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How to nullify the negative edge (for free)

Started by outsider, Jan 25, 07:27 PM 2021

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outsider

Quote from: outsider on Jan 27, 10:40 AM 2021
0.4865+(1-0.4865*0.3243) = 0.6530


Sorry, i forgot to type a crucial thing here:   0.4865+((1-0.4865)*0.3243)

outsider

Quote from: Roulettebeater on Jan 27, 10:49 AM 2021
Sorry but u didn’t convince me - anyway the house edge is still far away from ur little 0.44%

Thank god...  at least now you have the doubt that MAYBE i'm right.
Now we can talk about my little 0.44% (for now)

Well, i have explained 2 examples...
The one where i play a dozen and 3 triplets instead of a simple chance like the red numbers.
And the one where i play a simple chance  and a dozen instead of 2 dozens.

Now, we can combine them, to reduce the edge even further!!!

in the last example about the 2 dozens, instead of playing a simple chance and a dozen... we replace the simple chance with my first example.

So we are going to play in this way:

1 Dozen, if we lose, we bet on 3 triplets, if we lose again we play on the dozen again (of course the units must be managed to garantee our initial goal, which is; i risk 2 units to make 1,  or 30 units to make 15)

Now let's check what's the new probability

p = 0,3243+  ( (1-0.3243)*0,2432)+  ( (1-0.3243)*(1-0.2432)*0.3243) = 0,6545

0.6545 instead of 0,6530... so we earned another 0.15% for a total of 0.59%

Have we reached the limit? Of course not! It's possibile to combinate even more alternative ways to lower the negative edge to zero.

For example i could replace the dozen with 3 triplets and a carrè.

But i dind't because maybe it would become a bit complicated to follow all the mechanisms at the beginning.
I'm sure that it's already a lot to comprehend.









Drazen

Thank you very much, outsider! Finally something worth reading.

My only hope is that members with programming skills might show interest in this and pull out a few sims maybe.

Everything laid out on a golden platter.

QuoteIts possibile to combinate even more alternative ways to lower the negative edge to zero.

Are you saying that using differential bets alone we can nullify negative edge, or by alternative ways, you mean something else maybe?

Cheers


thereddiamanthe

Yeah OK, so you are gaining probability progressing in risk  in exchange for progressing in units (more pretty much the same numbers  & so risking more units.

So where is the overall advantage?



& point two - I can't believe that in 2021 we are equating the 'house' edge' with the robability  &still using the term 'reducing house edge'.
House edge is 2.7 for european wheel & you gonna pay it every single time when your bet & the wheel outcome are matched - according to the payout. I see no way to avoid that - tax to play to the game provider. Can you reduce the house edge in poker?! No. Same in roulette.
Perhaps use or invent another word.

Bottom line, at best you are increasing the probability of the outcome not reducing the house edge.

outsider

Quote from: Drazen on Jan 27, 01:08 PM 2021
Thank you very much, outsider! Finally something worth reading.

My only hope is that members with programming skills might show interest in this and pull out a few sims maybe.

you're welcome, yeah some users might try to do some simulations... to show how this thing really works.

Quote from: Drazen on Jan 27, 01:08 PM 2021
Are you saying that using differential bets alone we can nullify negative edge?
In this case, yes! With this technique it's possibile to concatenate different bets in order to reduce the edge virtually, not to zero but almost!
you have to disassemble the bets... like i showed in my examples.
This method alone will not lead to beat the roulette... still, will allow you to play it in a more fair way.
The thing is: all the (known) methods ends with the failure because of the negative edge... with this atleast you will lose less or nothing...

ozon

Playing your modification of 2 dozens
So one EC and one dozen
bet selection follow the last for Hi-Lo and using the wheel with La Partage rule
House edge will be very small

plolp


as soon as you play a "DZ" you lose 2.70%
and you will not be able to reduce this rate despite all your mental pirouettes
ditto if you play a split a "square". or a number.

Steve

You're forgetting every bet is independent, with its own odds and payouts.
"The only way to beat roulette is by increasing the accuracy of predictions"
Roulettephysics.com ← Professional roulette tips
Roulette-computers.com ← Hidden electronics that predicts the winning number
Roulettephysics.com/roulette-strategy ← Why most systems lose

outsider

Quote from: Steve on Jan 28, 07:59 AM 2021
You're forgetting every bet is independent, with its own odds and payouts.

It’s all calculated... playing a simple chance and a dozen in 2 spins, have the same payouts as 2 dozens
Risk 2 to make 1
Just the odds are different because the first event (red+dozen in 2 spins) has a greater chance than the second event (2 dozens in one spin)

outsider

Quote from: plolp on Jan 28, 05:40 AM 2021
as soon as you play a "DZ" you lose 2.70%
and you will not be able to reduce this rate despite all your mental pirouettes
ditto if you play a split a "square". or a number.

Mental pirouettes? I’m talking about math!
I showed numbers and formulas... not throwed random numbers

outsider

Quote from: plolp on Jan 27, 04:44 AM 2021

I will answer for you.
There are 700 loss scenarios
So why only talk about the positive scenarios?
I’m still waiting for an explanation about your “700” made out of nowhere!

outsider

Quote from: ozon on Jan 27, 07:17 PM 2021
Playing your modification of 2 dozens
So one EC and one dozen
bet selection follow the last for Hi-Lo and using the wheel with La Partage rule
House edge will be very small
The partage... nice observation!

Drazen

Quote from: outsider on Jan 28, 09:24 AM 2021
Just the odds are different because the first event (red+dozen in 2 spins) has a greater chance than the second event (2 dozens in one spin)

How do you define "event" in the sequence of roulette outcomes?

How is it different than the spin? If so?

ego


Drazen - way back in time - you mention trading Tennis - are you still active?

Outsider - a dozen has 50% probability to hit once within two spins - binomial probability!

Cheers
Denial of gamblers fallacy is usually seen in people who has Roulette as last option for a way to wealth, debt covering and a independent lifestyle.  Next step is pretty ugly-
AP - It's not that it can't be done, but rather people don't really have a clue as to the level of fanaticism and outright obsession that it takes to be successful, let alone get to the level where you can take money out of the casinos on a regular basis. Out of 1,000 people that earnestly try, maybe only one will make it.

outsider

Quote from: ego on Jan 28, 01:48 PM 2021
Drazen - way back in time - you mention trading Tennis - are you still active?

Outsider - a dozen has 50% probability to hit once within two spins - binomial probability!

Cheers

54.34% probability to hit at least once within two spins but... what's the point of this?

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