My new creation

[Advantage.Player]

Okay, this is my last try a roulette "System" seeing as out of all the roulette forums there are and despite all posts made and all the "Systems" designed and tests done – we still have yet to find a way to win consistently (I've stuck to my trend play which now after my Dublin bet tests has made 11.9K profit over 33 sessions with an average session time of 22mins...).

The only thing coming close to profitable in terms of a "System" is Jobsabobas neat little application which is good until you need to bet in the neighborhood of around 1000 units on the odd occasion.

So this is what I've been thinking. Walk into the casino. Play the 4 numbers shown in Rapid Roulette as "cold" with a progression, profit, then leave. So before any testing I'm going to run some maths through this to see if I in fact could make money with this simple method.

Lets get this out of the way. In order for a 4 numbers to be cold It has to NOT be hit for a while. How longs the while then? Well 33 spins. This is the least amount of spins it could take. This would require ALL but 4 of the numbers to be hit in 33 spins, which is near impossible it nearly is always way more than 33 spins. But nonetheless 33 spins is the minimum required for 4 cold numbers. So lets work the maths like as follows to ensure that even if the totally wild, "once in a million spins thing" does happen we kind of have it covered (plus doing the following will make things easier).

Lets assume that we will always win on the FOURTH step of the progression (more on this later). This is because on this step I make the LEAST profit! So if the maths proves I make money while winning the "least" amount of money then when It comes to playing I should make even more. Further more a lose will always be on the last step of the progression – costing me the MOST money possible. Once again this is all to do with the "extreme" case example – if anyone knows some maths here they would be familiar with the Rigging method used to prove convergence and divergence...well this is kind of what I'm going here – solving for worst case scenarios to make sure I am right.

So now. My progression is 23 steps long and the minimum possible amount of spins for 4 cold numbers is 33. Thus on my last step of the progression, should I lose it would mean my numbers didn't hit for a minimum of 56 spins (33 + 23). And for 4 numbers to NOT hit in 56 the probability of this is 0.165%. This means after trying my system 100 times (not for 100 spins but playing the progression 100 times) I will lose 0.165 times. And with my progression, on a loss it will cost me $2540. Which then means after 100 tries of the system I will lose $419.10 ($2540 * 0.165).

Lets assume that when ever I win its at the worst possible step.... Number 4. Now, on the fourth step of my progression I only make $40 and the chance of winning on the fourth step is 98.55%. So after 100 games I will win $3942 ($40 * 98.55). Meaning that after 100 trials of my system I will lose $419 and win $3942 making me a total of $3523 profit.

The progression I made is designed to profit moderately and use the max Rapid Roulette bet here. It goes as follows...

Spin     Bet PER Number (4 Numbers Covered!)
1   1.5
2   2
3   2.5
4   3
5   4
6   5
7   5
8   7.5
9   9
10   11
11   13
12   15
13   17.5   
14   20.5
15   24
16   28
17   32.5
18   37.5
19   73.5
20   50
21   57.5
22   66
23   76
24      87
25      100

EDIT: Tried to "Smooth" out progression for more linear profit, and ended up extending it slightly too.

The system is played until our 4 numbers start hitting AT expected value. So like this...

Spins Played       Hits Required To Stop Play
10         1
20         2
30         3
40         4
50         5
60         6
70         8
80         9
90         10
100         11

EDIT: I keep going back and forth between expected value and 1.645SD over - Ive had good results in all my tests with both but Ill keep my choice to "At Expected Value" to make things "safer".

Heres a test I did. I don't have rapid roulette at home so I manually crossed of numbers of a roulette wheel picture at home from Dublin bet numbers.

Heres the qualifying numbers (until only 4 numbers were left).

7
22
17
28
8
27
7
6
36
30
22
0
33
6
4
8
6
14
25
3
5
7
34
1
14
8
21
14
0
0
26
4
3
8
9
17
6
6
28
8
24
11
34
10
13
4
3
17
16
11
13
20
16
36
16
9
1
12
1
32
19
22
23
9
0
14
4
30
1
10
22
1
0
20
18
9
30
9
9
26
12
24
28
8
17
32
17
17
11
17
5
24
30
23
1
7
32
7
32
4
5
9
20
1
18
3
26
16
23
30
20
30
20
13
0
28
32
17
23
4
26
8
9
35

Yep it took that many spins to only have 4 numbers left – that is why I'm using the already available Rapid Roulette numbers. Also this proves how "pessimistic" my maths was above – it was ALL worst-case scenario (yet my system still shows profit). So now my 4 unhit numbers are 29, 31, 15 and 2. The rest of the spins are as follows. A win on my number is represented by HIT with the profit made next to it.

