as I'm in a good mood for how my method is going here... today i want to teach something very interesting imo.
How to manipulate the bets and the probabilities... to reduce the edge, until it (almost) completely disappears.
I will just do a simple example to let everyone understand.
If i have 6 units and i wanna make 6 new units... the common way is to bet the 6 units on the red, or black... with a probability of 0.4865.
BUT, there is another way to double your 6 units, with a greater probability... (0,4892 in this case)
In fact if i bet 3 unit on the first dozen, i can win 6 new units. But if i lose, i still have 3 more units that i can put on 3 triplets. And if i win this shot, i get a total of 12 units!
So in both scenarios we try to double or units but there is a difference.
1) simple chance (black, red, odds, even, manque, passe) P = 0.4865 (18/37)
2) half budget on the dozen, if i win i doubled my budget! If i lose, i put the second half on 3 triplets to double my initial budget P = 0.4892 (this number is the probability of winning the first shot + the probability of losing the first shot but winning the second...)
As you can see, same goal with two different probabilities! With the second way we have reduced the casinò edge (from 0.0135 to 0.0108)!!!
Now you can think, this is not a big deal... well, this is just a simple example of how to look things in a different way and open you mind.
Like i said above, It's possibile to manipulate the bets through different combinations that heavily crushes the edge.
This method is designed just to let the gambler play in a fair way, so it doesn't give any concrete advantage... otherwise i wouldn't give it for free just like that! :)
It's stressful for me to type in english so i will end it here... enjoy the gift and let your fantasy go wild. Find your way to nullify the edge (if you are interested in doing so).
Quote from: outsider on Jan 25, 07:27 PM 2021
as I'm in a good mood for how my method is going here... today i want to teach something very interesting imo.
How to manipulate the bets and the probabilities... to reduce the edge, until it (almost) completely disappears.
I will just do a simple example to let everyone understand.
If i have 6 units and i wanna make 6 new units... the common way is to bet the 6 units on the red, or black... with a probability of 0.4865.
BUT, there is another way to double your 6 units, with a greater probability... (0,4892 in this case)
In fact if i bet 3 unit on the first dozen, i can win 6 new units. But if i lose, i still have 3 more units that i can put on 3 triplets. And if i win this shot, i get a total of 12 units!
So in both scenarios we try to double or units but there is a difference.
1) simple chance (black, red, odds, even, manque, passe) P = 0.4865 (18/37)
2) half budget on the dozen, if i win i doubled my budget! If i lose, i put the second half on 3 triplets to double my initial budget P = 0.4892 (this number is the probability of winning the first shot + the probability of losing the first shot but winning the second...)
As you can see, same goal with two different probabilities! With the second way we have reduced the casinò edge (from 0.0135 to 0.0108)!!!
Now you can think, this is not a big deal... well, this is just a simple example of how to look things in a different way and open you mind.
Like i said above, It's possibile to manipulate the bets through different combinations that heavily crushes the edge.
This method is designed just to let the gambler play in a fair way, so it doesn't give any concrete advantage... otherwise i wouldn't give it for free just like that! :)
It's stressful for me to type in english so i will end it here... enjoy the gift and let your fantasy go wild. Find your way to nullify the edge (if you are interested in doing so).
By reading the classy introduction of your thread i thought you have knocked the roulette and want to share you multimillionaire money making machine but the joy didn’t last long, unfortunately it’s much said but the output is still 0
the content is not different than the usual bullshits we usually read
Quote from: Roulettebeater on Jan 25, 11:32 PM 2021
By reading the classy introduction of your thread i thought you have knocked the roulette and want to share you multimillionaire money making machine but the joy didn’t last long, unfortunately it’s much said but the output is still 0
the content is not different than the usual bullshits we usually read
hahahahahahaha the output is still 0? ...not different than the usual bullshits????? Do you realize what this means? Reducing the casinò edge... i'm sure that this thread is more useful than the 100% of the discussions here in this forum (no offense). THIS is something CONCRETE. I SHOWED how it's possible to do it!!! FACTS!!!
Why i'm saying that this thread is more useful than the 100% of the discussions here?
Because any known method here ends to lose because of the 2.7% tax! And at least i showed (with this simple example) how to lose less!!! FACTS!!! Who wants, can use this technique to play a FAIR game! If this is bullshit for you... you wasted a lot of time here, mate.
