Beating the even money

[Mister Eko]

Guys please share with me ideas too, I working on too 

[jono1167]

Guys please share with me ideas too, I working on too 

Mister Eko

I'm still working my way through your numbers. I've gone through seven different methods. All have eventually failed against your numbers. The current method is the strongest and its still going. If this method works I'll let you know. I'll need a few days yet. So many numbers to get through....

If this method fails, I have another one to test. With every failure, the method improves slightly.  Leave it with me.

Cheers

[Winner]

Guys please share with me ideas too, I working on too 

I will

[boyd30]

Mister Eko

I'm still working my way through your numbers. I've gone through seven different methods. All have eventually failed against your numbers. The current method is the strongest and its still going. If this method works I'll let you know. I'll need a few days yet. So many numbers to get through....

If this method fails, I have another one to test. With every failure, the method improves slightly.  Leave it with me.

Cheers

Haven't been here for a litte while. I wonder if it's worth going on with the D'Alembert testing? If It fails eventually no idea. Best regards!

[Mister Eko]

Thanks jono and winner.

I have another methid in my pocket. What if we could combinate this method with the VDW method?

[Firefox]

VdW offers no advantage in roulette bet selection in any form.

Either an arithmetic progression is not guaranteed, in which case, either colour can win. Or if an arithmetic progression is due, it can be formed by either colour in which case,  either colour can win.

The probability that either partition/colour can win the next spin is still 50%, or 48.65% if you care to include zero.

VdW is an example of a mis-application of a valid mathematical principle in order to justify a fallacy argument ie Trying to predict the result of the next spin from previous ones.

[Mister Eko]

VdW offers no advantage in roulette bet selection in any form.

Either an arithmetic progression is not guaranteed, in which case, either colour can win. Or if an arithmetic progression is due, it can be formed by either colour in which case,  either colour can win.

The probability that either partition/colour can win the next spin is still 50%, or 48.65% if you care to include zero.

VdW is an example of a mis-application of a valid mathematical principle in order to justify a fallacy argument ie Trying to predict the result of the next spin from previous ones.

Everyone, who is not ap or vb player, trying to predict the resukts of the next spin from previous ones.

[Firefox]

Fair enough. But you may as well predict randomly in that case.  Just trying to save you some time, and anyone else new who happens to be reading.

For what it's worth, I think Downtown's colour matrix is good. It doesn't offer any advantage, but psychologically takes the pressure off even chance selections.

[Mister Eko]

Why, have you some tstrategy in your pocket, which offers any advantage ? Expect vb ap shits

[Roulettebeater]

Fair enough. But you may as well predict randomly in that case.  Just trying to save you some time, and anyone else new who happens to be reading.

For what it's worth, I think Downtown's colour matrix is good. It doesn't offer any advantage, but psychologically takes the pressure off even chance selections.

I want that you tell me a comprehensive approach to the evaluation of accuracy !

In simple words, How do you evaluate the performance of a bet ? In the past I heard that if your bet falls within the result’s range (< 11 numbers) then your bets seems to have somehow accuracy.

[Firefox]

There's no strategy which offers any advantage in the selection of an even chance bet. They are distributed evenly around the wheel and as such, any prediction of a sector will contain a number of both partitions of any even chance.

If you are talking about the prediction of a number, then if you predicted randomly, your mean error would be +/- 9.25  pockets. So any system (computer, VB, DS, or rotor defect) which gives an average error of less than about +/- 9 pockets will give you an edge on the house.

If you are predicting randomly (or from immediate  past results - it comes to the same thing) even chances are better to play as the house edge is less (European wheel with partage only). If you are predicting randomly on US wheel, it doesn't matter if you play EC or numbers. All bets are equally bad, apart from the 5 number bet (0 00 1 2 3) which is even worse.

[Roulettebeater]

There's no strategy which offers any advantage in the selection of an even chance bet. They are distributed evenly around the wheel and as such, any prediction of a sector will contain a number of both partitions of any even chance.

If you are talking about the prediction of a number, then if you predicted randomly, your mean error would be +/- 9.25  pockets. So any system (computer, VB, DS, or rotor defect) which gives an average error of less than about +/- 9 pockets will give you an edge on the house.

If you are predicting randomly (or from immediate  past results - it comes to the same thing) even chances are better to play as the house edge is less (European wheel with partage only). If you are predicting randomly on US wheel, it doesn't matter if you play EC or numbers. All bets are equally bad, apart from the 5 number bet (0 00 1 2 3) which is even worse.

So a bet (1 number) that has an average <9 pockets from result will lead to an edge ?

[Firefox]

So a bet (1 number) that has an average <9 pockets from result will lead to an edge ?

Correct. See attached, courtesy of Bigbroben.

It could take a while to realise though. Variance is very high on just one number, so ideally you want to be be betting on 5-7 numbers in your prediction area to keep the cash flow up 

[Anastasius]

Does mean bet numbers hitting 9x above average...

[Firefox]

Nope, it means your prediction is less than 9 pockets from the result.

If it was a 9x frequency number, I think we'd all be on it regardless of method!