28
27
25
26
19
31 HIT $85

I stopped playing here because I went past the stop play condition (it hit 1 times in 6 spins not the afore mentioned 1hits in 10). Such rapid hits are due to the long "sleep" period of the numbers. However I continued to watch the numbers (not shown here) and it turned out that my chosen 4 numbers started to hit more than 3 standard deviations from the normal! Put a smile to my face (as possibly to fellow mathematicians).

So I played 6 spins and made $85 profit. In "real life" all you would do is go to the Rapid roulette table and play – not spend the 2 hours charting like I did on Dublin Bet for this test!

Also, once the cold numbers you played before are replaced by different ones, you could then go ahead and play them until they start hitting as expected.

So feel free to test; but I doubt any one will manually due to the ungodly charting. I'm hoping for someone to possibly code this in RX and let it run or a while to prove my initial maths was right and to ultimately test the system and shows its effectiveness.

[vandijkerwin]

hi sorry to interrup but I have a comment on the following

And for 4 numbers to NOT hit in 56 the probability of this is 0.165%. This means after trying my system 100 times (not for 100 spins but playing the progression 100 times) I will lose 0.165 times. And with my progression, on a loss it will cost me $2540. Which then means after 100 tries of the system I will lose $419.10 ($2540 * 0.165).

You say that after 100 tries you lose $419.10. In my opinion you make 2 mistakes

1. In 100 tries You will lose $2540! and not (2540*0.165)=419,10
2. 0.165% of 2540 is not 419.10, but 4,19 (You calculate 2540*0.165, which is actually 16.5%)

[vandijkerwin]

with this rates AND I MEAN IF you're lossrate is really 0.165% and not 1.65% you're profit would be the following. I will easen this mathematical project and rise the 0.165% to 0.2% which is easier to explain.

Loss 0.2% loss is the same is 2*2540= 5080 in 1000 progressions (you will lose 2 times in 1000 sessions)
(least) profit is 40 in the remaining 998 progressions = 998*40= 39920
39920>5080 so A very good profit

however if the percentage loss is 2% instead of 0.2%

Loss 2% loss is the same is 2*2540= 5080 in 100 progressions (you will lose 2 times in 100 sessions)
(least) profit is 40 in the remaining 98 progressions = 98*40= 3920
3920<5080 so unfortunately a loss.

So my question to you is

Is the lossrate 0.165% (1.65 losing sessions in thousand sessions) or
Is the lossrate 1.65% (1.65 losing sessions in hundred sessions)

[Advantage.Player]

In 100 tries I lose $419.10 THEORETICALLY SPEAKING. My loss rate at such a low number of trials is a decimal so yes, If I lose anything it is $2540 at a time (one progression) but I "scaled down" the example thus the decimal value. However the maths is fine. Put it this way.

I try the system 1 million times.

I lose 0.165% of the time. A Loss cost me $2540. So In total I lose $419,100,000

I Win (in the worst case) $40 which happens 99.835% of the time. So after 1 million tries I make $3,993,400,000

So a total profit of $3,574,300,000 (3.6 Billion...yeah I wish LOL)

Also I didn't calculate my loss rate but instead my chance of losing (Same thing isnt it though?). Simple maths gives me this. I covered 4 numbers out of a possible 37 for 56 spins (33 minimum needed for 4 sleepers an 23 for my progression) so my probability to lose is (33/37)^(33+23) which is 0.00165 which THEN equals 0.165 PERCENT.

So to me the maths check out fine

Further more Ive made one error which still doesn't really effect much. I said the probability of winning the lowest profit stage (step 4) in my progression is 99.835% (100 - % of losing LAST stage) this is in fact wrong. The proper percentage is 95.45% meaning in 100 tries of the system I lose $419.10 and win $3818 making a net profit of $3398.30 (still good)

PS I used 100 spins as an example in the original post because then i can use my percentage with out having to convert any numbers - ie; 95.45% of winning means 95.45 wins out off 100 times.

