Effectiveness of System 3/2 in Roulette - Pausing after 6 wins and commencing after an L

I am starting to explore gambling and have been researching on various systems/strategies.

Can someone please tell me the efficiency of this strategy for the American Roulette (00) with a $10 minimum bet and a table limit of $3000. If someone can simulate this, it will be fantastic.

Betting System/Strategy:
Step 1. Bet $6 on 1-18 and Bet $4 on 3rd 12 (i.e. 25 - 36). For every win, you win $2. When you lose, you lose $10.
Step 2. As soon as you lose, switch the IMMEDIATE FOLLOWING bet as follows: Bet $12 on 19-16 and Bet $4 on 1st 12 (i.e. 1 - 12).
If you lose this bet, PAUSE from the game.
If you win this bet, continue betting as follows TILL YOU LOSE: Bet $6 on 19-16 and Bet $4 on 1st 12 (i.e. 1 - 12). Once you lose, PAUSE from the game.
Step 3. Re-enter the game, when any of the numbers 19-24 appear ONCE. Go to Step 1.

It won't lose more per $ wagered than anyone elses roulette system on this planet.

Quote: affypburgh

I am starting to explore gambling and have been researching on various systems/strategies.

Can someone please tell me the efficiency of this strategy for the American Roulette (00) with a $10 minimum bet and a table limit of $3000. If someone can simulate this, it will be fantastic.

Betting System/Strategy:
Step 1. Bet $6 on 1-18 and Bet $4 on 3rd 12 (i.e. 25 - 36). For every win, you win $2. When you lose, you lose $10.
Step 2. As soon as you lose, switch the IMMEDIATE FOLLOWING bet as follows: Bet $12 on 19-16 and Bet $4 on 1st 12 (i.e. 1 - 12).
If you lose this bet, PAUSE from the game.
If you win this bet, continue betting as follows TILL YOU LOSE: Bet $6 on 19-16 and Bet $4 on 1st 12 (i.e. 1 - 12). Once you lose, PAUSE from the game.
Step 3. Re-enter the game, when any of the numbers 19-24 appear ONCE. Go to Step 1.

I can help you out with this. After mathematically reviewing your system I have determinal it will lose in the long run.

ZCore13

Thanks for your reply. Is there any strategy/system that I can use once a week? If I can score $100-$200 per weekend gambling, that will be awesome. Any suggestions will be welcome...

PS: I definitely know the house edge gets you and all betting systems lose money in the long run.

Quote: affypburgh

Thanks for your reply. Is there any strategy/system that I can use once a week? If I can score $100-$200 per weekend gambling, that will be awesome. Any suggestions will be welcome...

PS: I definitely know the house edge gets you and all betting systems lose money in the long run.

I think your best option would be to learn to count cards playing blackjack or play full pay video poker. There are no betting systems that can beat a game that has a house advantage, considering all casinos have table limits.

ZCore13

You want an efficiency analysis? Here's one:

Divide the system into three steps
Step 1 - bet 1-18 and 25-36 until you lose
Step 2 - bet 1-12 and 19-36 until you lose
Step 3 - skip spins until 19-24 comes up

First, a bit of math (aw, man):
The expected number of consecutive wins for a bet with probability p is:
0 x (1 - p) + 1 x p x (1 - p) + 2 x p² x (1 - p) + ...
= (1 - p) x p x (1 + 2p + 3p² + 4p³ + ...)
= (1 - p) x p x (1 + p + p² + p³ + ...)²
Since p < 1, this = (1 - p) x p x (1 / (1 + p))²
= p x (1 - p) / (1 - p)²
= p / (1 - p)

In Step 1, p = 15/19, so you are expected to win 15/4 times before a loss; that's 15/4 + 1 = 19/4 bets of 10.
In Step 2, p = 15/19 again; that's 19/4 bets of 16
In Step 3, you skip spins until 19-24 comes up; treat this as a "win" if 0, 00, 1-18, or 25-36 comes up, so p = 16/19, and you are expected to "win" 16/3 times before losing, so that's 16/3 + 1 = 19/3 bets of zero
Each cycle is expected to have 19/4 + 19/4 + 19/3 = 95/6 spins, with a total amount bet of 19/4 x 10 + 19/4 x 16 + 19/3 x 0 = 247/2.
That is an average bet per spin of 741/95 = 7.8
In other words, it is as "efficient" as betting 7.8 on each spin. It doesn't matter how the 7.8 is bet, as long as you don't bet on 0-00-1-2-3, since, except for that bet, the house edge in roulette is the same for all bets.
You get the same "efficiency" if you bet 10, 10, skip, 19, skip on any of the even-money bets.

