vertical analysis/strategy for 649

[CFLam]

Recently I saw someone use vertical sums to analyse Powerall/Mega Million, however, the analysis use 3 draws only (since the method was evolve from pick 3). I am curious why not use 3,4... even 49 draws to perform this analysis. Also, stimulate from this, I am considering the following strategy:

(strip from my local 649 results, 1 represent odd and 2 represent even)
2,5,13,37,42,44
7,26,30,33,34,38
12,15,17,30,31,43
1,5,11,23,34,47
4,21,25,29,32,37
14,17,23,24,25,37
1,18,22,27,45,48
6,8,12,16,32,49
2,6,10,20,38,40

when converted to odd/even, it become
2,1,1,1,2,2
1,2,2,1,2,2
2,1,1,2,1,1
1,1,1,1,2,1
2,1,1,1,2,1
2,1,1,2,1,1
1,2,2,1,1,2
2,2,2,2,2,1
2,2,2,2,2,2

if we sum odd/even code for each number vertically, we got:
15,13,13,13,15,13
and the average is
1.67,1.44,1.44,1.44,1.67,1.44
if odd/even is 50/50 (percent), then the expected value for odd/even value for each number shoule be 1.5
and the sum for 9 draws should be 9x1.5=13.5.
Can we find the right number of draws for this analysis ? e.g. if we use 49 draws to perform
this analysis, the total sum is 70 and if the "moving sum" of 49 draws become stable i.e. always 70
we can make use this information to filter/generate number to play.

My analysis is up to this part now and I have the following questions:
1) how many draw(s) to use to perform this analysis? 2? 3 or even up to 98?
2) Apart from the above "stable moving sum" strategy, anyone can give more insight ?
3) This analysis can apply to low/high, skip value, deltas, decades, last digit, total sum etc anyone give a try on these (or others) aspects of 649?

CFLam

 

[CFLam]

Little bit amendment:
I saw the vertical sums analysis for PowerBall/Mega Millions from Lottery Post, and I am not very good in statistics and have some difficulties to follow the post.

 

[Icewynd]

I have used the vertical analysis 3-sum in Pick 3 games. What I do is sum the last 3 games by positon (Low, Middle, High). There are 27 potential sums for each position, although only 10 are possible at any given time, and really only 3 or 4 need to be considered by position. Each sum has a probability and a skip value. As with most lotto games, there is a strong tendency for the same sum to repeat twice in a row (about 20% of the time) and about 50% of the time a sum with a skip of 7 or less will hit.

For example: Last 3 games: 754, 854, 185

457
458
158

3 sum at the low position is 4+4+1=9. For the next game the 3 sum will be 4+1+x=5 or higher. Low values of 0, 1 or 3 would hit a sum with a skip <7, so I would consider these. 5 at low would give sum 10 which hasn't hit in 54 games and is due for a hit, so that might be another possibility. Another 4 at the Low position would give sum 9 again, which would be a repeat sum, with a lower probability, i.e. 80% of the time you will get a different sum at a given position, so I would eliminate 4xx.

As you can see, there are no hard and fast answers with this technique, just clues.

 

[CFLam]

Hello Icewynd,

Is there any maths reason to use 3 draws instead of 4,5,6... ? Or just because its pick 3 ? If so, does this imply should use 6 draws in 649 ?

CFLam

 

[Icewynd]

I think the sum was restricted to the last 3 results to keep things manageable. With digits from 0-9 the 3-sum has a possible 27 sums. If this was raised to, say the last 5 results, you would have 45 possible sums to consider.

 

[jack]

hello, cflam you can also nso four possible groups
pp == 02. 44 .42
ip = 0
pi = 05
ii = 13.37
at pp mean pair / pair
ip = odd and even
pi = odd / even
ii = odd / odd
ip = mean odd / even sample = 72
so are noticed the two digits of the number
digit inicial and last digit 72 ex = 7 2

 

[CFLam]

Hello Jack,
Can you explain a ore little bit ? Youconsider the 2 digits of the result draw ?
12,15,17,30,31,43

OE,OO,OO,OE,OO,EO
?
And sume each digit's odd/even value ??

CFLam

 

[jack]

yes, it's simple, it's considered the two digits of each number
Type start and also the final digit
often not repeat the last for the next draw
therefore you can filter by groups. you have
make 4 lottery groups to use groups,
you can also do high / low. inside Outside. etc
all split into 5/5 10 digits from 0 to 9

 

[CFLam]

Hello Jack,

See if I understand correctly,assume:
OO=1
OE=2
EO=3
EE=4

Use these 2 draws as an example
2,5,13,37,42,44
7,26,30,33,34,38
and they converted to:
EE,EO,OO,OO,EE,EE
EO,EE,OE,OO,OE,OE
and then
4,3,1,1,4,4
3,4,2,1,2,2

Is this the case ? Also, what is inside/outside ? How to categorize the number as inside/outside ?

CFLam

 

[jack]

yes, yes, 1,2,3,4 convert corret"!!!
FILTERS DEFAULTS SUMS
INSIDE / OUTSIDE .. BALANCED PAIRS / ODD HIGH / LOW
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

 

[jack]

yes correct
and they converted to:
EE,EO,OO,OO,EE,EE
EO,EE,OE,OO,OE,OE
and then
4,3,1,1,4,4
3,4,2,1,2,2

 

[jack]

Hello cfllam, you are correct about 1,2,3,4 In addition to the p / i, we have other patterns of 3 5/5 the problem is how to assemble these 4 standards. !!!!

 

[CFLam]

Hello Jack,

After converted the O/E into 4 groups. You filter by pattern e.g.
431144
342122
or by sum
4+3+1+1+4+4=17
3+4+2+1+2+2=13

also for low/high, how to convert to your pattern?
0~4 is low (L) and 5~9 is high (H)?
2,5,13,37,42,44
7,26,30,33,34,38

LL,LH,LL,LH,LL,LL
LH,LH,LL,LL,LL,LH

Can you illustrate by an example about the other patterns of 3 5/5 and
about
INSIDE / OUTSIDE .. BALANCED PAIRS / ODD HIGH / LOW
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

Thx

CFLam

 

[jack]

Hello, it's easy, is equal to the odd and even, as we have 5/5
We also have 4 5/5 standards. The problem is how to combine the 4 patterns?

 

[jack]

Hello
You can do so = join four numbers (8 digits) deployed in 15 positions
example =
1st 2nd 3rd 4th 5th 6th
02 05 13 37 42 44
1 2 3 4 are the possible 15 100%
1 2 3 5
1 2 3 6
1 2 4 5
1 2 4 6
1 2 5 6
1 3 4 5
1 3 4 6
1 3 5 6
1 4 5 6
2 3 4 5
2 3 4 6
2 3 5 6
2 4 5 6
3 4 5 6
This list you have put in a horizontal topic 1 2, 3, 4, 3,4,5,6 .......... 15 positions
Example = positin=2,4,5,6= 05,37,42,44 = = = digits 0,5,3,7,4,2,4,4 = sum 29

 

[jack]

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

 

[jack]

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

 

[jack]

1 2 3 4 are the possible 15 100%
1 2 3 5
1 2 3 6
1 2 4 5
1 2 4 6
1 2 5 6
1 3 4 5
1 3 4 6
1 3 5 6
1 4 5 6
2 3 4 5
2 3 4 6
2 3 5 6
2 4 5 6
3 4 5 6

 

[jack]

02 05 13 37=1,2,3,4 x x
02 05 13 42=1,2,3,x 5 x
02 05 13 44 =1,2,3,x x 6
........

 

[jack]

hello ice, excel 2010 excel please