[TwoCatSam]

This, then, would be an adjunct system which you ran alongside some other system, right?  No one would want to sit that long doing nothing.

You could easily use bj's "last six" software and just look for four. 

I'm really unclear on how you know when to stop and start seeking again.


Sam

[Lanky]

26
8
9
35

Yep it took that many spins to only have 4 numbers left – that is why I'm using the already available Rapid Roulette numbers. Also this proves how "pessimistic" my maths was above – it was ALL worst-case scenario (yet my system still shows profit). So now my 4 unhit numbers are 29, 35, 15 and 2. The rest of the spins are as follows. A win on my number is represented by HIT with the profit made next to it.

28
27
25
26
19
31 HIT $85
31 HIT $70
5
7
30
16
12
15 HIT $85
36
2 HIT $60

Well My Mate there is problem here somewhere.

So now my 4 unhit numbers are 29, 35, 15 and 2.

Yet 35 shows as a hit number before you came up with the 4 Unhit Numbers.

And 31 Hit twice for You which is not one of the unhit Numbers ??

Unless you meant 2, 15 ,29 ,31 were the 4 unhit numbers in the First Place.


Lanky

[bjb007]

As Sam said you could use my "Last Six" programme
(in the Members Download area) and wait for the
last 4.

But from my tests the last four is too unreliable. The
last six is many times better.

[Advantage.Player]

Well spotted Lanky >.< I got too mixed up in TWO hours of charting, MSN, Forums and music LOL

But you worked it out - swap the 35/31 around. Ill edit that. Thanks for spotting it.

[Advantage.Player]

@ 2Cat

It was designed to use the readily available Rapid Roulette data (they tell you the 4 most hits numbers and the 4 least hit numbers in between spins). This would eliminate long charting times, however using it along side another system could be done if you dont already have the data that is available from Rapid Roulette.

Also stop according to this chart...

Spins Played     Times your 4 numbers were hit
10                                      1
20                                      2
30                                      3
40                                      4
50                                      5
60                                      6
70                                      8
80                                      9
90                                      10
100                                     11

So say I played for 34 spins and was hit for the third time on spin 35. I'm still ok, because I've got more than 30 spins (according to the chart I need more than 3 hits in less than 30 spins to stop). Say 3 spins later I'm hit again. This means Ive played for 37 spins and have been hit 4 times. Now I have 4 hits in 38 spins - now I stop. Think of it like this. Stop playing if you have played less spins than the corresponding amount of hits suggests; or, stop playing you have more hits than the corresponding amount of spins says.

To start play again you need a new set of 4 sleepers. So either chart again, but my original idea was to let the Rapid Roulette do that for me and all I had to do was to watch when the 4 sleepers from last time were ALL replaced. Then with the new set I could play again.

@ bjb

I know what you mean. Even thou mathematically its the same - it just isn't. But using 6 numbers means charting where as I has hoping to skip that and use the Rapid Roulette data.

EDIT: Updated example above.

[ebrand]

I like the sound of this

I'll check it out sometime soon just to make sure it's not flawed

[ebrand]

so my probability to lose is (33/37)^(33+23) which is 0.00165 which THEN equals 0.165 PERCENT.

So to me the maths check out fine

Sadly I think here is the error. The 33 spins are a past event so your losing chance is just (33/37)^23

[Advantage.Player]

Hmmm, I don't think i matters. Because every spin is a past event once it occurs and the probability shown (0.165%) is the chance of the 4 numbers not hitting for 56 spins - whether they "not hit" while we have bets down or while we are watching doesn't matter.

[TwoCatSam]

Well, brain fart here!!

Rapid Roulette is that pit at the casino where there is one wheel and 15 terminals, right?   I was thinking it was a program you ran beside the wheel on your computer like Roulette Raid.

I'll go back and read it again in the right frame of mind.

Sam

[ebrand]

Hmmm, I don't think I matters. Because every spin is a past event once it occurs and the probability shown (0.165%) is the chance of the 4 numbers not hitting for 56 spins - whether they "not hit" while we have bets down or while we are watching doesn't matter.

From a mathematical point of view it does matter - however from a reality point of view I don't know LOL

[cps10]

I would love to see someone program this into Rx and provide a bankroll graph. I think that maybe it would work.

[cps10]

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