 single spin chance of winning, or 33.33333%
The Martingale has a constant 18/37 single spin chance of winning, or 48.6%
2.) In addition to having a higher risk of failure, the goof-off system also loses more actual money on a failure. It loses $213 more for a $500 Table Max failure and $338 more for a $1,000 Table Max failure.
The Point
There's no point in really trying the goof-off system, except for fun. The wheel and the ball have no concern nor control over what happened before or what will later happen in relation to previous events. Because this is the case, you could just start it off with seven individual random numbers as opposed to waiting for six consecutive non-recurring numbers, at least then you don't run the risk of being bored waiting for non-recurring numbers.
The best way for this system to succeed would require a ridiculously high limit and a crapload of money. I may do the math on it at a later time. However, what you would do is basically just start picking one number (because that is the cheap end, anyway) and then increasing your bets and picking additional numbers in a system simillar to the one described above. The fail point would be a thirty-five number pick bet losing 35/37 and that's 94.6% to win by itself with nothing else considered. It would be almost impossible to lose, but the results would be catastrophic if you did.
Quote: NotnabHas there been anyone throughout the history of roulette ever seen ALL 37 numbers (0 - 36) land within exactly 37 spins of the wheel?
It will not happen in your lifetime.
Is it required that someone be at the table to record it the entire time? Just because nobody saw it doesn't mean it never happened, the Earth came into being (I make no claims on knowing how) at some point, nobody was around to record that happening.
I want you to try to think about your statement literally, how can you say what will or will not happen in my lifetime? It could happen in the next two hours. It could be happening now. If you start thinking of non-impossible things as impossible, I fear for your success rate as a gambler and HOPE that you never actually try to employ a system. I say that with all due respect, despite what I perceive (possibly inaccurately) as a certain presumptuousness and dogmatism, regarding this one issue only, you seem like a nice guy, so no disrespect intended.
Quote: Mission146Is it required that someone be at the table to record it the entire time? Just because nobody saw it doesn't mean it never happened, the Earth came into being (I make no claims on knowing how) at some point, nobody was around to record that happening.
I want you to try to think about your statement literally, how can you say what will or will not happen in my lifetime? It could happen in the next two hours. It could be happening now. If you start thinking of non-impossible things as impossible, I fear for your success rate as a gambler and HOPE that you never actually try to employ a system. I say that with all due respect, despite what I perceive (possibly inaccurately) as a certain presumptuousness and dogmatism, regarding this one issue only, you seem like a nice guy, so no disrespect intended.
I'll repeat: it will not happen in YOUR lifetime while YOU are playing roulette.
What are the odds that my prediction will be wrong? Please calculate. (LOL)
Quote: NotnabI'll repeat: it will not happen in YOUR lifetime while YOU are playing roulette.
What are the odds that my prediction will be wrong? Please calculate. (LOL)
It is incredibly likely that it will not happen in my lifetime while I am playing Roulette. Don't close the Universe, though, my friend.
I cannot possibly calculate the odds of you being wrong because there are entirely too many variables. Did you like my post about the system? I went ahead and slapped that together for you, it took a few minutes.
Quote: Mission146It is incredibly likely that it will not happen in my lifetime while I am playing Roulette. Don't close the Universe, though, my friend.
I cannot possibly calculate the odds of you being wrong because there are entirely too many variables. Did you like my post about the system? I went ahead and slapped that together for you, it took a few minutes.
My strategy is much better and unique. ;-)
I'm now just fine-tuning.
Quote: mustangsallyfrom guido111 binomial distribution tables
to bet on 17 numbers you have a 0.0853715 chance of NOT winning at least 1 time
to bet on 18 numbers you have a 0.0357075 chance of NOT winning at least 1 time
to bet on 19 numbers you have a 0.0132564 chance of NOT winning at least 1 time
Looks right to me.
Great response! Very much appreciated! This is the sort of quality info I'm looking for.