Remember: the probability of red coming up 10 times in a row in American roulette is about 1 / 1758, but the probability of red coming up 9 times in a row followed by black is also about 1 / 1758.

The best non mathematical explanation I ever heard saying why the martingale won't work is this: The odds might be 10 billion to one against getting run over and killed by an elephant charging down the street. But if you saw one charging down the street straight at you would you stand still? Another one is it's like picking up nickels in front of a bulldozer.

Quote: lilredrooster

The best non mathematical explanation I ever heard saying why the martingale won't work is this: The odds might be 10 billion to one against getting run over and killed by an elephant charging down the street. But if you saw one charging down the street straight at you would you stand still? Another one is it's like picking up nickels in front of a bulldozer.

Here's the one I use:

If the casino offered you a bet that had a 1 in 613 chance of winning but paid off 1023-1, would you bet it?
Because that's how a 10-step double-zero roulette Martingale (5 minimum, maximum anywhere from 2560 to 5115) works - except that you are the casino.

You probably already know this but Jean le Rond d'Alembert (d'Alembert progression) was actually a brilliant 18th century mathematician who unfortunately had one very wrong theory.


"While he made great strides in mathematics and physics, d'Alembert is also famously known for incorrectly arguing in Croix ou Pile that the probability of a coin landing heads increased for every time that it came up tails. In gambling, the strategy of decreasing one's bet the more one wins and increasing one's bet the more one loses is therefore called the D'Alembert system, a type of martingale."

Hi Affy,

Sincerest welcome to the forum.
Be aware that American roulette won't be 'beaten' by any such gambling 'system.' It has a dreadful house advantage of 5.26% and that means pure and simple that you can expect to tend towards losing 5.26% of the sum of all of what you put on the table. No system changes that.

You'll find the regulars here pretty helpful in putting straight anyone who thinks they have the makings of a winning system. That may come across as hostility, but it's never intended to offend, just to advise.

Systems such as yours can change the 'experience' of losing playing, such that you might find a system that wins more often than it loses ( like Martingale) but then loses BIGGER than it wins, or maybe a system (inverse Martingale) that loses more often than it wins, but when/if it wins, it wins big.

It's only the shape of the downward graph of bankroll that changes, not the long term prospect.

Quote: affypburgh

I am starting to explore gambling and have been researching on various systems/strategies. . . .

If someone can simulate this, it will be fantastic.

I'd be delighted to create a simple simulator for this or any similar roulette system, using nothing more complex than Excel. I'd even show you how to customise it.

What could you do for me in return $:o)

To save you some deep analysis.....
It makes absolutely no difference which half or third, or whatever you place your wagers on. You will be on a path towards giving 2/38 times whatever action you place on the table as your contribution to the casino's profits.
When you plot your rolling bankroll as a chart, for any system such as this, you get a characteristic shape which will illustrate your path to probable ruin.
In the case of Martingale it would be a shape something like this. . .

In the case of D'alembert, it will be something like this. . .


The common factor is that you will lose your bankroll more often than you double it. Simple as.

Quote: affypburgh

Thanks for your reply. Is there any strategy/system that I can use once a week? If I can score $100-$200 per weekend gambling, that will be awesome. Any suggestions will be welcome....

Great news!
Sure.
I can give you a system with which you can achieve a 99% probability of winning $200 for each and every session, whether you play one session per week or one session per hour.

The system will be crushingly simple to execute and you would not need to have many sessions beyond 10 minutes to achieve that. Some sessions would only take one spin before you'd be able to stand up, cash out, and take the journey home.