If you conclude this, you have made a miscalculation.
a lower percentage (2.70%) played twice in a row cannot give a higher probability than a single shot played at 1.35%.
Quote from: plolp on Jan 26, 12:00 PM 2021
If you conclude this, you have made a miscalculation.
a lower percentage (2.70%) played twice in a row cannot give a higher probability than a single shot played at 1.35%.
Exactlyyyyyyyyy!!!!!!
Quote from: plolp on Jan 26, 12:00 PM 2021
If you conclude this, you have made a miscalculation.
a lower percentage (2.70%) played twice in a row cannot give a higher probability than a single shot played at 1.35%.
please... read again (carefully what i explained
I urgently invite uncle Steve to leave a comment here
Ousider :
if I play 3 coins on the 0 and if I lose I play 3 more coins on the zero, will I lose more than you do?
Quote from: plolp on Jan 26, 12:00 PM 2021
If you conclude this, you have made a miscalculation.
a lower percentage (2.70%) played twice in a row cannot give a higher probability than a single shot played at 1.35%.
I told each probabilty above...
normal way to double the capital 0.4865
alternative way to double the capital 0.4892
Is math, and you can not run away from it... test it if you want.
I want to do another example, 2 units to make 1
normal way: you play 2 units on 2 dozens
alternative way: you play 1 units on red, and if you lose, 1 units on 1 dozen
both ways leads to the same result, but the alternative way has a greater chance.
I repeat it's math... try by yourself or give this to a math teacher, an expert... or whoever can do a clear objective analysis, if you don't trust me. (the numbers that i SHOWED)
You are summing two independent probabilities for two different bets !!
I call this an elementary error, I always warn my pupils from making such errors in maths...
Outsider :
How many losing possibilities in your first scenario?
In other words, how many times are you going to lose your two bets?
If you have considered all the possibilities you should respond instantly.
I will answer for you.
There are 700 loss scenarios
So why only talk about the positive scenarios?
@plolp
Yea, the forum is full of gambler fallacies
Maybe Steve should create a section called „fallacies“ for such threads :thumbsup:
Quote from: plolp on Jan 27, 04:44 AM 2021
Outsider :
How many losing possibilities in your first scenario?
In other words, how many times are you going to lose your two bets?
700? What this even means?
Look, it's simple.
1 unit on the red ( p = 0.4865)
if we lose the first bet we play 1 unit on a dozen ( p = 0.3243)
What's the probability to win at least one of those bets? ----> 0.4865+(1-0.4865*0.3243) = 0.6530
This means that the probabiliy to lose both bets is: q = 1 - 0.6530 = 0.347
Instead... if you play 2 units on 2 dozens = the probabilty would be only 24/37= 0,6486
So with this trick we "earned" 0.6530 - 0.6486 = 0.0044 (0.44%)
Quote from: outsider on Jan 27, 10:40 AM 2021
----> 0.4865+(1-0.4865*0.3243) = 0.6530
i'm pretty sure that here, i have teached an unkown formula to the 99,9999% (maybe 100%?) of the users in this forum.
Sorry but u didn’t convince me - anyway the house edge is still far away from ur little 0.44%
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)
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.
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
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.
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...
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
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.
You're forgetting every bet is independent, with its own odds and payouts.
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)
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
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!
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!
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?
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
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?
Quote from: Drazen on Jan 28, 01:24 PM 2021
How do you define "event" in the sequence of roulette outcomes?
How is it different than the spin? If so?
well, if we play a total of 2 units on 2 dozens (at the same time! For those who have not understood) we have played a single spin.
If we play in the alternative way... like 1 unit on red... and if we lose, 1 unit on a dozen, we played 2 spins and those 2 spins are just a single event.
event = multiple spins needed to reach the same goal as if we play in the normal way.
I hope i was clear! My english is very rusty at the moment.
Quote from: ego on Jan 28, 01:48 PM 2021
Drazen - way back in time - you mention trading Tennis - are you still active?
Hi ego
Not at the moment.
Drazen if you don't have acces to Betfair Exchange you can use a Befair Clone and open up account with Matchbook Exhance for a fee.
Cheers
Quote from: outsider on Jan 27, 11:36 AM 2021
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.
Can you please continue as far as you can?
Quote from: Drazen on Jan 29, 01:01 PM 2021
Can you please continue as far as you can?
unfortunately today i don't have much time.
