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POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!

Started by Blue_Angel, Jul 24, 05:46 AM 2016

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Steve

omg you killed the forums recent posts with capital letters
"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

RouletteGhost

The idea is to bet 2 dozens

Then use the casinos/other players money to continue profiting on a streak

But you know how that goes. Sometimes it works and sometimes it doesnt
the key to winning with systems : play for a statistically irrelevant number of spins

link:[url="s://m.youtube.com/watch?v=nmJKY59NX8o"]s://m.youtube.com/watch?v=nmJKY59NX8o[/url]

Blue_Angel

QuoteSometimes it works and sometimes it doesnt


Like everything else

Blue_Angel

Quote from: Steve on Jul 27, 07:44 AM 2016
omg you killed the forums recent posts with capital letters


I always express my opinion, some might like it, while others not.

I don't find it necessary to agree with everyone in everything.

Steve

The question is does your method of bet selection change the odds, to be better than 1 in 37. If yes, then it wins long term. If not, then the opposite.

The law of a third does sound interesting at first, but Ive never known a way to make use of it. I know of chaos theory that suggests it may be possible one way or another, but as far as I know its just theory without real proof. BA i dont know exactly what you have, but your focus really needs to be on bet selection accuracy. In a nutshell, can you explain why you method of bet selection changes the odds? Or did you not want to disclose this?

ps about the capital letters, its not necesarry to get seen. We all see it just fine.
"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

Turner

Quote from: Blue_Angel on Jul 24, 05:46 AM 2016the only thing which I'm not going to reveal you here is the criteria which I'm using to select the betting number.
Isnt that the only thing here that would make a difference? Improve the accuracy of your bet selection, and you arent saying what that is.

There are a lot of assumptions here.


Blue_Angel

Quote from: Steve on Jul 27, 08:03 AM 2016
The question is does your method of bet selection change the odds, to be better than 1 in 37. If yes, then it wins long term. If not, then the opposite.

The law of a third does sound interesting at first, but Ive never known a way to make use of it. I know of chaos theory that suggests it may be possible one way or another, but as far as I know its just theory without real proof. BA i dont know exactly what you have, but your focus really needs to be on bet selection accuracy. In a nutshell, can you explain why you method of bet selection changes the odds? Or did you not want to disclose this?

ps about the capital letters, its not necesarry to get seen. We all see it just fine.

It's like I've shown you a nice big house but didn't give you the keys...I'd like to keep it this way.

I'm not going to reveal what it does but what it doesn't, so the numbers' selection doesn't depend on "hot" or "cold" numbers.

Dear Steve, we could make a test to validate the accuracy of my method, you could provide a bunch of numbers in portions, then I would provide you my prediction and so on.
My only requests is that those numbers to be from European wheel and not to disclose the steps of the method.

Blue_Angel

Quote from: Turner on Jul 27, 08:19 AM 2016
Isnt that the only thing here that would make a difference? Improve the accuracy of your bet selection, and you arent saying what that is.

There are a lot of assumptions here.


Nothing can be more clear than the law of thirds, how one could expose it is another thing.

RouletteGhost

"The only thing which I'm not going to reveal you here is the criteria which I'm using to select the betting number."

This happens a few times a year

Expected

Moving on.
the key to winning with systems : play for a statistically irrelevant number of spins

link:[url="s://m.youtube.com/watch?v=nmJKY59NX8o"]s://m.youtube.com/watch?v=nmJKY59NX8o[/url]

Blue_Angel

Quote from: RouletteGhost on Jul 27, 08:52 AM 2016
"The only thing which I'm not going to reveal you here is the criteria which I'm using to select the betting number."

This happens a few times a year

Expected

Moving on.

Goodbye

nottophammer

Quote from: RouletteGhost on Jul 27, 08:52 AM 2016
"The only thing which I'm not going to reveal you here is the criteria which I'm using to select the betting number."

This happens a few times a year

Expected

Moving on. yeah they'll be a few
How do you win at roulette, simple, make the right decision

nottophammer

Quote from: Blue_Angel on Jul 24, 05:46 AM 2016
Dear readers,

this is a thread which requires your full attention in order to fully comprehend the hidden opportunity.

Let's start with the basic about the Law Of the Third:

In every 37 spins cycle there are 24 numbers which have appeared once or more and 13 sleepers
(numbers which haven't appeared within the last 37 spins)
Those are average numbers and deviations exist, for example I've seen up to 30 different numbers to show up within the 37 last spins and the least were 18 different numbers.
Those extremes are from my experience during gambling sessions and not from simulations.
In order to find the average we should determine the extremes,or in other words the limits, in that case 18 and 30 are the limits.