Can we take one more step?
What are the chances on winning / losing based on ...
... the same set of 17 numbers bet 4 times (to 37 spins) ?
... the same set of 18 numbers bet 5 times (to 37 spins) ?
... the same set of 19 numbers bet 6 times (to 37 spins) ?
Thanks in advance.
- Notnab
You can do this yourself. There is a lot of math to be done or you can use an online calculator.Quote: NotnabGreat response! Very much appreciated! This is the sort of quality info I'm looking for.
Can we take one more step?
What are the chances on winning / losing based on ...
... the same set of 17 numbers bet 4 times (to 37 spins) ?
... the same set of 18 numbers bet 5 times (to 37 spins) ?
... the same set of 19 numbers bet 6 times (to 37 spins) ?
Thanks in advance.
- Notnab
http://stattrek.com/online-calculator/binomial.aspx
For the first one Probability of success on a single trial:
use (a decimal) 0.45945945945945945945945945945946
(17/37) I like to just copy and paste
Number of trials: 4
Number of successes (x): 0
click the calculate button.
The result you want is:
Binomial Probability: P(X = 0)
Have fun.
We are patiently waiting to see your results for your system in a new Betting System post
Quote: mustangsallyHow does that help you in choosing winning Roulette numbers?
You want to believe that a Xspin distribution will always turn into a X+Y distribution.
How much are you willing to lose to prove you are correct?
Your betting methods can be easily simulated.
I see also that guido111 shows you to be waiting many,
many spins for just the right combos of numbers to show and not show before you even bet one time.
What kind of bankroll do you have an idea to use?
and
How many times do you expect to make these bets before your bankroll is ruined?
Even I would like to know those numbers.
Sally
Suffice it to say: my strategy involves 3 elements of information lining up perfectly (like a jigsaw puzzle falling into place)
I have a method for tracking these 3 elements and letting the table "decide" which numbers I should bet on.
And the strategy is based on flat betting. No fear of decimating a bankroll with nonsensical progressions.
I'm not at liberty to share more than that. ;-)
Quote: thecesspitIt's not a reliable stat. It's an emergent phenomena from the fact there's a 1/38 chance of each number. If I see 1-36 in the last 36 spins... the chance of hitting 0 or 00 is still exactly the same as it was.
I've heard what you've said, but never met a weatherman who can describe the nature of the snow or sun.
It's not a perfect stat, but it IS a reliable stat, which clearly indicates that it is very ... very ... very ... very unlikely to happen. It's just common sense.
A weatherman understands the basic operations of various elements of the weather. He knows that when he sees and tracks a particular type of cloud, it's likely to rain, so he's likely to know when to take shelter.
Quote: NotnabIt's not a perfect stat, but it IS a reliable stat, which clearly indicates that it is very ... very ... very ... very unlikely to happen. It's just common sense.
What is the difference between 'perfect' and 'reliable'.
Does it change the next spin of the wheel if I've seen 1-35 come up in the last 35 spins?
Quote:A weatherman understands the basic operations of various elements of the weather. He knows that when he sees and tracks a particular type of cloud, it's likely to rain, so he's likely to know when to take shelter.
Well, yes, thanks. What I mean is I have no idea what this means in relation to a roulette wheel. EITHER every spin of the wheel is independent... or it isn't.
Quote: mustangsallyNo it is not.
weather predicting is about dependent events.
The probabilities are always changing and can never be exactly calculated.
Roulette spins, either one or a sequence of many, are independent events.
The probabilities are known and are always the same.
The past results, as a pointer to what numbers to bet in the next upcoming results,
in no way affects the next spin or a number of next spins.
If you are choosing to bet 17 numbers that did not show in the last 33 spins
and I choose any 17 numbers at random or by a coin toss.
We both will win and lose the next sequences at the same rate, percentage or average.
You need to understand the difference between dependent events and independent events.
And to make you better than the rest, know the difference between independent events and mutually exclusive events.
