vertical analysis/strategy for 649

[CFLam]

Hello Jack,

How to convert
05,37,42,44
to
2,4,5,6
???
And why just convert 4 number ? Ignore 02 and 13 ?

CFLam

 

[jack]

seram fixed numbers, or other study you have to dismember 15 positions

 

[jack]

Hello, ice, I need your help to 49/6.
I have the perfect mother always hit up next
100% are, 337 formations, but attention is starting digit only
The final digit is the number that will close
Example = = 02,05,13,37,42,44 initial digit 0 0 1 3 4 4
Draw ascending order
So ice, there may be cheating, forging draw all
The initial type in 337 hits in 100% draw ascending order
Ic I need to reduce the 337, you get the percentages of each training
Example
000000 or 444444 is almost zero probability
You want to see the file and check if it is 100%? You have to separate the first and last digit
Donates raffle register

 

[CFLam]

Hello Jack,

What is the meaning of
D = 3,4,5,6,7
f = 0,1,2,8,9
b = bullet = 1,2,4,5,8
n = nced = 0,3,6,7,9
a = 5,6,7,8,9
b = 0,1,2,3,4
???

CFLam

 

[CFLam]

Hello Jack,
I get what you mean, for each draw result, you break the 6 numbers into 4 number combination i.e. 6C4
02,05,13,37
02,05,13,42
02,05,13,44
02,05,37,42
02,05,37,44
02,05,42,44
02,13,37,42
02,13,37,44
02,13,42,44
02,37,42,44
05,13,37,42
05,13,37,44
05,13,42,44
05,37,42,44
13,37,42,44
And then calculate the digit sume of the 4 number of the 15 combinations:
21
17
19
23
25
21
22
24
20
26
25
27
23
29
28

So, how to use these 15 sums values?

CFLam

 

[jack]

hello to be in ascending order in certain positions the value of the sum is always greater than the primeirar positions
there will be an upward vlaor sums of each 4 positions
you have to filter the 4 numbers in the 15 positions with these standards
D = 3,4,5,6,7
F = 0,1,2,8,9
b = bullet = 1,2,4,5,8
n = NCED = 0,3,6,7,9
A = 5,6,7,8,9
b = 0,1,2,3,4
p = 0,2,4,6,8
i = 1,3,5,7,9
05,37,42,44

 

[CFLam]

Hello Jack,

What is the meaning of "the sum is always greater than the primeirar positions" ?
Also, how to make use of
D = 3,4,5,6,7
F = 0,1,2,8,9
b = bullet = 1,2,4,5,8
n = NCED = 0,3,6,7,9
A = 5,6,7,8,9
b = 0,1,2,3,4
p = 0,2,4,6,8
i = 1,3,5,7,9
?
p and i are odd/even and use D F b n A b to convert the draw result ??

CFLam

 

[jack]

hello are patterns to filter the sum
is just another criteria can be used or not
is thus 15 positions 4 numbers
We have a reference to the sum

 

[CFLam]

Hello Jack,

Ok, after generate the 15 sums from the 4 number, how to apply them ? i.e. how to use them as filter ?
For total sums (e.g. sum all 6 numbers) we can use it as filter by range e.g. between 120 to 210
How to use the 15 sums in this case ?

CFLam

 

[jack]

yes, perfect, corret

 

[jack]