18 19 20 21 22 23 24 26 26 27 28 29 30 start by discarding the outer numbers from both sides:

First discarding 18 and 30
Second discard 19 and 29
Third discard 20 and 28
Fourth discard 21 and 27
Fifth discard 22 and 26
Sixth discard 23 and 25

And we conclude to 24 which is the average total of the appeared numbers within 37 last spins.
Another way to calculate the average is to add all the totals and then divide the total sum with the total of the averages,for example:

18 different numbers within the last 37 spins happened 1 time
19 different numbers within the last 37 spins happened 2 times
20 different numbers within the last 37 spins happened 3 times
21 different numbers within the last 37 spins happened 4 times
22 different numbers within the last 37 spins happened 5 times
23 different numbers within the last 37 spins happened 6 times
24 different numbers within the last 37 spins happened 7 times
25 different numbers within the last 37 spins happened 6 times
26 different numbers within the last 37 spins happened 5 times
27 different numbers within the last 37 spins happened 4 times
28 different numbers within the last 37 spins happened 3 times
29 different numbers within the last 37 spins happened 2 times
30 different numbers within the last 37 spins happened 1 times

[(1 x 18)+(2 x 19)+(3 x 20)+(4 x 21)+(5 x 22)+(6 x 23)+(7 x 24)+(6 x 25)+(5 x 26)+(4 x 27)+(3 x 28)+(2 x 29)+(1 x 30) / 49] =>

[(18+38+60+84+110+138+168+150+130+108+84+58+30) / 49]=>

1176 / 49 = 24

This means that if someone was betting one number for 37 spins, 24 times he/she would have won and 13 times he/she would have lost.
To calculate the total sum of lost bets is easy:
13 times multiplied by 37 (1 x 37) equals minus 481 units.

In order to find the total amount of won bets, we should again calculate the averages as we did above,let's see:
We start from the middle numbers within a 37 spin cycle and add 2 more numbers from both sides till we have a total of 24.
18th and 19th spins
17th and 20th spins
16th and 21st spins
15th and 22nd spins
14th and 23rd spins
13th and 24th spins
12th and 25th spins
11th and 26th spins
10th and 27th spins
9th and 28th spins
8th and 29th spins
7th and 30th spins
So we have 24 wins from different spins, now let's calculate the total amount of net profit.

29+28+27+26+25+24+23+22+21+20+19+18+17+16+15+14+13+12+11+10+9+8+7+6 = 420

But since we would lose 481 - 420 = -61 loss, therefore the definition negative expectation.

Do we agree so far?
Is everything clear?

Let's see how we could turn the negative expectation into positive without changing a thing in terms of probability, in fact with exactly the same results which have lead us to the negative balance above.

The average expectation is to get twice the wins for every loss, with my method you only need approximately 1 positive cycle for every negative cycle, which means that the proportion of 2 to 1 changed to 1 to 1.

Let me explain how this is possible, the only thing which I'm not going to reveal you here is the criteria which I'm using to select the betting number.  this is far as i have got,should i really read on as this part you are not going to show?will it end up like having a book with the important page missing
You could pick a random number, or your "lucky" number or anything else you like. Before you start flat betting are you collecting the spins, say looking for a sector of the wheel

I start flat betting 1 number with 1 unit for 36 spins.
When the betting number appears, no matter in which spin, I re-bet the same number plus its neighbour regarding the wheel layout (right or left doesn't matter)

So now I'm flat betting 2 numbers with 1 unit each for the next 18 spins.
When one of the two betting numbers hits within eighteen spins, then I re-bet those two numbers and adding the other wheel neighbour, thus in total three numbers for the next 12 spins, always flat bet 1 unit each.

When one of the three betting numbers comes, I add one more neighbour,this time from the table layout.
So far we have 4 numbers to flat bet with 1 unit each for the next 9 spins.
Let's see if you keep on winning what happens:

5 numbers flat bet with 1 unit each for the next 7 spins
6 numbers flat bet with 1 unit each for the next 6 spins
7 numbers flat bet with 1 unit each for the next 5 spins
8 numbers flat bet with 1 unit each for the next 4 spins
9 numbers flat bet with 1 unit each for the next 4 spins
10 numbers flat bet with 1 unit each for the next 3 spins
11 numbers flat bet with 1 unit each for the next 3 spins

Eleven is the maximum numbers you could bet, this may happen if your first number was 17 for example, because 17 has 2 neighbours on the wheel's layout (like every number) and 8 neighbours regarding the table layout 13,14,15,16,18,19,20,21 (yes,the diagonals too).

What if someone is lucky enough and after adding all neighbours one by one continues to win?
In this case you start adding 1 unit each time you win to only 1 of your betting numbers, you should start from your first selection and continue with the same order.
The minimum total of betting numbers are 6, for example 34 has 2 neighbours at the wheel's layout (17,6) and 3 neighbours at the table (31,32,35)
All together with 34 are six numbers.