These are basic statistical facts.
Then people will listen to you.
a good starting point is here
http://stattrek.com/probability/probability.aspx?Tutorial=Stat
They will not try to sell you anything :)
A computer simulation will easily show the differences between all of these.
I'm aware that roulette is a string of independent events. ;-)
My analogy of the weather and weatherman relates solely to my own roulette strategy:
I look for roulette conditions that I have personally defined, using the weather analogy.
When I see my personally defined rain signal, I decide to stay out of the rain; I stop betting, until I see my personally defined sun signal, which indicates that it's the right time to play once again. It requires patience, but it works.
That's the foundation of my personal strategy: when to play; when to track; when to stop (come out of the rain). ;-)
Quote: mustangsallyYou can do this yourself. There is a lot of math to be done or you can use an online calculator.
http://stattrek.com/online-calculator/binomial.aspx
For the first one Probability of success on a single trial:
use (a decimal) 0.45945945945945945945945945945946
(17/37) I like to just copy and paste
Number of trials: 4
Number of successes (x): 0
click the calculate button.
The result you want is:
Binomial Probability: P(X = 0)
Have fun.
We are patiently waiting to see your results for your system in a new Betting System post
I will have fun with this.
You have been very helpful.
Thank you.
Quote: thecesspitWhat is the difference between 'perfect' and 'reliable'.
Does it change the next spin of the wheel if I've seen 1-35 come up in the last 35 spins?
Well, yes, thanks. What I mean is I have no idea what this means in relation to a roulette wheel. EITHER every spin of the wheel is independent... or it isn't.
It's not perfect in the sense that the statistical number is suggesting with absolutely certainty that the event can't happen.
However, it's a reliable stat in the sense that the stat is suggesting that the event is very unlikely to happen, which of course is borne out in reality by the observation of roulette players throughout the entire globe -- both in the past, the present and I predict, also into the distant future.
When playing roulette, the frequency with which the numbers fall will affect any strategy used. On that basis, it's important to know when your strategy is likely to be negatively affected based on changing table conditions (like monitoring the weather and knowing when storm clouds are brewing so that you can take shelter at the right time). Afterall, the roulette wheel is a physical random number generator, which has its ebbs and flows in number distribution, just like changes in the weather.
Every spin is an independent event, but my common sense observation informs me that that truth can be taken too far and can be a stumbling block to developing a good strategy which combines observable facts with probability and statistics.
As a case in point: I can say with confidence that I'm unlikely to see all 37 numbers land in 37 spins. And that confidence is based on an observable fact as expressed in the above previous paragraphs.
Very true. But you have not shown how to make any money from knowing that fact. That is all that really matters.Quote: NotnabAs a case in point: I can say with confidence that I'm unlikely to see all 37 numbers land in 37 spins. And that confidence is based on an observable fact as expressed in the above previous paragraphs.
You say you watch the last 31 spins and bet only if there are 19 numbers not hit for the next 6 spins.
You bet on those 19 numbers.
I can randomly also choose 19 numbers to bet on.
from Nope27 blog
https://wizardofvegas.com/member/nope27/blog/
1-((1-(p/37))^n)
where n=# of spins and p=# of numbers
Your probability of winning at least 1 time in those 6 spins =1-((1-(19/37))^6) = 0.986743628
My probability of winning at least 1 time in those 6 spins =1-((1-(19/37))^6) = 0.986743628
Does not matter how we chose our numbers to bet, the winning probability is exactly the same.
Now,
You can not flat bet for 6 bets and show a profit at every bet.
Your bets have to go up. and so will your losses using a negative progression.
Bet#1 loss (1 unit on 19 numbers)
-19 units
Bet#2 win (1 unit on 19 numbers)
Now the probability of winning has also changed. you have only 5 spins to win. =1-((1-(19/37))^5) = 0.972750791
Looks like your probability of winning by the 6th spin has somehow gone down! Ouch!