I will just say that it's possible to go deeeeep.... for example, you can divide the dozen in 2 bets!
3 triplets+ 1 carrè = dozen but with a greater % of success!
So instead of bet 2 units to make 1, with 2 dozen in 1 spins, you can do this (the units needs to be splitted so they need to be more but the proportion remain risk 2 to make 1)
First alternative way: 1 EC, if we lose we bet on 1 dozen
Second alternative way: 1 dozen, if we lose we bet on 3 triplets, if we lose again we bet on 1 dozen.
Third and "best" way: 9 triplets, if we lose we bet on 1 carrè, if we lose again we bet on 3 triplets again... if we lose again we bet on 3 triplets again and if we lose again, the last bet will be on 1 carrè.
the units most be divided in order to always have the unit earned
Maybe it's a mess how i have explained it, but i'm in a hurry so, sorry!
Tomorrow MAYBE i will tell the best way to crush to about zero the edge!
Thanks very much, outsider!
Interesting indeed.
Please take your time, but am sure a few lines of practical examples might spark more interest.
Cheers
Third and "best" way: 9 triplets, if we lose we bet on 1 carrè, if we lose again we bet on 3 triplets again... if we lose again we bet on 3 triplets again and if we lose again, the last bet will be on 1 carrè.
Hi outsider!!!
9 triplets in one Spin?
Perhaps you want yo say 3 triplets in one Spin or 3 triplets during 3 spins.....
Thanks so much!!!!
Quote from: quos on Jan 30, 09:39 AM 2021
Third and "best" way: 9 triplets, if we lose we bet on 1 carrè, if we lose again we bet on 3 triplets again... if we lose again we bet on 3 triplets again and if we lose again, the last bet will be on 1 carrè.
Hi outsider!!!
9 triplets in one Spin?
Perhaps you want yo say 3 triplets in one Spin or 3 triplets during 3 spins.....
Thanks so much!!!!
oh thank you for let me know about this mistake... yeah it's 3 triplets in 1 spin (i was thinking about 9 numbers so i made a mix of the 2 things)
Tomorrow MAYBE i will tell the best way to crush to about zero the edge!
Hi outsider!!!
Please, continue when you can!!
Thanks again!!!
Ok... it's time to reveal the big thing! This could change the way you play and see things.
Honestly i wasn't sure about sharing this but... i will do it! This should be such a simple basic knowledge to who wants to live with the roulette... and yet just a few people knows it.
(I will use my terms because i don't like the american ones)
The best way to play a simple chance (18 numbers) is...................... to play a plein (1 number) until you win the first time... for a maximum of 25 spins! Why? Because you have the probability of 49.60% ! I repeat the """""49.60%"""""!!!! instead of the canon 48.65%.
Ok, first of all, i will explain how you can see this by yourself.
a plein has a chance of 2.70% 1/37 = 0,02702....
Let's take the plein number "3"... what's the probability that this number doesn't show up for 25 spins? it's the same probability as the other 36 numbers will come up 25 times in a row!
So, all the numbers, except the "3" have the probability of (36/37= 0,9729...) 97.3%
What's the probability that those 36 numbers come up 25 times in a row? 0.9729^25= 0.504 (50.4%)
This means that our plein "3" has the probability of 1 - 0.504 = 0.496 (49.6%) to come up at least 1 time in 25 spins.
Now that you know this... how we manage to play a plein as if it were a simple chance? With a correct and calibrated progression, of course!
But before i show you the progression... i want do a little modification.
We wont play with the 49.6% but with the 51% (50.95%) IN OUR FAVOUR! How? We extend the spins! instead of 25, we play for a maximum of 26 spins. (you can see why is 51% by doing the calculation by yourself now that you know how to do it).
I did this modification because i find pretty useless to play with a negative mathematical hope.
So, the progression can be a little bit tricky because of the difficulty to split the units so i made the best compromise possible in order to make it playable.
here are the units that you need to bet for each spin, until you win once. (the first time that you win... you have automatically reached the target and you stop playing that game)
spins units
1 1
2 1,1
3 1,1
4 1,1
5 1,2
6 1,2
7 1,2
8 1,3
9 1,3
10 1,3
11 1,4
12 1,4
13 1,5
14 1,5
15 1,5
16 1,6
17 1,6
18 1,7
19 1,7
20 1,8
21 1,8
22 1,9
23 1,9
24 2
25 2
26 2,1
Total units that we can use: 39.2, each event/game (max 26 spins)
total units that we can won each event/game : 36.2 (in average).