I'd be happy to guarantee my system and I'm sure that we could get strong mathematical analysis and/or simulations to prove conclusively that my claims are true.

Quote: affypburgh

Thanks for your reply. Is there any strategy/system that I can use once a week? If I can score $100-$200 per weekend gambling, that will be awesome. Any suggestions will be welcome...

For roulette, bet $100 - $200 on either red or black one time each week. You will win between $100 and $200 dollars 18 out of every 37 weeks.

Dear sabre, Zcore13 & lilredrooster: Thanks so much for your quick and insightful replies.

Dear ThatDonGuy & OnceDear: You guys made my day. Thank you so much for taking the pain to crunch up numbers, do the math and explain me the about the premise. I am overwhelmed. A BIG Thank you. If you guys had explained this to me in person, I would have treated you both to a round of drinks.

I will continue to learn the tricks of the trade and learn more about numbers/probabilities/odds before stepping in a casino.

Happy Betting!!

Quote: affypburgh

Dear ThatDonGuy & OnceDear: You guys made my day. Thank you so much for taking the pain to crunch up numbers, do the math and explain me the about the premise. I am overwhelmed. A BIG Thank you. If you guys had explained this to me in person, I would have treated you both to a round of drinks. !

You are welcome.
I suspected for a moment that you were being sarcastic in that reply. . . but only for a moment.

There are many helpful members here that have already either done the maths, better than I ever do,, or who know where to go straight to for analysis. Rest assured that every roulette system that you could come up with would fare no better than any of the previous ones from the past few hundred years.

But don't be disheartened, though. There are systems which can make that game and others more fun and less dangerous. It's mostly a matter of style and bankroll.

There are also experts here who know how to turn certain games into long term winning games that you could actually earn a living from. But not roulette!

And numbers and probabilities can be fun.

Quote: TomG

For roulette, bet $100 - $200 on either red or black one time each week. You will win between $100 and $200 dollars 18 out of every 37 weeks.

OK. Here's my system. You'll need an obliging casino with high limits. ( It's a theoretical exercise )

I offered >99% probability of leaving the casino with $200 profit.

Seriously ( Well theoretically ).

Are you ready?

American double zero roulette.

Start session.

Bet $200 on each of the numbers 2 to 36. That's total funds of $7,000 invested.

Most times (35/38=92.1% of the time) the ball won't land in 1, 0, or 00 and you will win on one of your numbers. You will pick up your 36 x $200 = $7200 chips and cash out with your $200 profit. Simple.

But don't worry if you lose with a 1 a 0 or a 00. Simply place a second set of wagers of $7,200 on each of the numbers 2 to 36. Which 35 numbers makes no difference.

You are likely to win this wager. You are likely to pick up 36 x 7200 = $259,200. That will mean you've invested a total of $7,000+ $252,000 = $259,000 and so you've made a session profit of $200, so again you cash out and walk away.

Do you start to see the problem?

To work out your probability of a winning session, we first work out the probability of a losing session.
The probability of losing wager 1 is 3/38 = 0.079 = 7.9%
The probability of losing wager 2 is 3/38 = 0.079 = 7.9% BUT you only place wager 2 if you already lost wager 1

So the probability of losing wager 1 AND losing wager 2 is (3/38)x(3/38) = 0.006233 = 0.62%
The probability of NOT losing both of wager 1 and wager 2, That is the probability of a winning session, is (100-0.6233) = 99.3767%

I.e. greater than 99% probability of having a winning session where you walk out with a profit of $200.

Do that every day for 3 years with a potential profit of 3 x 365 x 200= $219,000

But of course, occasionally, about once every couple of years, you'll lose $259,000. See the problem?

A bit like russian roulette with a single bullet in a six chamber revolver. Most times (5/6=83% of the time) you would survive a single round.

But hey, until that happens, every week, you can say "Look guys I continue to win $200 every week with OnceDear's great system."

If I was on commission, I'd have you mail me $10 dollars after each winning session with which to buy my beer. When no check arrives I'd shrug and sell my system to the next guy.