Quote from: 8OR9 on March 14, 2017, 10:07:58 AM
"Then suppose that we have 16,26,21 and write on the opposite side sequence:

232 (1)"

Not sure what the 232 and the (1) mean...can you explain ?

I am preety sure, that it is the first set od 3 different dozens, ..as you keep tracking, lets say, taht nex 3 dozens are 112 and you write on you list number 2 next to them..and so on..next three you write 3, next three you write 4
those are the numbers you cen see on posts 1 and 2 form Azymax

Quote from: Azymax on March 13, 2017, 02:04:13 PM
hi, i use flat betting method mixed with progression if you lose 3 times in a row.

So lets say last 3 DZ were 3-3-1, then you flat bet againts it, and if you loose, you do martingale than???Or some other progression??

Hi,  how do you bet?? any progressions(like 1-3-9) or flatbet??

I have to agree with Still for 100% with his statment....until moglizu give us proof - his worksheet-the way he plays, Still is right here...

I am back for a whille...and I think I understand finally, how to write all 4 events all together...The only another quetstion is, how many RUNS or Changes  a one single Event must have??? Moglizu wrote": Event no 1 has 2 or less RUNS...so we need Run....and so on. And also he wrote that for example Event nr.4 is balanced-so this is my question...if any event. can be balanced, it means, taht it has to be 1-1, or 2-2, 3-3, and so on....So I ques that number of RUNS or CHANGES in every EVENT must be an even number(2,4,6,8 and maybe more). Can somebody give me his opinion about this?? Moglizu would be the best but I know he is going to tel me, that I need to read the whole thread again....but I think I did not find an answear for that....

I think we all here are glad for what you write here and we all need to find the way, what you thought by your writings.All I asked his been written here, and I ll try to do my best to figure that out the way, you play.

Thank you and I wish you the best.

Ok.

The only puzzle is what you writte here: let me copy  this from yours posts:
Post nr. 1 ..............So 4 events = 5 last ECs

and then, another post:

EVENT NO1
   R    C
   X

The next spin is R R (B) ... so we have to update the chart so it will be

EVENT NO1
  R   C
  X   X     ( We are puting X under the R(Run) or the C(Change)

next spin is R R  B (R)

EVENT NO1

  R  C
  X  X
      X

So until now we have 1 RUN and 2 CHANGES by comparing in every new spin the last 2 outcomes.

when we have RR=RUN  when we have BB=RUN when we have RB or BR = CHANGE.

Now this was ONLY the EVENT NO1



I Have to writte this again: NOW THIS WAS ONLY THE EVENT NO1.

There is only one question--why you use 4 last spins for only event. nr.1 and then next-FITH spin for another 3 EVENTS??  I don't see the way, how it is possible....or  I get it wrong.....and I think that another majority of users as well. Your worksheet would solve everythink I quess...

But as I said..thank you anyway...It is something different.....

Moglizu write:  add 3 more events of the R/B.(Read my previous posts)
That all of them are being updating with the new spin...

So if I understand this-we need 4 spins for event nr. 1, next.fifth spin for event. 2, another spin-sixth for event. 3, and seveth spin for event 4.

There are 2 possible answears..the first-I am stupid or the second..Moglizu is able to make out of 4 spins one event and then with one more spin another 3 events--which does not make any sense to me.(but maybe it is the way to go)...

I know that Moglizu tried his best to answear all the questions here, but If he just scaned or showed his paper-(Make 4 different tables on your paper--this is his first post )  with the  way he playes, I think everything would be crystal clear..

But I have to thank him anyway, because I like to solve any kinds of puzzle....

Ok, I will try. R-R-R-B-B..lets put the numbers so we understand better...1-2-3-4-5....We compare last spin, which si numer 5(B) with number 4(B)..so we have X on runs in event. 1. Than we compare spin before that, whis number 3(R), witth numer 5, which gave us X on CHANGES in event. 2....and so on...

EVENT 1
============

R(uns)   C(hanges)
  x

EVENT 2
=============
R(uns)   C(hanges)
                x

EVENT 3
=============
R(uns)   C(hanges)
                 x

EVENT 4
=============
R(uns)   C(hanges)
                x

I think that user Badger did this allready...and I think this is the only way we can get 4 events out of 5 spins..Of course we can get 1 event out of five spins, but I think it would be wrong...


Can Atlantis say something about my post?  English is not my native language, so i hope  everybody understand.