We have seen how it might proceed if you are lucky and win frequently, this is not so rare because sometimes the croupiers are hitting specific sectors/numbers frequently (more than their probability)
Personally I've reached two times the 6 numbers bet selection after not many trials, of course this is not always the case.
Let's see what happens when we eventually lose, when you lose during 2 numbers bet selection or more, then you just pick another number and start from scratch, which means flat bet 1 number for the next 36 spins.

There is only one exception, if you lose during 1 number bet selection, in that case you pick another number but this time the betting wouldn't be flat, each and every time our number fails to hit we would add some interest by adding 1 unit.
So the bet on our number would be like this:
1st spin 1 unit
2nd spin 2 units
3rd spin 3 units
4th spin 4 units and so on till your number eventually appears or till you have lost for 37 spins in a row.

If you lose for second consecutive time your 37 bet cycle,this means that somewhere ahead are 4 more winning cycles/rounds but you would only need 2 of them in order to overcome the 2 negative ones.
You change again the number and continue with the dynamic progression till your winning cycles/rounds equals your losing ones.

Let's calculate the losing and winning totals by betting with the dynamic progression instead of the flat bet.
Remember that the average expectation is to lose 13 times within 37 cycles/rounds, so the total loss will be:
[13 x (1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37)]=>
13 x 703 = -9139 units

Now let's calculate the total amount of profit:

[(7 x 35) - 21] + [(8 x 35) - 28] + [(9 x 35) - 36] +[(10 x 35) - 45] + [(11 x 35) - 55] + [(12 x 35) - 66] + [(13 x 35) - 78] + [(14 x 35) - 91] + [(15 x 35) - 105] + [(16 x 35) - 120] + [(17 x 35) - 136] +[(18 x 35) - 153] + [(19 x 35) - 171] + [(20 x 35) - 190] + [(21 x 35) - 210] + [(22 x 35) - 231] + [(23 x 35) - 253] + [(24 x 35) - 276] + [(25 x 35) - 300] + [(26 x 35) - 325] + [(27 x 35) - 351] + [(28 x 35) - 378] + [(29 x 35) - 406] + [(30 x 35) - 435] =>

(245 - 21) + (280 - 28) + (315 - 36) + (350 - 45) + (385 - 55) + (420 - 66) + (455 - 78) + (490 - 91) + (525 - 105) + (560 - 120) + (595 - 136) + (630 - 153) + (665 - 171) + (700 - 190) + (735 - 210) + (770 - 231) + (805 - 253) + (840 - 276) + (875 - 300) + (910 - 325) + (945 - 351) + (980 - 378) + (1015 - 406) + (1050 - 435) =>

224 + 252 + 279 + 305 + 330 + 354 + 377 + 399 + 420 + 440 + 459 + 477 + 494 + 510 + 525 + 539 + 552 + 564 + 575 + 585 + 594 + 602 + 609 + 615 = +11080

We deduct 9139 from 11080 and we find 1941 net profit, therefore the negative expectation has been turned into positive!
But wait, there are more good news!

Since always the first losing cycle/round costs us 36 units instead of 703, that's why you don't need the exact probability to happen (2 winning cycles for every 1 losing cycle), you need approximately equal winning/losing cycles.
I say approximately because you never know in which spin your number is going to appear, thus the net gain differs.
Let me put it this way,if after 2 losing cycles you have 2 winning ones, then you are in profit and re-start from 1 number flat bet with 1 unit for 36 spins.

But after 3 or more consecutive losing cycles you would need the same amount of winning cycles plus 1 more.
Even in such situation, you can be in the positive with worst results than what probability theory supports.
For example in order to overcome 3 losing rounds, I just need 4 instead of 6 which probability dictates as average.
Of course,the results could also be better than average!:-)

Just remember that in the first phase, while we flat bet, we try to find the possibility to win by frequent repeaters.
We build slowly and safely a net gain and we don't limit the possibility for more profits (sky is the limit!) but we limit the loss.

During any stage of the 1st phase (flat bet) the maximum amount we risk is 36 units or less, if you don't lose during the 1st (1x36) or 2nd stage (2x18) of the first phase, then you would be in the positive.
The longer you keep on winning with the flat bets, the more would be the profit, it's all about what happens first.

Under any circumstance you will know what to do next, I'd recommend a total bankroll of 2000 minimum, 6000 average and 10000 units maximum in order to overcome occasional distribution anomalies.

Angelo A.
How do you win at roulette, simple, make the right decision

Blue_Angel

@ nottophammer

Just a few observations, usually up to 9 spins

nottophammer

Thanks.
Is it live wheel only, as have plenty of rng numbers
How do you win at roulette, simple, make the right decision

Blue_Angel

Quote from: nottophammer on Jul 27, 10:50 AM 2016
Thanks.
Is it live wheel only, as have plenty of rng numbers

My method cannot be applied to RNG, ALSO i'M ONLY PLAYING eUROPEAN SINGLE ZERO, THUS CANNOT SAY IF IT WORKS ON aMERICAN 00 BECAUSE i'VE NOT TRIED IT THERE.

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