Increasing Odds

[GLC]

Here's a fun system to test that will win a handful of chips more often than it loses.

We're going to start betting on an even chance until a win or 3 losses.

Then we will bet 2 times on 2 dozens until 2 wins or 2 losses.

Then we will bet on 5 lines until 3 wins or a loss, and then again on 5 lines until 4 wins or a loss.

We will start betting that the last number to spin will not hit before we win our unit.  This is not exactly true, but close enough.

Here's a suggested progression:
Even chance bets:
1-2-4 limited martingale.  A win gives us +1.  3 losses cost us 7 units

1st Double Dozen bet:
4-4, must win 2 times.  1 loss costs us 8 units and 2 wins nets us +1 over the 7 previous lost units.  If we lose at this level we will be down 15 units (11 if we win once then lose).

2nd Double Dozen bet:
8-8, must win 2 times.  1 loss costs us 16 units and 2 wins nets us +1 over the 15 previous lost units.  If we lose at this level we will be down 31 units (give or take 8-12 units).

1st 5 line bet:
11-11-11-11-11, must win 3 times.  1 loss costs us 55 units and 3 wins nets us +2 over the 31 previous lost units.  If we lose at this level we will be down 86* units.

2nd 5 line bet:
22-22-22-22-22-, must win 4 times.  1 loss costs us 110 units and 4 wins nets us +4 over the 86 previous lost units.  If we lose at this level we will be down 196* units.  Take the loss and call it a day.

Bet selection:

The first even chance bet is on the opposite Hi/Lo of the number that just spun.  We will bet 3 times for the opposite Hi/Lo.  Let's say the last number to spin was B15.  We will bet for Hi to spin for the next 3 spins and we will bet 1-2-4 progression.  Any win nets us +1 and  we can bet for the opposite Hi/Lo of the number we just won on.

If we lose the 3 even chance bets, we go to the 1st double dozen bet.
We will bet 4-4 on the 2 dozens our number is not in.  So in our example of B15, we will bet 4 units on the 1st dozen and the 3rd dozen.  If we win, we will bet 4-4 on the same 2 dozens.  If we win again, we will be +1 overall and will re-set to 1 unit on the opposite Hi/Lo.

If we lose the 1st double dozen bet, we will bet 8-8 on the same 2 dozens for a loss or 2 wins.  If we get 2 wins we will be at +1 overall and will reset to 1 on the opposite Hi/Lo.

If we lose the 2nd double dozen bet, we will bet the first 5 lines bet at 11 units each.  We will bet on the 5 lines that don't contain our number.  In this case B15.  We must win 3 times in a row on this bet to reach +2.  If we lose we will be down 86 units.

If we lose the 1st 5 lines bet, we will go to the 2nd 5 lines bet at 22 units on each line.  We will bet again on the 5 lines that don't contain our number, B15.  We must win 4 times at this level to fully recover and we will be at +4; re-set.

*Now, when I said above that if we lose a bet we will be at a certain number of units in the hole, that's not exactly true because if we win and then lose on the double dozen bets, we will not be down the total of the double dozen bet because we will have won either 4 units or 8 units before we lost either 8 units or 16 units.  The same is true for the line bets.  We could win 1, 2 or 3 times before losing and it could reduce the number of units lost substantially.

Like I said, this is a pretty solid method and with a little luck, you can win a fistfull of dollars without having a total loss.  A maximum loss will be -196 units.  An almost impossibility.  In order to lose everything, we will have to lose 7 times in a row.  This means the Hi/Lo, dozen or line our number is in will have to come up 7 times in a row.  It can happen, but with a little luck, you won't see it too often.

Naturally, if you can play on non-zero roulette, you will have the best odds of winning.

Enjoy,

GLC

P. S.  One safety measure you can take is if you lose the 1-2-4 bets and win the 1st 4-4 bet on double dozens, you will only be at -3 units and can drop back and bet the 4 unit bet on the even chance.  That way if you win you are fully recovered and if you lose, you are only back to -7 and can start the 4-4 double dozen bet again.

A similar situation develops if you lose at 4-4 and win the 1st 8-8 bet, you could drop back down and replay the 4-4 double dozen bet line.

