How to nullify the negative edge (for free)

[outsider]

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]

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.

Its 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]

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.

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.

[outsider]

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]

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]

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]

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]

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

[outsider]

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?