Can roulette be beat by volatility extraction?
March 4th, 2026 at 7:54:04 PM
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I need to be clear here…
I do not mean in an EV sense, I do not expect the wheel to change. EV is based on total wagered…kinda a shit metric and stupid premise in general. More important for the casino’s bottom line than mine.
But, after tons of monte carlos, 100k round sims etc
it does appear that certain strategies are stable enough to bounce up and down enough times
that you can extract more than your initial starting balance on average.
Roulette can’t be beaten in a static odds sense, but I am quite confident various bet types and progressions can create enough fluctuations that they total FAR GREATER than your starting balance.
and any gambler probably knows this or has witnessed this sort of thing. You have $100 and play and get the notification on ur phone “you have wagered $603 in the past 30min”
meanwhile ur balance is $45 lol
lets use $200 bankroll, 3 streets as an example
$1 a pop. Double every 4 losses, reset on win.
• 4 losses → $12
• 8 losses → $36
• 12 losses → $84
• 16 losses → $180
• 17th loss → +$24 = $204
You bust on the 17th straight loss.
Bust Probability
q^{17} = 0.7632^{17}
≈ 0.98%
you have a 99+% chance at profiting any given cycle… as it gets deeper the profit increases.
avg cycle profit ~ $15
regardless… on average how many cycles could you do this til bust?
well… the MEDIAN is 78, the mean is ~ 113
so lets super lowball here…
if you withdraw $15 each successfully completed cycle
and survive 30 cycles
is that more than $200..??
sounds closer to $450
and so lets look at that, what’s the chance of hitting just 30 cycles before bust?
is exceptionally below the avg expectation but It makes an important point.
~ 76%
TLDR you can more than double your money at a rate of 70+%
but the vacuum math nerds (likely casino hired) look at everything per spin and talk static odds instead of probability windows.
they lie to your face claiming “the game can’t be beat”
actually it definitely can
what you ACTUALLY mean is…
…you cannot alter the static odds per spin.
yeah uhh.. everyone already knew that
shocker!
even if my math isn’t perfect, I know for a fact this along with many other strategies, create enough fluctuations in the bankroll… that you can extract 2x your starting balance
more times than not
P({survive } { cycles}) = q^n
• Survive 10 cycles: 0.9912^{10} \approx 0.914 → 91.4%
• Survive 20 cycles: 0.9912^{20} \approx 0.835 → 83.5%
• Survive 30 cycles: 0.9912^{30} \approx 0.768 → 76.8%
• Survive 50 cycles: 0.9912^{50} \approx 0.642 → 64.2%
• Survive 100 cycles: 0.9912^{100} \approx 0.412 → 41.2%
• Survive 200 cycles: 0.9912^{200} \approx 0.170 → 17.0%
even just 20 cycles gets u over $200 profit.
please someone tell me im wrong.
I do not mean in an EV sense, I do not expect the wheel to change. EV is based on total wagered…kinda a shit metric and stupid premise in general. More important for the casino’s bottom line than mine.
But, after tons of monte carlos, 100k round sims etc
it does appear that certain strategies are stable enough to bounce up and down enough times
that you can extract more than your initial starting balance on average.
Roulette can’t be beaten in a static odds sense, but I am quite confident various bet types and progressions can create enough fluctuations that they total FAR GREATER than your starting balance.
and any gambler probably knows this or has witnessed this sort of thing. You have $100 and play and get the notification on ur phone “you have wagered $603 in the past 30min”
meanwhile ur balance is $45 lol
lets use $200 bankroll, 3 streets as an example
$1 a pop. Double every 4 losses, reset on win.
• 4 losses → $12
• 8 losses → $36
• 12 losses → $84
• 16 losses → $180
• 17th loss → +$24 = $204
You bust on the 17th straight loss.
Bust Probability
q^{17} = 0.7632^{17}
≈ 0.98%
you have a 99+% chance at profiting any given cycle… as it gets deeper the profit increases.
avg cycle profit ~ $15
regardless… on average how many cycles could you do this til bust?
well… the MEDIAN is 78, the mean is ~ 113
so lets super lowball here…
if you withdraw $15 each successfully completed cycle
and survive 30 cycles
is that more than $200..??
sounds closer to $450
and so lets look at that, what’s the chance of hitting just 30 cycles before bust?
is exceptionally below the avg expectation but It makes an important point.