A similar situation develops if you lose at 8-8 and win the 1st 11 unit line bet.  You will have recovered 11 units of the 16 units loss on the 8-8 level, so you could drop back down to the 8-8 bet and if you win, you will be fully recovered and can start over.

These are just examples of how you can buffer the losses a little bit.

[vladir]

Hey, nice one.
Try this tweaks:

First, wait for 2 hits on the same hi/lo section to start betting (a virtual loss on the first bet).

Then, in the first progression, while playing hi-lo, instead of 1-2-4 , do only 1-2. A win gives us +1.  2 losses cost us 3 units (instead of previously 7)


Then continue with your plan, adjusting the bets accordingly for the dozens:

1st Dozen bet:
2-2 

2nd Dozen bet:
4-4

(After loosing the 2nd dozen bet, we will be in the worst case scenario with -15 units)

And now I suggest 2 options:

OPTION A

1st 5 line bet:
16-16-16-16-16 (we just need to win ONCE, instead of 3 times to be in profit - 1 unit +)

2nd 5 line bet:
25-25-25-25-25 (must win 4 times to be in profit - 5 units +)

(Maximum loss following this is: 1+2+2+2+4+4+16+16+16+16+16+25+25+25+25+25 = 220)

OPTION B

1st 5 line bet:
6-6-6-6-6 (must win 3 times to be in profit - 3 units +)

2nd 5 line bet:
12-12-12-12-12 (must win 3 times to be  in profit - 3 units +)

(Maximum loss following this is: 1+2+2+2+4+4+6+6+6+6+6+12+12+12+12+12 = 105)

[GLC]

Vladir,

I like it.  I don't think the numbers are necessarily that critical.  It's the idea that can give us a shot at winning a few chips.

Every suggestion you made is valid.  It's a good example of how people should approach systems.

1.  Understand the system.
2.  Adjust triggers to suit yourself.
3.  Adjust progressions to suit your bankroll parameters.
4.  Test thoroughly before risking money.
5.  Always have a stop loss in case the unfortunate happens.
6.  Never play with money you can't afford to lose.
7.  Repeat above until you find the "mother lode" or die a grizzly, old prospector.

LOL,

GLC

[Bayes]

Nice little system, George. The chance of losing all 7 bets is 0.06% on a single zero wheel, pretty good. (a loss every 1,672 or so attempts, on average) 

Regarding vladir's tweak, it certainly keeps the outlay down but you don't really gain anything by waiting for two hits on the hi/lo section before betting against it, because you lose out on all the chops. Unfortunately, in any apparent advantage there's always a trade-off. However, you pays your money and takes your choice. 

This kind of system is better played on a non-zero wheel. It's often not appreciated what effect the zero has on these kind of high coverage bets. Re-running the maths but using fair odds (no zero) gives a probability of losing of 0.04% or a loss every 2,524 attempts, which is a % increase of 51%!!

Details for single zero:

probability of loss on an EC = 19/37
probability of loss on 2 dozens = 13/37
probability of loss on 5 lines = 7/37

The sequential probability of 7 losses is then - (19/37)³ × (13/37)² × (7/37)² = 0.0005982

1 / 0.0005982 = 1,672

And for no-zero:

(18/36)³ × (12/36)² × (6/36)² = 0.0003962

1 / 0.0007716 = 2,524

Percentage difference = (2,524 â€" 1,672) / 1,672 = 0.51 or 51%

[GLC]

Bayes,

Thanks for your mathematical analysis of this system.  Yes, since testing systems on no zero wheel, it makes playing on a double zero wheel feel like suicide.

Cheers,
GLC

[GLC]

Nice little system, George. The chance of losing all 7 bets is 0.06% on a single zero wheel, pretty good. (a loss every 1,672 or so attempts, on average) 

Regarding vladir's tweak, it certainly keeps the outlay down but you don't really gain anything by waiting for two hits on the hi/lo section before betting against it, because you lose out on all the chops. Unfortunately, in any apparent advantage there's always a trade-off. However, you pays your money and takes your choice. 

This kind of system is better played on a non-zero wheel. It's often not appreciated what effect the zero has on these kind of high coverage bets. Re-running the maths but using fair odds (no zero) gives a probability of losing of 0.04% or a loss every 2,524 attempts, which is a % increase of 51%!!