Clfam, these formations are always 337 hits in 100% in any draw
Including the next course of 49/6
But attention is the initial digit
Example
02 08 15 32 45 48 digits = 0.0, 1, 3, 4, 4
Will hit 100% always
No flat terminations or last digit the most problematic
The ice may look to reduce to 40 to 60 and set
000000
000001
000002
000003
000004
000011
000012
000013
000014
000022
000 023
000 024
000033
000034
000 044
000111
000,112
000113
000,114
000,122
000,123
000124
000,133
000,134
000,144
000,222
000,223
000,224
000,233
000,234
000244
000,333
000,334
000,344
000,444
001,111
001,112
001,113
001,114
001122
001,123
001,124
001,133
001,134
001,144
001,222
001,223
001,224
001,233
001234
001244
001,333
001,334
001,344
001,444
002,222
002,223
002,224
002,233
002,234
002,244
002,333
002334
002,344
002,444
003333
003,334
003,344
003,444
004,444
011111
011,112
011,113
011,114
011,122
011,123
011,124
011,133
011,134
011,144
011,222
011,223
011,224
011,233
011,234
011,244
011,333
011,334
011,344
011,444
001122
012,222
012,223
012,224
012,233
012,234
012,244
012,333
012,334
001,344
012,444
001,233
013 333
013 334
013 344
013 444
001,344
014,444
001,222
022 222
022 223
022 224
022 233
022 234
022 244
022 333
022 334
022 344
022 444
002,233
012,233
023 333
023 334
023 344
023 444
002,344
012,344
024 444
002,333
012,333
033 333
033 334
033 344
033 444
003,344
013 344
223344
003,444
003,444
013 444
023 444
044 444
011111
111111
111112
111,113
111114
111,122
111123
111,124
111,133
111,134
111,144
011,112
011,122
111222
111223
111224
111,233
111,234
111,244
011,133
111,333
111,334
111344
011,144
111444
011,122
011,222
112222
112223
112224
112233
112,234
112244
011,223
011,233
112,333
112334
112344
011,244
112444
011,233
011,333
113333
113334
113344
011,334
113444
113444
011,344
011,444
114,444
011,222
012,222
122222
122223
012,224
122,233
122234
122244
012,223
012,233
122,333
122334
122,344
012,244
122,444
012,233
112233
012,333
123333
123334
123344
012,334
012,344
013 444
012,344
112344
012,444
124444
012,333
112,333
013 333
133,333
133334
133344
013 334
013 344
133,444
013 344
113344
123344
013 444
134,444
013 444
113444
123444
014,444
144444
012,222
022 222
122222
222222
222223
222224
222233
222234
222,244
022 223
012,223
022 233
122,233
222,333
222,334
222344
022 244
122244
222444
022 233
122,233
022 333
122,333
223333
223334
223344
022 334
122334
022 344
122,344
223444
022 344
122,344
022 444
122,444
224444
022 333
122,333
023 333
123333
233333
233334
233,344
023 334
123334
023 344
123344
133,444
023 344
123344
223344
023 444
123444
234444
023 444
123444
223444
024 444
124444
244444
023 333
123333
033 333
133,333
233333
333333
333334
333344
033 334
133334
233334
033 344
133344
233,344
333444
033 344
133344
233,344
033 444
133,444
233444
334,444
033 444
133,444
233444
034 444
134,444
234444
344444
034 444
134,444
234444
044 444
144444
244444
344444
444444

 

[jack]

clfam but attention hit 100% but it is only the initial digit
Step 1 put all casdrato sweepstakes in ascending order, then separate the initial digit
example =
02,03,15,26,46,48 = 0,0,1,2,4,4 separate this formçao had hit
can confeir when winning remember me
lack clear the last digit that each position will 0-9
the 1st part already resolved
the ice can see the most probalidades

 

[jack]

hello cflam sure 100% the next draw will take one of the 337 ok
but I repeat is the initial digit is half the other half are the endings or the final digit of the number is only an okay part
the icewynd, can do the percentages of probability
to reduce and so choosing a 20 to 30 and fixing them

 

[CFLam]

Hello Jack,

I tried to summarized the methods you mentioned:
1) use 4 patterns to convert each digit of each draw (odd/even, low/high etc)
e.g.
02,03,15,26,46,48
EE,EO,OO,EE,EE,EE
let's assume
OE=1
EO=2
OO=3
EE=4
and the draw can convert to:
4,2,3,4,4,4
and we can collect patterns for previous and tried used them as filter to predict future draw
or even use the sums of the converted strings:
4+2+3+4+4+4=21

2) select 4 numbers for each draw, therefore each draw can generate 15 4-numbers combinations
and get the sum of 15 combinations to further filter/generate draws. However, how is this related to
total sums ? And how to make use of the followings on the 15 combinations?
D = 3,4,5,6,7
F = 0,1,2,8,9
b = bullet = 1,2,4,5,8
n = NCED = 0,3,6,7,9
A = 5,6,7,8,9
b = 0,1,2,3,4
p = 0,2,4,6,8
i = 1,3,5,7,9

