TurboGenius Advantage of Repeaters explained - part 4 - 2021

The best way *to date* to approach this is that the unit size = the "x"

1x = 1 unit, 2x = 2 units, etc.

The bet on any dozen can be removed only once it's won equal times to it's "x"

Using the original chart above -
Dozen 1 won on the 1x bet and won on spin 14 so it's removed.
Dozen 2 went to 4x and would be removed on spin 42 (2 wins were required to offset
the 2 losses which the betting amount easily covered)
Dozen 3 *same as Dozen 2* removed on spin 41 after 2 wins to offset 2 losses.

This isn't a problem because the wins always outnumber the losses so you are assured of an exit
point for each dozen. The unit size being equal to the "x" means as you win more often, you win
at higher amounts and that covers the amount lost along the way.
For example, a loss at 1x, 2x, 3x is easily won back with a win on 4x, 5x, 6x.
If a loss happens in the middle of the wins, you of course have to calculate that loss in and
it requires another win to balance out, which also always happens.
Cheers.

 

A wise man also said that there is a Horse race analogy,
so when 3 things are happening at random and they have a task to perform,
1 will win much faster than expected, 1 will win on average and the last will take a
long time to happen.
This means the other thing the wise man said - Never bet on the last of something -waiting for it to happen.
So exit after 1 dozen has completed, or 2 dozens - never play the last location
because it will be hundreds of spins on average to appear and deplete your profits.
Just throwing out advice.

 

Hi turbo, thank you for your explanation.

Could you explain with an example on one dozen the next?

The bet on any dozen can be removed only once it's won equal times to it's "x"

And

If a loss happens in the middle of the wins, you of course have to calculate that loss in and
it requires another win to balance out, which also always happens.

Thanks so muc in advance.
Regards!!!!

 

a very wise man also said a horse race has a finish line...i really like that quote and miss talking to that guy

 

I think this one is best played when you cover the 6 best performing numbers on the board, with 2 best numbers from each dozen. All in all, very playable in a bm setting behind a terminal with pen and paper. Also many exit points where there is profit made. Profit = start over = handle greed.
Cheers.

 

The only problem with betting on the 2 best performing numbers in each dozen is when to let a number go and when a new one takes it place!

for example, when we have 1 number that has hit 4 times and one has hit 2 times while the rest stands at 1 and zero hits
when another one hits for the second time, do we switch to that number and remove the other (oldest) one? or do we remove both 2 hits and bet on the 4th hit number only until there is a second stand alone number again, in this case when a number hita for the 3rd time?

 

What if we do it this way......

best # D1 2nd best # D1

1x 1x
2x 2x
3x 3x
4x 4x
5x 5x
etc etc

We count the same for dz 2&3.

 

Some of that is covered here in the first post (ie. 1st, 2nd and 3rd to reach 1x, 2x, etc on the whole table) :
https://www.gamblingforums.com/threads/advantage-of-repeaters-explained-part-3-2020.18024/

 

Thanks.
But here comes the problem.
Like you said, betting the top 3 or like i said betting the top 6 (top 2 from each dozen) won't cut it in the end. There have to be entry and exit points in betting otherwise our bankroll will be gone in no time. Ex, you can have the top 3 numbers at all time, but like you said it won't work, why? well because even though they are the leaders, it can take a while before the next hit will come, when all below numbers are getting the hits instead.

 

So there's a crystal clear A to Z method above here....

I'll give it a go on RX
Thx TG

(The naysayers could do the same and prove it wrong )

 

It's clear...
For those who doubt Turbo, here is a take on lines instead of numbers.
Here i played only the top performing line.

Line 12 won 12 out of 13.

It was the first line to..
reach 2* 3* 4* 5* 6* 7* 8* 9* 10* 11* 12* and first to 13 x.
now if this still is hocus pocus, let sir anyone explain it then.
Bet he can't, because he doesn't understand random.

 

The basic starts here...
all numbers appear on average 1/37 spins.
all numbers that don't, are over due.
everyone who say it isn't true, then they don't know the math to this game and should play poker instead of roulette. There is a bell curve, but somehow everyone refuses to see it, a shame really.

 

One example...
when a number hasn't apeared 1/37 it will do what ever it takes to get back or close to it's average point as time goes by. yes it can take 2000+ spins agree, but it will happen 100%
you will never see a number run further and further away from that point, it will always and i mean always run home, no matter how long it takes, but it will. end of story.
have a nice day.

 

Cheers buddy.

I've been busy testing his method from above.....honestly im waaaaaay up in profit without 1 losing session. But hey, each to their own...

 

Hi Denzie and DutchCrown!!

When you say the best two numbers per each Dozen. What do you mean? bet the two numbers with more hits per dozen? or the two best numbers to reach 1x, 2x, 3x .... for every dozen)



That is, in my example, what are the two best numbers to bet at each moment?

Thanks so much in advance!!!

 

Quos, i'm playing/testing the method above...( the best way *to date*) .

Works great so far. But i'm not gonna advertise it too much looool

Make sure if you play more numbers that your progression is good enough to cover some "losses" between the hits though

 

Yes a cold number will get closer to its expected average over thousands of spins. It can do this - surprise - by hitting at 1/37 or even slightly less.

 

Data begins when you start charting it.

Every number that appears greater than 37 spins will tend to balance in the future.
There are limits - the lower limit being 1 spin and the max limit is however long it takes
for a number to appear. This makes for a bell curve that spikes up rather quickly because
the 37 will be the peak and 1 is the least it could be... and the right side of the curve
will stretch out for a possibly very long time. But - any number that is averaging to the
right side of 37 peak will have a win, and multiple wins on the left side (1-37 spins) which
is clearly where the player has the advantage because they need to win better than
the payout of 35 to 1.

Look at this is a math problem. We know the lower limit is 1, we know the average is 37.
The unknown variable is the upper limit number. We know the answer is 37.
From there it's much easier to visualize.

 

I like to call it (LAW OF RANDOM)
According to this law, everything that goes down, must go up and visa versa.
So a number or a group of numbers that are behind their balance point of 1/37 won't stay behind or won't stray further and further behind, they will catch up to the rest of the batch, until they are all even out. This is not fiction, but a FACT!

 

This is the basis for all those who want to explore roulette, heck, we should start a school How's your book coming turbo by the way?