System based purely in Maths

[gingermolloy]

OK gents, I feel an impatience in here. Apologies, but to understand this system I have to go through all this stuff. Apologies again If you guys know this lot already.

So we have the Martingale system as our starting point.

Instead of just winning 1 unit per win I have changed the progression so that we win 1 unit per spin,  whether we win or loose on that spin. In order to achieve this we must increase the bet by an amount that covers all the previous bets made plus the number of spins in that we are.

Example I think!!!:

So.
£1 on Red    Win!

If we win we go back to the start so

£1 on red   Lose!
£3 on red (if we win we win £2, £1 for each spin, plus the £1 we have lost)   Lose
£7 on red (if we win we win £3, £1 for each spin, plus the £4 we have lost)   win
back to the start.

The progression therefore is as follows: 1,3,7,15.

What is very important here (for the maths to work) is that if we lose after the £15 bet, we must stop and go back to the start.

If you run this system on any evens bet you have a 93% chance of winning and a ratio of 0.93 giving us a 7% advantage. (again, i will elaborate on the maths later).

The rest of this system is all about applying this theory to different bets in order to balance your chance of winning with the ratio you can achieve. 

The theory could be put into a statement as follows:

"A progression where the first bet will win 1 unit and each successive bet will win an amount that covers all loses sustained since its beginning plus the amount of spins since its start. In the event of a win, go back to the start, and we must have a cut of point where, even if we have lost all the way, we go back to the start"

Hope thats clear.

So, if we consistently win during a session, we will win an amount of units equal to the amount of spins since its start. 10 spins equals 10 units whether we have to go 3 spins for a win or we win every one.


From this point we can go one of two ways:

- We can reduce our chances of a loss, by making the bet a better than evens bet (like betting on 2 dozens)
or
- We can reduce the ratio, and increase our advantage, but increase the chance of a loss.

Basically, with this system, you may be able to reduce the ratio to a lower value but in doing so we increase the chances of loosing.

It is kind of like ending up with odds of 40/1 on a 1 in 20 chance of winning, you still only have a 5% chance of winning, but you are getting great odds on it.

I suppose it comes down to value. The great thing about roulette is it is a game of fixed mathematics that cannot be tampered with, and therefore we can define this value (i.e. good value is a ratio below 1)

These are the 2 extremes of the system ( as far as i have discovered)
- There is the example above that gives us a ratio of 0.93 and a 7% chance of a loss
- and the other extreme example is one i will explain later that give a ratio of around 0.6 but a 50% chance of a loss. (like getting odds of around 3.30 on an evens shot)

Any questions before I go on?

[gingermolloy]

Okay i have a question.

How do you know if you are going to play red or black?

LS

This system is based purely in maths and therefore, as there is no difference between any of the evens payout bets, you can play this with Red, Black, Odd, Even, 1-18 and 19-36. Always betting on the same or changing when ever you like.

It makes no difference what has come before. It has no effect on what is to come.

[redhot]

please continue ginger this is very interesting

[Tucktuckster]

i think it took winkel about 15 pages and he left the forum about 4 times before he finished explaining GUT.

Shall we let the full system out before we decide that it can or cannot work.

the thing is - that most systems are subject to the house edge. this we know. its the way it is implemented that decides if it wins or if it lose.

The LW is a great method. I bet that there would be plenty that would lose with it and plenty that would win with it. Its down to personal variations based on discipline and the like in how it is implemented

[VLSroulette]

We are gentlemen here, so I know we all vouch for:

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So, let's both sides elaborate in peace.

We work on the accepted condition no mechanical roulette system work, but it doesn't mean one can't elaborate on one's take to roulette at this message board. This is why we are all here for: sharing ideas

Ginger, please go on mate.

Dear fellows, remember you can always open a thread at the dark side section explaining why any specific system doesn't work, reminding about the math of the game, performing the traditional 1.000.000-spin tests and displaying the stats; it is even okay to politely link it right from the thread; there is space for everyones' views, but not space for interruptions when a fellow poster is elaborating.

I think it is also we can agree on.

Kind regards.

Your friend,
Victor

[VLSroulette]

Certainly there's room for criticism! But not particularly here at the Full Systems library.

There is a section for "straight talking about roulette systems and -2.7%" and it is the dark side section:
nolinks://vlsroulette.com/the-dark-side/

I'm splitting and pasting non-system-related posts to a thread there & link to it. Lucky, I know that you will be okay with setting the example to keep the Full System library a place for systems and have the debating at proper section.

Cheers & best regards to you dear friend.

Your mate,
Victor

[kompressor]

I cant understand you im afraid. You make no sense.

Ill ignore your contributions from now on as you are being rude without reason

just lock this thread...and start a new one for questions and answer....so INTERESTED people can see what people have to say about your system

[winkel]

This system is based purely in maths and therefore, as there is no difference between any of the evens payout bets, you can play this with Red, Black, Odd, Even, 1-18 and 19-36. Always betting on the same or changing when ever you like.

