System based purely in Maths

[gingermolloy]

Hi Guys.

I an new here and would like to share my mathematical system for winning at roulette.

I have been develping a system for use on the roulette wheel for a while now.

I am very intersted to see that so many people have systems which take into account previous spins of the wheel.

I mean no disrespect to these people, but I am of the opinion that it is a fact that previous spins do not have any impact at all on the next spin at all.

So for example if Zero was to come in 3 times on the trot, the probability of the next spin bringing a Zero is still a 1 in 37 chance. (European 1 Zero wheel).

As a scientist I have a very good understanding of maths, and I set out to develop a system that was undesputable. That is, it was based in mathamatics and not on the previous behaviour of the wheel.

Of course there is the exception of the rigged wheel, or a wheel with imperfections, but for my system I am assuming that it is a perfect roulette wheel that has not been rigged. This basically makes the roulette wheel a Random number generator that produces numbers from 0 - 36. (1 Zero)

So if we are dealing in the mathamatics of the roulette wheel, the first thing we find is that the roulette wheel cannot be beaten on any single spin.

The house always has an advantage of about 2.78% (1 Zero). It doesn't mater if you bet on a safe bet (Red or Black) or if you go for a long shot (say a single number). The odds offered by the house are always weighted in their favour.

For example, a bet on Red pays at even money (or 2.00). This would be fair odds if we had a 50% chance of winning, but we dont, we have a 48.65% chance of winning. So fair odds would be 2.055556. This descrepency in the odds gives the house a 2.78% advantage.

So lets get down to it. In the example I just gave the given odds (in decimal form) were 2.00, and the fair odds of winning were 2.055556. We can therefore say the ratio of fair to given odds was 2.055556/2.00000 = 1.0277778. For any given single spin of the roulette wheel you will get this ratio.

So in order to beat the house we must take this ratio and distort it to give a number below 1.0000000. Because if the ratio were 1.000000 we would be getting fair odds and it would be a draw between us and the house.

HOW DO WE DEFINE FAIR AND GIVEN ODDS??

I will awnser this question and elaborate further in my next post.

Does any one have any questions so far?

Back soon!

[winkel]

As a scientist I have a very good understanding of maths

May I ask what kind of science you are involved in?

and

I mean no disrespect to these people, but I am of the opinion that it is a fact that previous spins do not have any impact at all on the next spin at all.

how do explain the binomial distribution then`?

br
winkel

[gingermolloy]

I'm an Analytical Chemist.

[gingermolloy]

binomial distribution? Where? Explain?

[winkel]

Thx for your answers.

pls go on explaining your strategy.

br
winkel

[gingermolloy]

HOW DO WE DEFINE FAIR AND GIVEN ODDS??

Given odds are easy. If it pays even money it is 2.00. We will have to always deal in decimal odds so that a proper fair to given odds ratio can be calculated.

A bet on a single number pays at 36.00 (35/1). You get the drift!

Fair odds are a little more difficult as they are complicated by the persence of the Zero.

Lets work with the example of a bet on Red/Black.

The odds of loosing are 19/37. 19 of the 37 numbers will mean a loss. We can convert this into a decimal as follows 19/37 = 0.5135 (i.e 1.00 is a certainty)

If this is the odds of loosing the odds of winning are simply 1-0.5135 = 0.4865. And to convert this into a decimal odds (where evens = 2.00) we take its recpricle (or 1 divided by it) i.e. 1/0.4865 = 2.0555. These are the fair odds.

Let us do another example to drive this home.

Take a bet on the low dozen (1 - 12) The odds of loosing are 25/37 = 0.6757, so odds of winning are 1-0.6757 = 0.3243 so fair odds are 1/0.3243 = 3.0836.

Given odds are 2/1 or 3.00. So again our ratio is 1.0278.

Is this clear???

[Carlitos]

Intresting threat Gingermolloy But how do you convert this all into making an bett or select an bett which works??


But continue your explanation.......



Carlitos

[VLSroulette]

Gingermolloy,

Thanks for your contribution, please go on as we sure want to know your take on this.

Also, please remember you can always lock the topic till you are fully done elaborating and then re-open it for comments.

Kind regards,
Victor

[Marven]

Gingermolloy,

Welcome to the forum.
Keep posting mate. We're interested in what you have to say.

Regards,
Marven

[pighead]

Thx for your answers.

pls go on explaining your strategy.

br
winkel

Winkel already got the answer.. so did I ..

[gingermolloy]

So, it is now clear that we need a ratio of less than 1 to be on the upper hand and to beat the house.

Lets take a simple version of my system!

This system can take the ratio down to around the 0.6 mark. That gives us an advantage of 40%. But for now I think i need to start with a simpler version of the system that is based on the same idea

( I am not trying to say people are to thick, i am just going through this step by step to reduce questions)

The systems roots are in the martingale system, which I'm sure you are all familiar with.

In the Martingale system we would make a bet and if we loose we make another on the same thing but with double the stake until we win. i.e a progression as follows: 1,2,4,8,16,32 and so on.

What this system does is make it so we win a unit for each set of bets. What i mean is, if we double up 5 times to 32 units we are still only 1 unit up.

In my system we would use the following progression if betting on Red/Black: 1,3,7,15,31,63.

Using this progression we ill win 1 unit per spin even for the spins we loose. and it is this that gives us the edge.

It seems maths is not that welcome here so i will go into that later if people would like me to, but for now i will just get to the meat of it.

The secret of this type of progression is to have a cut off point where you quit and go back to the start even if you have lost. The longer you go on the better because it will diminish your chances of a loss, but it also make the bank roll a lot higher.

Lets get to the real system then.

When this type of progression is applied to a more safe bet it becomes very powerful, because you can continue to win 1 unit per spin but reduce the chance of loosing to an outside chance.

Ill go into this later!

[Ulysses]

Hi Ging

Variations of the Martingale progression are as bad as the original in my view. They work for a while then murder all the gains you made then your bankroll. If you have made a good profit so far I would stop now before you get burned.

Ulysses

[kompressor]

Hi Ging

Variations of the Martingale progression are as bad as the original in my view. They work for a while then murder all the gains you made then your bankroll. If you have made a good profit so far I would stop now before you get burned.

Ulysses

gingermolloy.....you can lock this thread and start another for questions if you want

[gingermolloy]

Hi Ging

Variations of the Martingale progression are as bad as the original in my view. They work for a while then murder all the gains you made then your bankroll. If you have made a good profit so far I would stop now before you get burned.

Ulysses

My system is not a variation!

I said it has its roots in the martingale! im starting there as a method of explaining it.

[gingermolloy]

Let us consider a different type of progression.

Instead of betting on a evens bet and then increasing that bet until you win. We bet on a better than evens bet.

For instance let us consider a bet of £1 on two dozens. If we win we still win £1 if we win and if we loose we increase the bet on both dozens.

We want to win £1 for each spin plus we need to cover what has been lost. So if we loose we need to up the bet to £4 on each dozen.

The progression would look like this: £1 on two dozens. If we loose. £4 on two dozens. If we loose. £13 on two dozens.

The critical thing is that if we loose again, we must stop and go back to the start.

So the bank roll is, £2 + £8 + £26  =  £36.

The odds of loosing are really small and we will win £1 a spin.

I will go into the maths more if people are interested but the bottom line is that for this progression the ratio of odds is 0.965.

This is in our favor but not enough.

My best variation gives a ratio of 0.6 or so.

Ill go into that next

[gingermolloy]

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