37 spins, 1 Dozen Will Always Have 2 Repeating Numbers In It

[Proofreaders2000]

Glad you could get back to break even Ati 

I'm a big fan of betting right away so here is a proposal.  Look at the last 12 spins (or however many spins are on the marquee.)  Note the hottest two dozens and bet continuously the newest number from both dozens--until a new hottest dozen emerges then bet the newest two numbers again from those two.

[Nickmsi]

Hi All . ..

Thanks AMK for alerting me to this thread.  I have attached an Excel Tracker that shows the numbers that repeat in a 37 spin cycle and shows them by their respective Dozen and Columns.

Hope this helps.

Nick

[Buffster]

Here's an idea...

Repeat numbers...repeat dozens

As soon as you start getting repeat numbers...always bet the last dozen hit.

Ex: say we have our first repeat number ... # 35

Everytime doz 3 hits bet # 35 hoping it wil hit # 35 for the third time.

Do this will all repeats ... play them when their particular dozen hits.

Yes this will probably take longer to win and you will surely miss some 3peats because we bet on the wrong doz. But hey aren'nt we looking for a way to play this without breaking the bank.


Just a thought.

B

[Proofreaders2000]

Check out this system

Maison One: A Single Number Thirty-Six Spin System

Procedure: Note the last 10 spins on
the marquee.  Which dozen is the hottest?

Bet the last number to show from that dozen for up to 18 spins. 
Stop on a win or after 18 consecutive misses.  Repeat procedure
once more for 18 spins.

Goal is two wins per session.

Bankroll suggestion: 300 units per one unit
base bet.  Six-hundred units per two unit base...

Win-Target/Stop-Loss: +10%/-50% of the original bankroll amount.
------------------------------------------------------------------------------------------
Test: Celtic Casino American Wheel
Monday, November 11, 2013 @ 8:00pm CST USA

...36,9,35,11,22,17,3,26,32,29

Third dozen is dominant.  Bet #29 and
any new third dozen number that shows.

Bet #29:    1.) 6(x)-5               2.) 25(x)-5*new trigger number active

Bet #25:    3.) 33(x)-5*new trigger number active

Bet #33:    4.) 36(x)-5*new trigger number active

Bet #36:    5.) 14(x)-5               6.) 34(x)-5*new trigger number active

Bet #34:    7.) 23(x)-5               8.) 25(x)-5*new trigger number active

Bet #25:    9.) 30(x)-5*new trigger number active

Bet #30:    10.) 20(x)-5               11.) 15(x)-5                12.) 34(x)-5*

New trigger number active.    Bet #34:    13.) 22(x)-5

14.) 11(x)-5               15.) 00(x)-5               16.) 10(x)-5

17.) 10(x)-5               18.) 34(win)+175
------------------------------------------------------------------------------------------
break even
====================================
Test: Celtic Casino American Wheel-
Monday, November 11, 2013 @ 8:46pm CST USA

...35,1,27,15,7,6,8,31,12,22,31

First dozen is dominant.  Bet #12 and
any new first dozen number that shows.

Bet #12:    1.) 28(x)-5               2.) 6(x)-5*

New trigger number active.  Bet #6:    3.) 32(x)-5

4.) 23(x)-5               5.) 6(win)+175
---------------------------------------------------------------------
+155*(two hits within 37 spins)

[amk]

Like that idea as well Buffster.

Will do some thinking about so that I see all angles clearly.

Can get tricky : )

Anybody out there remember the Dyksexlic clues about his method for catching one repeater in 37 spins? He kept people focusing on just one repeat in 37 spins, but this statement was not accurate as there will always be 4 repeats.

[amk]

That looks good Proof!!

I am always very impressed with the roulette work you do.

I am sure if I just played all the methods you have presented only once for a few games with the advised win/loss percentages, I'd do fine.

One a day sounds good : )

When I start doing this I will make a thread for it.

Thanks again Proof.

[vladir]

Wait for 12 unique numbers without repeats then start betting them with a progression (it must be constructed assuming we will get 4 hits in next 25 spins after start betting...)...  not sure its possible, due to table limmits....

Subsequente uniques should be added to our bets....its tricky to get a progression for this....

[GLC]

Wait for 12 unique numbers without repeats then start betting them with a progression (it must be constructed assuming we will get 4 hits in next 25 spins after start betting...)...  not sure its possible, due to table limmits....

Subsequente uniques should be added to our bets....its tricky to get a progression for this....

I'm not saying this will work perfectly, but Oscar's Grind is always worth a try on any system.

Just play 1 unit on each number until a win.  If you're in profit, stay at 1 unit on each number for the next attack.

If you're down after a hit, add a unit to your next attack.  So, you'll play 2 units on each number until a hit.

Continue in this fashion on until you have a hit that puts you in profits.

GLC

[Carsch]

A thought,

Let's assume that within the 37 spins, we'll have repeaters of both colors (Red & Black), and not just one color. Thus we could wait till we have one or two repeaters of a same Color (Red, let's say), and from there use all the Black numbers that have shown till the end of the 37th spin.

[gorki]

two repeaters ?

[Skakus]

Let's assume that within the 37 spins...

First thing to do is drop the 37 spins cycle. It's meaningless and useless.

[ugly bob]

First thing to do is drop the 37 spins cycle. It's meaningless and useless.

The most important thing is what's happening now / not happening now....not what happened 20 spins ago or what's going to happen in 20 spins!

I am going of track a bit...BUT....if you think about different forms of winning gambling...they more or less concentrate on the here and now!

The 'true count' in a game of Blackjack tells you your advantage now...not what it's going to be in 10 hands time.

Why should it be any different when coming up with a winning strategy for roulette!

....................and now back to our programme. 

[Carsch]

First thing to do is drop the 37 spins cycle. It's meaningless and useless.

But isn't the whole idea here about repeaters.........what has happened and what is going to happen?

What 'has happened' is a count of repeaters..........and if we know that we can only have a minimum of 4 repeaters within 37 spins, knowing how many repeaters we've had in the past, isn't it right to assume that we can tell more or less how many more repeaters we will have within the next few spins within the 37 spin limit?

[Skakus]

There are two certainties here,


1) There will always be repeaters.

2) The 37 spin window will always shift along 1 spin at a time blurring into infinity.

[Carsch]

There are two certainties here,


1) There will always be repeaters.

2) The 37 spin window will always shift along 1 spin at a time blurring into infinity.

In other words, you don't agree with the test which says that "after millions of spin tests it has been shown that there will always be a minimum of 4 repeaters in a 37 spin cycle." Am i getting you right?

Cause that is exactly what i suggested in my propose thought above.........that if we have had 2 repeaters, it is ok to assume that within 37 spins (including the 2 repeaters) we should have a least 2 more repeaters. Unless there is something about the test that i'm missing here.