3) seperate the 2 digits and get the pattern code. e.g.
02,03,15,26,46,48
become:
1st digit code
0,0,1,2,4,4
2nd digit code
2,3,5,6,6,8
after this, how to come up the 337 hits ?
since for 1st digit code, can range from 0,0,0,0,0,0 to 4,4,4,4,4,4
and each position can range from 0 to 4, there should be 15625 combinations
from 0,0,0,0,0,0 to 4,4,4,4,4,4

CFLam

 

[CFLam]

Sorry, should not be 15625 combinations, since there should not be
1,0,0,0,0,0
i.e. 1st number > 2nd number. According to my calculation, should only has 210 combinations
0,0,0,0,0,0
0,0,0,0,0,1
0,0,0,0,0,2
0,0,0,0,0,3
0,0,0,0,0,4
0,0,0,0,1,1
0,0,0,0,1,2
0,0,0,0,1,3
0,0,0,0,1,4
0,0,0,0,2,2
0,0,0,0,2,3
0,0,0,0,2,4
0,0,0,0,3,3
0,0,0,0,3,4
0,0,0,0,4,4
0,0,0,1,1,1
0,0,0,1,1,2
0,0,0,1,1,3
0,0,0,1,1,4
0,0,0,1,2,2
0,0,0,1,2,3
0,0,0,1,2,4
0,0,0,1,3,3
0,0,0,1,3,4
0,0,0,1,4,4
0,0,0,2,2,2
0,0,0,2,2,3
0,0,0,2,2,4
0,0,0,2,3,3
0,0,0,2,3,4
0,0,0,2,4,4
0,0,0,3,3,3
0,0,0,3,3,4
0,0,0,3,4,4
0,0,0,4,4,4
0,0,1,1,1,1
0,0,1,1,1,2
0,0,1,1,1,3
0,0,1,1,1,4
0,0,1,1,2,2
0,0,1,1,2,3
0,0,1,1,2,4
0,0,1,1,3,3
0,0,1,1,3,4
0,0,1,1,4,4
0,0,1,2,2,2
0,0,1,2,2,3
0,0,1,2,2,4
0,0,1,2,3,3
0,0,1,2,3,4
0,0,1,2,4,4
0,0,1,3,3,3
0,0,1,3,3,4
0,0,1,3,4,4
0,0,1,4,4,4
0,0,2,2,2,2
0,0,2,2,2,3
0,0,2,2,2,4
0,0,2,2,3,3
0,0,2,2,3,4
0,0,2,2,4,4
0,0,2,3,3,3
0,0,2,3,3,4
0,0,2,3,4,4
0,0,2,4,4,4
0,0,3,3,3,3
0,0,3,3,3,4
0,0,3,3,4,4
0,0,3,4,4,4
0,0,4,4,4,4
0,1,1,1,1,1
0,1,1,1,1,2
0,1,1,1,1,3
0,1,1,1,1,4
0,1,1,1,2,2
0,1,1,1,2,3
0,1,1,1,2,4
0,1,1,1,3,3
0,1,1,1,3,4
0,1,1,1,4,4
0,1,1,2,2,2
0,1,1,2,2,3
0,1,1,2,2,4
0,1,1,2,3,3
0,1,1,2,3,4
0,1,1,2,4,4
0,1,1,3,3,3
0,1,1,3,3,4
0,1,1,3,4,4
0,1,1,4,4,4
0,1,2,2,2,2
0,1,2,2,2,3
0,1,2,2,2,4
0,1,2,2,3,3
0,1,2,2,3,4
0,1,2,2,4,4
0,1,2,3,3,3
0,1,2,3,3,4
0,1,2,3,4,4
0,1,2,4,4,4
0,1,3,3,3,3
0,1,3,3,3,4
0,1,3,3,4,4
0,1,3,4,4,4
0,1,4,4,4,4
0,2,2,2,2,2
0,2,2,2,2,3
0,2,2,2,2,4
0,2,2,2,3,3
0,2,2,2,3,4
0,2,2,2,4,4
0,2,2,3,3,3
0,2,2,3,3,4
0,2,2,3,4,4
0,2,2,4,4,4
0,2,3,3,3,3
0,2,3,3,3,4
0,2,3,3,4,4
0,2,3,4,4,4
0,2,4,4,4,4
0,3,3,3,3,3
0,3,3,3,3,4
