To All Math Guys!

[jekhb76]

Let's pretend we have a bankroll of â,¬30.000 euro and are playing in a casino where we have a Table limit of also â,¬30.000 euro and with this in mind that in a 37 spin cycle we will have at least 4 repeaters min. We will have at Most 33 unique numbers in 37 spins (well that's the Most i Have ever seen). We will be betting as soon as one number repeats itself, and then start betting the numbers that Have fallen and are unique, and Keep adding whenever a unique numbers Comes into play, hoping that the next spin would produce a repeat. Knowing that a min of 3 repeats are still waiting for us before the 37 spin set is over.
Everytime one of our unique gets hit, we will be deleting it and only play the numbers that Have fallen and still are unique. Is there a math behind this, that we will always be in profit of min. â,¬1 euro till the cycle is over?

Let's say our First repeat Comes at spin 12.

5
29
3
16
4
21
6
11
24
33
9
4 (First repeat)

We now start betting the 11 unique numbers and be adding a new unique as they come. Once a unique from our list is hit, we remove that number and continue with the rest.

Is this possible with math?

[Roulettebeater]

Unfortunately you didn't provide enough information so I can model/plot a trigonometric function over an interval of time, are you saying that you will bet all uniques (altogether) previous spun numbers that have appeared waiting for one to be a repeater ?

[jekhb76]

Unfortunately you didn't provide enough information so I can model/plot a trigonometric function over an interval of time, are you saying that you will bet all uniques (altogether) previous spun numbers that have appeared waiting for one to be a repeater ?

Yes, as soon as the First repeater Comes, we start to bet All the numbers that came except the repeating number, with such a progression that we will always be in profit when the next repeater shows. With keeping in mind that after the First repeater has shown, there Will be atleast 3 More.
The First repeater can come on spin 2 or on spin 19. We only start once the First repeater has shown at start to bet All the uniques that came before that and after a spin loss we add the unique that came . And on a hit we remove the hitted one, and continue. The hitted one can't return in our list, because we are chasing the normal repeaters. Hope it is clear. I once saw on a Forum how this can be done, but that's years ago and i never saved it. Just wondering if it is possible with such bankroll and Table limit.

[jekhb76]

Example play Without any progression, because i don't know it, that why i asked.

37 spin cycle.
Total of 10 repeaters in this example cycle . 9 chances to bring us profit.
But how? That's the question.

23
0
8
4
25
5
7
26
4 (First repeat - start betting All uniq.)
(23-0-8-25-5-7-26)

14 (lost add #14 to the list)
16 (lost add #16 to the list)
30 (lost add #30 to the list)
30 Win remove from the list
3. (lost add #3 to the list)
28 (lost add #28 to the list)
35 (lost add #35 to the list)
17 (lost add #17 to the list)
21 (lost add #21 to the list)
15 (lost add #15 to the list)
19 (lost add #19 to the list)
22 (lost add #22 to the list)
26 Win remove from the list
8  Win remove from the list)
7  (lost add #7 to the list)
12 (lost add #12 to the list)
30 Lost Already repeated. No adding
8. Lost Already repeated. No adding
15 Win remove from the list
32 (lost add #32 to the list)
12 Win remove from the list
0 Win remove from the list
35 Win remove from the list
7 Win remove from the list
7 Lost Already repeated. No adding
19 Win remove from the list
26 lost already repeated. No adding
12 lost already repeated. No adding

[jekhb76]

If we take a min. Of 4 repeaters in our mind, with possible profit points of 3, because we start after the First repeater Comes.
How is it possible to Always profit if one of those 3 repeaters come? No matter the spingap between them.
What progression can we use to Always show a profit of min. â,¬1 with a â,¬30.000 Table limit.
The more repeaters we get the better ofcourse but let's asume we get only 3. Because i never saw More then 33 unique numbers in 37 spins in my 25 roulette years. Meaning that a min. Of 4 repeaters came be it doubled, tripple or quads.
There must be some great math Guy around here how can fix this puzzle?

[The General]

Jekh,

You're still trying to side step probability.  You can't.  You will never find just the right progression to make it work.

Focus on finding a method to beat the wheel, NOT the random game. 
Once you've done that, the progression and money management is very simple.

[Nimo]

On average, 24 unique, 8 doubles and the rest are trebles or quads within 37 spins.  With there being a minimum of 4 doubles, you could always bet until you have hit 4 doubles then reset. 

With the progression you are asking, it would have to change as numbers are added or removed.  It is possible to do this as you go, create a spreadsheet and enter the amount of numbers plus the units to always be at a minimum profit.  It can't be a static progression, too many variables.