~ 76%
TLDR you can more than double your money at a rate of 70+%
but the vacuum math nerds (likely casino hired) look at everything per spin and talk static odds instead of probability windows.
they lie to your face claiming “the game can’t be beat”
actually it definitely can
what you ACTUALLY mean is…
…you cannot alter the static odds per spin.
yeah uhh.. everyone already knew that
shocker!
even if my math isn’t perfect, I know for a fact this along with many other strategies, create enough fluctuations in the bankroll… that you can extract 2x your starting balance
more times than not
P({survive } { cycles}) = q^n
• Survive 10 cycles: 0.9912^{10} \approx 0.914 → 91.4%
• Survive 20 cycles: 0.9912^{20} \approx 0.835 → 83.5%
• Survive 30 cycles: 0.9912^{30} \approx 0.768 → 76.8%
• Survive 50 cycles: 0.9912^{50} \approx 0.642 → 64.2%
• Survive 100 cycles: 0.9912^{100} \approx 0.412 → 41.2%
• Survive 200 cycles: 0.9912^{200} \approx 0.170 → 17.0%
even just 20 cycles gets u over $200 profit.
please someone tell me im wrong.
March 4th, 2026 at 8:14:51 PM
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@wizard hook me up on that “nobody can beat roulette” money 😂
actually im willing to believe it was a farce proposition and yall expect people to beat the wheel itself which everyone knows isnt possible
actually im willing to believe it was a farce proposition and yall expect people to beat the wheel itself which everyone knows isnt possible
March 4th, 2026 at 11:19:44 PM
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Quote: cheddachaser$1 a pop. Double every 4 losses, reset on win.
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Thread moved to Betting Systems.
Quote:please someone tell me im wrong.
All betting systems are equally worthless.
May the cards fall in your favor.
March 4th, 2026 at 11:25:38 PM
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Quote: Dieter...
All betting systems are equally worthless.
link to original post
If no progression can work at all, neither can card counting.
Prove me wrong!
March 4th, 2026 at 11:35:42 PM
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Quote: AutomaticMonkeyQuote: Dieter...
All betting systems are equally worthless.
link to original post
If no progression can work at all, neither can card counting.
Prove me wrong!
link to original post
Monkey,
The wheel gets shuffled every hand.
May the cards fall in your favor.
March 4th, 2026 at 11:45:53 PM
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Quote: DieterQuote: AutomaticMonkeyQuote: Dieter...
All betting systems are equally worthless.
link to original post
If no progression can work at all, neither can card counting.
Prove me wrong!
link to original post
Monkey,
The wheel gets shuffled every hand.
link to original post
Aw shucks, I can't come up with some pedantic possibility where a progression might give you something on a wheel too.
I'm at a loss for words!
March 5th, 2026 at 12:41:38 AM
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$1 a pop. Double every 4 losses, reset on win.
• 4 losses → $12
• 8 losses → $36
• 12 losses → $84
• 16 losses → $180
• 17th loss → +$24 = $204
You bust on the 17th straight loss.
$1 per street, 3 streets. Streets pay 11:1 or 12 for 1.
So you bet $3 four times and wind up at -$12. A win would put you back at even or above.
2nd set, you bet $6 four times and wind up at -$36. A win on the last of those four spins would get you back $24, you'll still be -$12 short. You could just start over those 4 spins at 3X $2 again.
3rd set, you bet $12 four times and wind up at -$84. A win on the last spin could put you back up $48, you'll still be $36 short. You could just start over with those 4 spins at 3X $4 again.
4th set, you bet $24 four times and wind up at -$180. A win on the last spin could put you back up $96, you'll still be $84 short. You could just start over with those 4 spins at 3X $8 again.
5th set, you bet $24 one time and wind up at -$204. A win on the last spin could put you back up $96, you'll still be $108 short. You could just start over with the last 4 spins at 4X $8 again.
For a $3 roulette machine, you need to multiply all values by $3. For a $5 machine, multiply all values by $5. For a $10 table, multiply all values by $10. For a $15 table, multiply all values by $15.
$3 machine buy-in = $612
$5 machine buy-in = $1,020
$10 table buy-in = $2,040
$15 table buy-in = $3,060
$20 table buy-in = $4,080
If you were starting at $5 machines and doubling your bets every time you exceeded the next doubled buy-in, you'd need 13.6 wins ahead to double up your buy-in. (16X $75 = $1,200) (16X $150 = $2,400), (16X $300 = $4,800; and winning $180 extra + $360 extra + $720 extra equals a $1,260 extra for a restart or a session up. So at 48 cycles, you'd be at around a 2 in 3 chance of making it if you reset the way your numbers did. If you double up your bets on 16 wins, you could cash out $1.2K + $2.4K+ $4.8K = $8.4K + $1K buy-in = $9.4K and still be under a CTR.
I'd have to see more about how your actual betting turns out when you reset your bets on a win. It's vague to me as it relates to profits. But be sure, the automatic plunger on these roulette machines can evade your betting choices to the extreme. Get to a real table after you graduate to the table minimums.
I personally prefer to bet two dozens bets and climb up a half spin bet per win.