It makes no difference what has come before. It has no effect on what is to come.

So you say the bet-selection has no effect and the outcome of the spins have no effect and what has come and what is going to come has no effect

then you can bet about 10 times using this progression and have lost 4083 units
so you have to win 4084 times in the first bet or
2042 times in the second bet
to equalize

How do you do this by a chance of 50:50??

br
winkel

[kompressor]

are you gonna finish your explanation ??

[gingermolloy]

If I gave you odds of 3.00 (2/1) on getting heads when you flip a coin, would you not take the bet because you only had a 50:50 chance of winning?

Of course you would, because over time, if you were to repeat the bet over and over, you would come out on top.

That is pure mathematics and cannot be argued against.


Think of the roulette wheel as a 37 sided coin flip.

If I gave you odds of 40/1 on Zero coming in instead of the 35/1 offered by the house you would take the bet because you are getting better odds than is fair.

If you bet on Zero with odds of 40/1 for ages and ages, over time you would come out on top. Irrespective of the fact that you only have a 1 in 37 chance of winning each individual bet.

My system makes this possible, it gives odds of above 2.00 on a 50:50 chance. Over time you will win more money than you bet. Fact!!!!!

Thoughts?

[TwoCatSam]

gingermolly

Please continue with your system.  We who have played the game a while are skeptical.  What you're describing is called the "Grand Martingale".  But you may have a new twist no one has thought of before, so please continue.

kompressor is right.  You--if you can--or Victor should lock your thread until you are finished.  You can open a Q & A thread in the "Child Board" as it's called.

Don't be hard on Lucky Strike.  He's a good guy; just impatient.  

When it's all said and done, if you get it past winkel and Kon-Fu-Sed (KFS) you've got a winner!  

Carry on!

Sam

[redhot]

Yes please continue gingermolly, maybe provide an example of how you play to backup your explanation

[gingermolloy]

gingermolly

Please continue with your system.  We who have played the game a while are skeptical.  What you're describing is called the "Grand Martingale".  But you may have a new twist no one has thought of before, so please continue.

kompressor is right.  You--if you can--or Victor should lock your thread until you are finished.  You can open a Q & A thread in the "Child Board" as it's called.

Don't be hard on Lucky Strike.  He's a good guy; just impatient. 

When it's all said and done, if you get it past winkel and Kon-Fu-Sed (KFS) you've got a winner! 

Carry on!

Sam

Im sure I do have a new twist. If I new you guys were as clued up as this I would have just started at the Grand Martingale and taken it from there.

Will go on tomorrow now.

Night

Ginger

[Proofreaders2000]

"Skeptical"-TwoCat

Yes Ginger, after playing this game for a while, one can easily get skeptical and find it harder to be enthused about new ideas. If Einstein said (in essence) maths can't beat the wheel... 

The house edge requires a typical system to win at least 60% of the time to stay viable.

Looking forward to "blowing the doors off", Ginger. (as the hype builds to a fever-pitch)

-PR

[gingermolloy]

OK let get into the maths for a minute.

Because this is the bit that I am most worried about being discredited as a load of coblers.

Let us take the Grand Martingale Progression on any evens payout bet. It goes 1, 3, 7, 15.

The mathematical odds of winning are indisputable! The course of evens that leads to us loosing our bankroll of £26 is that we must lose 4 times on the trot. (If we are betting on red, then it must be a black or Zero 4 times).

The odds of this happening are easy to calculate. We take the odds of it happening once and multiply it by itself by as many times as it must happen. (i.e. (19/37) x (19/37) x (19/37) x (19/37) = 0.07)

So we have a 7% chance of loosing and a 93% chance of winning.

To convert this into decimal odds we take 1 / (1 - 0.07) = 1.075

So the fair odds for this progression failing would be 1.075.

Now the given odds are a bit harder and are up for dispute.

We need the given odds in decimal form so first we must define how to calculate odds given by the house.

Decimal odds may be defined as the amount payed out if the bet comes in including your stake divided by the stake itself.

i.e. if you take a 2/1 shot. This is stated as 3.00 as decimal odds,

Take a £3 on this 2/1 shot, total payout = £9 and stake = £3 So 9/3 = 3.00.

Yes I know basic stuff, but this is just showing the theory.

So for our Grand Martingale bet.

We could say that our stake is out bankroll and our payout is £4 (£1 for each spin)

So total payout = £30
Stake = £26

Given odds therefore equal 30/26 = 1.154

These given odds are better odds than the fair odds of 1.075 so over time we will have an advantage over the house. Ratio = 1.075 / 1.154 = 0.93.

Thoughts? , before I go on?

[gingermolloy]

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