Let's put it in this way... each 100 games, we win 36.2 units, the 51% of the time
and we lose 39.2 units the 49% of the time.
So at the end of the day 36.27*51=1850 units earned
and 39.2*49=1920 units lost
the total is 1850-1920 =.......ONLY and i repeat O-N-L-Y -70 units each 100 games (100 games could mean about 2000 spins).
Now, to show how great this thing is... i want to tell you how much i would lose, if i would play in a classic way!
I could have played 19 numbers with probability of 51.35% against 18 numbers (48.65%) for 100 games (in this case 100 games are 100 spins).
for each spin i would have betted 39.2 units on the 19 numbers (about 2 units each number) and my gain could have been 35.07 units.
So 39.2*48.65= 1907
35.07*51.35 = 1800
So each 100 spins you would LOSE about 106ish units.
It's time for a fnal paragon and the fake experts will not gonna like it at all!
alternative way: normal way:
you only lose 71 units each 100 games/about 2000 spins. you lose 106 units each 100 spins, so 2140 units each 2000 spins.
AW NW
-71 vs -2140
How does it feel now? The tiny little difference has turned in a MASSIVE, HUGE difference!!!
We have crushed/destroyed the edge.
now, if you wanna play it... you need to face some problems like the splitted units in order to do the correct bets and you should use a money management strategy that forces a little bit the game in your favour, (not too much though).
Before someone asks: is it possibile to earn with this?
I mean, if you do all the things that i just said above, you could make some units, the only sure thing is: you (almost) wont pay the edge, so the main goal is not to earn something with this.
Like i said at the very beginning, this is a gimmick... something that alone, it's not enough to beat the roulette, otherwise i wouldn't have shared it for free. Still, i'm sure that is the best system ever made public.
My main goal in this thread was to show how the average users (or players) in the roulette forums (or tables) doesn't know a shit about their preferite hobby (obsession/reason to live) but they (try to) criticize who wants to give them something more.
The only way to beat the roulette, is with the math, and if you don't know well those 2 things... you are screwed!
With that being said, i have prooved what i claimed at the beginning and now i can quit my teacher clothes, to go back to the true reason why i am here.... which is: to sell!
Quote from: quos on Jan 30, 01:29 PM 2021
Tomorrow MAYBE i will tell the best way to crush to about zero the edge!
Hi outsider!!!
Please, continue when you can!!
Thanks again!!!
Finally done! I don't know exactly but... it took me some hours. So, enjoy it :)
Quote from: outsider on Jan 30, 08:38 PM 2021
to go back to the true reason why i am here.... which is: to sell!
To sell what?
there is another thread close to this one.
Quote from: outsider on Jan 30, 08:38 PM 2021
Ok... it's time to reveal the big thing! This could change the way you play and see things.
Honestly i wasn't sure about sharing this but... i will do it! This should be such a simple basic knowledge to who wants to live with the roulette... and yet just a few people knows it.
(I will use my terms because i don't like the american ones)
The best way to play a simple chance (18 numbers) is...................... to play a plein (1 number) until you win the first time... for a maximum of 25 spins! Why? Because you have the probability of 49.60% ! I repeat the """""49.60%"""""!!!! instead of the canon 48.65%.
Ok, first of all, i will explain how you can see this by yourself.
a plein has a chance of 2.70% 1/37 = 0,02702....
Let's take the plein number "3"... what's the probability that this number doesn't show up for 25 spins? it's the same probability as the other 36 numbers will come up 25 times in a row!
So, all the numbers, except the "3" have the probability of (36/37= 0,9729...) 97.3%
What's the probability that those 36 numbers come up 25 times in a row? 0.9729^25= 0.504 (50.4%)
This means that our plein "3" has the probability of 1 - 0.504 = 0.496 (49.6%) to come up at least 1 time in 25 spins.
Now that you know this... how we manage to play a plein as if it were a simple chance? With a correct and calibrated progression, of course!
But before i show you the progression... i want do a little modification.