It's important to find a high limit roulette wheel with only one zero. With your street bets counting as inside bets, a $100 table minimum could have 3X $35 or $40 bets to start. With $40 street bets, you'd need $8,160 buy-ins. You could double up your street bets on 16 wins to $80 bets, and $160 bets and still be under the $500 table maximum.
You could win (16X $600 = $9.6K, 16X $1.2K = $19.2K, $16X $2.4K =$38.4K) or a total of $67.2K for 48 wins.
But where I live, the double 0 wheel has a limit of $2,500 per spin. I could eventually get my street bets to $200, then $400, then $800 on a $40.8K buy-in and cash out at $376,800.
Really, all it takes is 3 winning sessions in a row to get from $1K to $400K+. After that, you'll have 400 buy-ins to try it all over again.
• 4 losses → $12
• 8 losses → $36
• 12 losses → $84
• 16 losses → $180
• 17th loss → +$24 = $204
You bust on the 17th straight loss.
$1 per street, 3 streets. Streets pay 11:1 or 12 for 1.
So you bet $3 four times and wind up at -$12. A win would put you back at even or above.
2nd set, you bet $6 four times and wind up at -$36. A win on the last of those four spins would get you back $24, you'll still be -$12 short. You could just start over those 4 spins at 3X $2 again.
3rd set, you bet $12 four times and wind up at -$84. A win on the last spin could put you back up $48, you'll still be $36 short. You could just start over with those 4 spins at 3X $4 again.
4th set, you bet $24 four times and wind up at -$180. A win on the last spin could put you back up $96, you'll still be $84 short. You could just start over with those 4 spins at 3X $8 again.
5th set, you bet $24 one time and wind up at -$204. A win on the last spin could put you back up $96, you'll still be $108 short. You could just start over with the last 4 spins at 4X $8 again.
For a $3 roulette machine, you need to multiply all values by $3. For a $5 machine, multiply all values by $5. For a $10 table, multiply all values by $10. For a $15 table, multiply all values by $15.
$3 machine buy-in = $612
$5 machine buy-in = $1,020
$10 table buy-in = $2,040
$15 table buy-in = $3,060
$20 table buy-in = $4,080
If you were starting at $5 machines and doubling your bets every time you exceeded the next doubled buy-in, you'd need 13.6 wins ahead to double up your buy-in. (16X $75 = $1,200) (16X $150 = $2,400), (16X $300 = $4,800; and winning $180 extra + $360 extra + $720 extra equals a $1,260 extra for a restart or a session up. So at 48 cycles, you'd be at around a 2 in 3 chance of making it if you reset the way your numbers did. If you double up your bets on 16 wins, you could cash out $1.2K + $2.4K+ $4.8K = $8.4K + $1K buy-in = $9.4K and still be under a CTR.
I'd have to see more about how your actual betting turns out when you reset your bets on a win. It's vague to me as it relates to profits. But be sure, the automatic plunger on these roulette machines can evade your betting choices to the extreme. Get to a real table after you graduate to the table minimums.
I personally prefer to bet two dozens bets and climb up a half spin bet per win.
It's important to find a high limit roulette wheel with only one zero. With your street bets counting as inside bets, a $100 table minimum could have 3X $35 or $40 bets to start. With $40 street bets, you'd need $8,160 buy-ins. You could double up your street bets on 16 wins to $80 bets, and $160 bets and still be under the $500 table maximum.
You could win (16X $600 = $9.6K, 16X $1.2K = $19.2K, $16X $2.4K =$38.4K) or a total of $67.2K for 48 wins.
But where I live, the double 0 wheel has a limit of $2,500 per spin. I could eventually get my street bets to $200, then $400, then $800 on a $40.8K buy-in and cash out at $376,800.
Really, all it takes is 3 winning sessions in a row to get from $1K to $400K+. After that, you'll have 400 buy-ins to try it all over again.
Last edited by: ChumpChange on Mar 5, 2026
March 5th, 2026 at 1:09:11 AM
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Quote: Dieter
All betting systems are equally worthless.
link to original post
Untrue. There's a long history of successful betting systems in various forms of sports betting which yielded a significant positive expectation. You need to add a qualifier "in a game of independent trials".
For roulette specifically the answer is not technically accurate either. On a wheel with any physical bias creating serial correlation, progression systems will have a second-order effect on expected value — positive for anti-martingale-type systems, negative for martingale-type systems — proportional to the degree of bias. Most wheels do in fact have some kind of bias though not generally < p=1/34. Games like blackjack and baccarat are similar: there are second-order effects due to card removal which have a very minor effect on expectation.
As a practical matter at roulette though you would be looking to exploit the source of any bias directly rather than use a progression.