-18 units total loss + 35 units total win: Net +17
You still have a -2 unit loss to over come.
You win 50% of your system bets and have a loss to show for it.
This can not be a flat bet progression. It will fail quickly.
Sally
Quote: mustangsallyVery true. But you have not shown how to make any money from knowing that fact. That is all that really matters.
You say you watch the last 31 spins and bet only if there are 19 numbers not hit for the next 6 spins.
You bet on those 19 numbers.
I can randomly also choose 19 numbers to bet on.
from Nope27 blog
https://wizardofvegas.com/member/nope27/blog/
1-((1-(p/37))^n)
where n=# of spins and p=# of numbers
Your probability of winning at least 1 time in those 6 spins =1-((1-(19/37))^6) = 0.986743628
My probability of winning at least 1 time in those 6 spins =1-((1-(19/37))^6) = 0.986743628
Does not matter how we chose our numbers to bet, the winning probability is exactly the same.
Now,
You can not flat bet for 6 bets and show a profit at every bet.
Your bets have to go up. and so will your losses using a negative progression.
Bet#1 loss (1 unit on 19 numbers)
-19 units
Bet#2 win (1 unit on 19 numbers)
Now the probability of winning has also changed. you have only 5 spins to win. =1-((1-(19/37))^5) = 0.972750791
Looks like your probability of winning by the 6th spin has somehow gone down! Ouch!
-18 units total loss + 35 units total win: Net +17
You still have a -2 unit loss to over come.
You win 50% of your system bets and have a loss to show for it.
This can not be a flat bet progression. It will fail quickly.
Sally
Hey Sally,
Hmmm ... nice precise calculations based on probability.
The thing is ... my strategy combines probability and real world stats based on my own real play. So, I don't rely solely on probability.
Also, choosing numbers randomly will produce different results compared to my number selection method and roulette strategy. I wait until 3 specific elements line up.
I appreciate your calculations, but unfortunately they have no bearing on my method and are unable to explain my results. ;-)
I just came to this forum today to ask a question, which has now been answered, together with a useful binomial tool thrown in for good measure.
Thanks again Sally. You're a darling. ;-)
"
Aw, stick around. It's refreshing to have a forum member who does not let logic and/or facts get in the way of expressing his views.
SOOPOO
guido111
mustangsally
... for assisting me with my question.
May you be blessed with discovering your own successful roulette strategies.
Don't let anyone tell you that roulette can't be beaten.
Probability is only one aspect of the triangular roulette puzzle. ;-)
Quote: NotnabI express thanks to ...
SOOPOO
guido111
mustangsally
... for assisting me with my question.
May you be blessed with discovering your own successful roulette strategies.
Don't let anyone tell you that roulette can't be beaten.
Probability is only one aspect of the triangular roulette puzzle. ;-)
I can't let this one slide.......
Unless you are able to identify a biased wheel, roulette cannot be beaten.
I'm telling you roulette can't be beaten.
'The triangular roulette puzzle' is gobbledygook.
I'll bet you that you can't beat roulette consistently.
You tell me to what extent you think you can beat roulette, and I'll propose a bet against you.
As with all previous posters who claim to beat a negative expectation game, you will likely just fade away.....
choosing numbers randomly will NOT produce different results compared to your number selection method.Quote: NotnabAlso, choosing numbers randomly will produce different results compared to my number selection method and roulette strategy. I wait until 3 specific elements line up.
I appreciate your calculations, but unfortunately they have no bearing on my method and are unable to explain my results. ;-)
Each time you make your 19 bets you have a 19/37 or 51.35% chance of success on the very next spin.
My randomly chosen 19 numbers have exactly a 19/37 or 51.35% chance of success on the very next spin.
You can not change the actual probability or the combined probability by some "empirical statistics" you have.
Neither can God herself
Good Luck to you in your studies!
Roulette can only be beaten by luck.