We wont play with the 49.6% but with the 51% (50.95%) IN OUR FAVOUR! How? We extend the spins! instead of 25, we play for a maximum of 26 spins. (you can see why is 51% by doing the calculation by yourself now that you know how to do it).
I did this modification because i find pretty useless to play with a negative mathematical hope.
So, the progression can be a little bit tricky because of the difficulty to split the units so i made the best compromise possible in order to make it playable.
here are the units that you need to bet for each spin, until you win once. (the first time that you win... you have automatically reached the target and you stop playing that game)
spins units
1 1
2 1,1
3 1,1
4 1,1
5 1,2
6 1,2
7 1,2
8 1,3
9 1,3
10 1,3
11 1,4
12 1,4
13 1,5
14 1,5
15 1,5
16 1,6
17 1,6
18 1,7
19 1,7
20 1,8
21 1,8
22 1,9
23 1,9
24 2
25 2
26 2,1
Total units that we can use: 39.2, each event/game (max 26 spins)
total units that we can won each event/game : 36.2 (in average).
Let's put it in this way... each 100 games, we win 36.2 units, the 51% of the time
and we lose 39.2 units the 49% of the time.
So at the end of the day 36.27*51=1850 units earned
and 39.2*49=1920 units lost
the total is 1850-1920 =.......ONLY and i repeat O-N-L-Y -70 units each 100 games (100 games could mean about 2000 spins).
Now, to show how great this thing is... i want to tell you how much i would lose, if i would play in a classic way!
I could have played 19 numbers with probability of 51.35% against 18 numbers (48.65%) for 100 games (in this case 100 games are 100 spins).
for each spin i would have betted 39.2 units on the 19 numbers (about 2 units each number) and my gain could have been 35.07 units.
So 39.2*48.65= 1907
35.07*51.35 = 1800
So each 100 spins you would LOSE about 106ish units.
It's time for a fnal paragon and the fake experts will not gonna like it at all!
alternative way: normal way:
you only lose 71 units each 100 games/about 2000 spins. you lose 106 units each 100 spins, so 2140 units each 2000 spins.
AW NW
-71 vs -2140
How does it feel now? The tiny little difference has turned in a MASSIVE, HUGE difference!!!
We have crushed/destroyed the edge.
now, if you wanna play it... you need to face some problems like the splitted units in order to do the correct bets and you should use a money management strategy that forces a little bit the game in your favour, (not too much though).
Before someone asks: is it possibile to earn with this?
I mean, if you do all the things that i just said above, you could make some units, the only sure thing is: you (almost) wont pay the edge, so the main goal is not to earn something with this.
Like i said at the very beginning, this is a gimmick... something that alone, it's not enough to beat the roulette, otherwise i wouldn't have shared it for free. Still, i'm sure that is the best system ever made public.
My main goal in this thread was to show how the average users (or players) in the roulette forums (or tables) doesn't know a shit about their preferite hobby (obsession/reason to live) but they (try to) criticize who wants to give them something more.
The only way to beat the roulette, is with the math, and if you don't know well those 2 things... you are screwed!
With that being said, i have prooved what i claimed at the beginning and now i can quit my teacher clothes, to go back to the true reason why i am here.... which is: to sell!
So now you only need to find an event that happens every 26 spins while betting max 19 numbers for 26 spins or less.
Quote from: RayManZ on Jan 31, 04:44 AM 2021
So now you only need to find an event that happens every 26 spins while betting max 19 numbers for 26 spins or less.
yeah... at this point even a little selection* it's enough to overcome the wheel
*the selection must really works... and the 99.999...% of those are trash!
Quote from: Clf7 on Jan 30, 09:54 PM 2021
To sell what?
here
link:s://:.rouletteforum.cc/index.php?topic=27691.new#new
If we've already reached one number.
The optimal bet selection will probably be last number for repeat.
If there is someone who is coding in RX, he can check how low the HE will be
If it turned out to be at least a little positive
I would use positive progression cycles of 25 spins
1-2-4
Quote from: ozon on Jan 31, 07:45 PM 2021
If we've already reached one number.
The optimal bet selection will probably be last number for repeat.
Unfortunately not. Each spin is new so there is no advantage (in this case and many others) because the probability for the last number will always be 1/37... like for the other ones.
Quote from: ozon on Jan 31, 07:45 PM 2021
Quote from: ozon on Jan 31, 07:45 PM 2021
1-2-4
the Paroli progression? it won't work.