Details for single zero:

probability of loss on an EC = 19/37
probability of loss on 2 dozens = 13/37
probability of loss on 5 lines = 7/37

The sequential probability of 7 losses is then - (19/37)³ × (13/37)² × (7/37)² = 0.0005982

1 / 0.0005982 = 1,672

And for no-zero:

(18/36)³ × (12/36)² × (6/36)² = 0.0003962

1 / 0.0007716 = 2,524

Percentage difference = (2,524 â€" 1,672) / 1,672 = 0.51 or 51%

To help my understanding and those of our readers, can you show what our expected win amount would be in 1,000 spins on a single zero wheel?

Saying that we should win 1671 times and lose once sounds like we should net 1671-196=1475 units every 1672 attacks.  I know that this isn't possible.

I know that when we work out the math, we will end up with a 50/50 shot on a non-zero wheel.  Can you show us the math?

Thanks,

George

[vladir]

The problem I see in the calculations is that only the event of loosing 7 in a row is being accounted to.

But in this system, we can loose not only with 7 consecutive losses, but also with some combinations of losses and wins in the midle (wich are not being considered in the way the calculation are being made I guess).

We can loose with, for example, this sequence:  3Losses, 1Win, 1 Loss, 1 Win, 1Loss, 2 Wins, 1 Loss, 3 wins, 1 Loss.

Cheers

[GLC]

Thanks Vladir

[vladir]

No problem. I hope bayes can give us exact numers for this kind of bet

[Bayes]

The problem is it's not easy to work out all the possible ways of losing (because there are so many of them), especially if there are different staking options which only increase the number of possibilities. The best solution is to do a simulation. But yes, if the final result comes out positive then it usually means some scenario or other hasn't been taken into account. Mathematically speaking, there is no possibility of getting a profit in the long term. 

Let me know a fixed staking plan and I'll do a simulation. 

I don't mean to sound negative, but this is why I don't usually bother with simulations any more. All a sim does is to apply 'the math' in another way.

[GLC]

Thanks guys,
It reminds me of something I read from the Wizard of Odds many years ago regarding overcoming the house edge with roulette using bet selection and progressions.  He stated that the best way to understand why no system devised can defeat the game in the long run is that the wheel doesn't know what you're doing.  All it's doing is spitting out random numbers.  Just because you put together some unique or complicated bet selection and/or vary the size of the units bet doesn't mean anything in the long run here's why.

Let's say we're betting follow-the-last on R/B with a D'Alembert progression.  Various sequences can be misleading and make us think that we're winning but if you take every bet of R at 1 unit and pull them out of the sequence into a pool, the final result will be a loss of 2.7%.  Then take out every bet of B at 1 unit and do the same thing and it will have the same result.  This applies to every system.  They're all made up of tens or hundreds or even thousands of individual bets and all have a negative expectation of 2.7%.  So eventually you will lose.

The wheel doesn't know in what sequence these individual bets are made.  All it knows is that you have placed a bet on number 23 of 4 units.  What you did before that is not remembered nor does it effect this bet in any way and what you are going to do for the next result is the same.

That's why I state that all my systems are devised with the idea in mind that rather than go to the casino and put $200 in a slot machine, use one of these systems and you may have a chance to win a few units on a regular basis.  In the long run, no.

The more complicated the system, the longer things may take for the averages to accrue, but eventually they will.  Although, with some luck, you may be able to win more than you lose in the long run.  Some people are just lucky.

I've been trying for years to prove him wrong, but so far I can't claim victory.  Others have, but not everyone can make their systems work, so it's still not conclusive.

Luck to all,

GLC

[flukey luke]

It reminds me of something I read from the Wizard of Odds many years ago regarding overcoming the house edge with roulette using bet selection and progressions.  He stated that the best way to understand why no system devised can defeat the game in the long run is that the wheel doesn't know what you're doing.  All it's doing is spitting out random numbers. 

Hello George,

It's correct that the wheel does not know what YOU are doing... BUT... you can see what the wheel is doing and play along accordingly. You only ever risk one chip to find out. Where many go wrong is chasing with more chips hoping the wheel will conform. Treat each spin as an individual event (which it is) and re-evaluate the position after every spin. Each and every chip should be treated as gold dust. Systems/methods that require large bankrolls and can wipe half of the bankroll in short order are not in my humble opinion the way to go.

cheers