0,3,3,3,4,4
0,3,3,4,4,4
0,3,4,4,4,4
0,4,4,4,4,4
1,1,1,1,1,1
1,1,1,1,1,2
1,1,1,1,1,3
1,1,1,1,1,4
1,1,1,1,2,2
1,1,1,1,2,3
1,1,1,1,2,4
1,1,1,1,3,3
1,1,1,1,3,4
1,1,1,1,4,4
1,1,1,2,2,2
1,1,1,2,2,3
1,1,1,2,2,4
1,1,1,2,3,3
1,1,1,2,3,4
1,1,1,2,4,4
1,1,1,3,3,3
1,1,1,3,3,4
1,1,1,3,4,4
1,1,1,4,4,4
1,1,2,2,2,2
1,1,2,2,2,3
1,1,2,2,2,4
1,1,2,2,3,3
1,1,2,2,3,4
1,1,2,2,4,4
1,1,2,3,3,3
1,1,2,3,3,4
1,1,2,3,4,4
1,1,2,4,4,4
1,1,3,3,3,3
1,1,3,3,3,4
1,1,3,3,4,4
1,1,3,4,4,4
1,1,4,4,4,4
1,2,2,2,2,2
1,2,2,2,2,3
1,2,2,2,2,4
1,2,2,2,3,3
1,2,2,2,3,4
1,2,2,2,4,4
1,2,2,3,3,3
1,2,2,3,3,4
1,2,2,3,4,4
1,2,2,4,4,4
1,2,3,3,3,3
1,2,3,3,3,4
1,2,3,3,4,4
1,2,3,4,4,4
1,2,4,4,4,4
1,3,3,3,3,3
1,3,3,3,3,4
1,3,3,3,4,4
1,3,3,4,4,4
1,3,4,4,4,4
1,4,4,4,4,4
2,2,2,2,2,2
2,2,2,2,2,3
2,2,2,2,2,4
2,2,2,2,3,3
2,2,2,2,3,4
2,2,2,2,4,4
2,2,2,3,3,3
2,2,2,3,3,4
2,2,2,3,4,4
2,2,2,4,4,4
2,2,3,3,3,3
2,2,3,3,3,4
2,2,3,3,4,4
2,2,3,4,4,4
2,2,4,4,4,4
2,3,3,3,3,3
2,3,3,3,3,4
2,3,3,3,4,4
2,3,3,4,4,4
2,3,4,4,4,4
2,4,4,4,4,4
3,3,3,3,3,3
3,3,3,3,3,4
3,3,3,3,4,4
3,3,3,4,4,4
3,3,4,4,4,4
3,4,4,4,4,4
4,4,4,4,4,4


However, this basically decade analysis and the 2nd digit combination is basiacally last digit analysis

CFLam

 

[jack]

Okay, okay okay, now you see the save percentages of probability of each formation
Objective use a 30 and 40 formations and repeat for the 2nd digit

 

[jack]

[hello to the 2nd digit you can use as if it were two pick 3 exe =
01 03 26 34 38 42 digtios = 2 = 1,3,6, 4,8,2

 

[CFLam]

Hello Jack,
How to apply the digit analysis to Low/High?
e.g.
01 03 26 34 38 42
for 1st digit, can range from 0 to 4, so we should treat 2 as low or high ?
for 2nd digit, can treat 0~4 as low and 5~9 as high
the above can convert to
LL,LL,LH,HH,HH,HL
or
LL,LL,HH,HH,HH,HL
since 2 can be low or high, what's your opinion?

CFLam

 

[jack]

ok, perfect

 

[jack]

Hello, the initial digits (210 formations) using no bottom, you have to make the percentages of probability of each formation, as the final digits yes you can use
The high low as two pick3