But here is a progression that has a minimum of $1 profit,(rounded up to nearest profit)  20 Step progression adding a number each time until you hit 20 numbers with a ban of $22382.  Adding a 21st number would go beyond table limit requiring a bankroll of $53735 as you can see from the sheet
Spin    UNITS   Total Bet   RETURN   LOSSES   PROFIT
1    $ 1.00    $1.00    $ 36.00    $1.00   $35.00
2    $ 1.00    $2.00    $ 36.00    $3.00   $33.00
3    $ 1.00    $3.00    $ 36.00    $6.00   $30.00
4    $ 1.00    $4.00    $ 36.00    $10.00   $26.00
5    $ 1.00    $5.00    $ 36.00    $15.00   $21.00
6    $ 1.00    $6.00    $ 36.00    $21.00   $15.00
7    $ 1.00    $7.00    $ 36.00    $28.00   $8.00
8    $ 2.00    $16.00    $ 72.00    $44.00   $28.00
9    $ 2.00    $18.00    $ 72.00    $62.00   $10.00
10    $ 3.00    $30.00    $ 108.00    $92.00   $16.00
11    $ 4.00    $44.00    $ 144.00    $136.00   $8.00
12    $ 6.00    $72.00    $ 216.00    $208.00   $8.00
13    $ 10.00    $130.00    $ 360.00    $338.00   $22.00
14    $ 11.00    $44.00    $ 396.00    $382.00   $14.00
15    $ 19.00    $285.00    $ 684.00    $667.00   $17.00
16    $ 34.00    $544.00    $ 1,224.00    $1,211.00   $13.00
17    $ 64.00    $1,088.00    $ 2,304.00    $2,299.00   $5.00
18    $ 128.00    $2,304.00    $ 4,608.00    $4,603.00   $5.00
19    $ 281.00    $5,339.00    $ 10,116.00    $9,942.00   $174.00
20    $ 622.00    $12,440.00    $ 22,392.00    $22,382.00   $10.00
21    $ 1,493.00    $31,353.00    $ 53,748.00    $53,735.00   $13.00

[nottophammer]

Its all about the prog for TG

So here's one to try. Bet all repeats as they come till you've used 36 units (1 units)
If no win, now bet the repeats and any more that come untill you've used 180 units (5 unit )
The same as above but now untill 900 units used (25 unit)  the 1,5,25




I used 1 units twice, the 1st gave me a bankroll of 2999, down 1 unit, so used 1 unit again then 5 unit and 25 unit. bank showed 3,396

[maestro]

@elvis..

you can try this
note as numbers are comming
you bet nothing yet

at  some point you will have numbers with 1 hit numbers with 0 hit and numbers with 2 hit

as soon as you get first number with 3 hits you will start beting this number with 3 hits ,numbers with 2 hits and any other number that get 2 hits

to sum it up

as soon as you get your first number with 3 hits you bet that 3timer plus all numbers with 2 hits and add up any other number that hits 2 times...your betting is only for 6 spins if no hit retrack with new numbers and discard all your bets

[nottophammer]

using the above Eddy

[jekhb76]

My math tells me, that as soon as i Have hit 4 repeats and in profit, it is best to restart. Or continue at own risk.

[nottophammer]

[jekhb76]

It's Sounds Crazy, but the longer it takes before our First repeater hits, the better it is for us.

Let's say we get our First repeater on spin 21 we now know that within 16 spins, we will get atleast 4 repeats (profit points) and because we need to bet 21# on spin 22 the chances we'll be getting a hit on the next spin is extremely huge!
What also crossed my mind is that after a hit we erase that number and when not in profit we raise All bets by one.

[jekhb76]

I've Just tested some extreme sessions where i did not stop when at First profit but continue.
What i saw is that there Will Always be a point where you Have a new High if you play like this.

1. When First repeater hits, start betting All unique numbers that came before it .
2. Continue to add new unique numbers to your list until you get a hit again.
3. When you get a hit on one of your numbers and are at a new High, restart.
4. When you get a hit on one of your numbers and you are not at a new High, look at your balance and how many numbers you need to bet. If you Have for example 14 numbers in play and your in debt of -12, you don't Have to raise your bets for the next spin, becaus if you get a hit, you Will be in profit even Without a raise. Do this everytime you get a hit and are in debt, check your balance if raising is needed. Prefer Always not to if not nessesary.
5. When you Have raised your numbers by +1 and Have a hit but not at a new High, don't Lower your bets, but Keep it in place. When a new numbers needs to be added give it the same chip amount as the others in play.
6. Continue this progress until a new High is reached and reset.

I've Just had a session that needed to be Extended to 42 spins before i was at a new High but i did Make it. 

[jekhb76]

The only problem i See playing like this is when you get An early repeater, before spin 5 and the second one in the late 20's.