comparing won combinations

[mirage]

sorry if i confused you...i was previously talking about gaps...
i have found you can use gaps quite reliably for the following;
all 6 numbers which create gap patterns of 5, and the "inside"
4 numbers ( omit column 1 and column 6 occurance ) just use
inside 4 numbers to get a gap pattern of 3.

..the stuff i mentioned more recently was a different 'angle'...
a different line of thought...it involves looking at the winning
results in terms of "columns" rather than rows,,,
for instance, sat night has 15 20 32 39 40 47 b30..
if you take all historic occurances, you will observe that
(obviously) #1 lands in column 1 all the time...never anywhere
else....also 49 lands in column 6 all the time..never anywhere else..ordinarily, when you do a frequency distribution (f/d) you'll
get 47, 31, 27, 32, etc...( these are by memory so they'll be
not exact)..so although 31 is 1st or second most common #,
it only ocurrs in column 4 - 107 x vs 01 which has fallen in col 1
285 x ! same with 49...falls in col 6 262 x !...i find that looking
at where a number usually or should fall ( ie column) will assist
in picking numbers...there are usually 2 or 3 in the top 4 of
any column which fall in the next draw...the set is relatively
stable..does not change much....looks like this..
col 1 col2 col3 col4 col5 col6
1 14 21 31 43 49
2 12 20 34 38 47
4 9 22 28 40 48
3 8 25 27 41 45
etc..
so use this as a filter...take the set of 24..play a ticket with
3 or more of them...
also...take last 3 draws..possible set of 21...and take any 2/21
as a filter as well...
these two work very well together

Ah, -still a bit opaque and number dense... But thought provoking. Is what you are doing cross-referencing "columns" and positionals...(?)

When you say "the top 4 of any column", which columns?-those printed on the cards, or the 'positional columns' which might be a list of winning #'s from past draws?

As for the previous 3 draws, use 2 out of possible 21 - it is usually less than 21 #s, yes I do that routinely now. There will be at least 2 or 3, or sometimes more, from past three draws, usually. 2 - can be conservatively, fairly certain of.

Thanks for your info!

 

[millsy]

re:top 4 of any column...meaning...for example...
last few draws look like ;
15 20 32 39 40 47 b 30
15 17 24 26 34 35 b 08
04 07 13 33 40 47 b 37
etc...
now, use only the 1st (lowest ) number to create your f/d for
"column 1" so use 15, 15, 04, (02, 06, 02, 08, 01, 02, etc)
creates a f/d ranked highest to lowest of;
01-285x, 02-237x, 04-213x, 03-208x etc...
do the same for col 2 numbers thru col 6 numbers (no bonus)

for instance, column5 - 43-124x, 38-123x, 40-122x, 41-122x
col 6 49-262x, 47-256x, 48-220x, 45-186x

use three or more of the top 4 in each column - presently
01 02 03 04 08 09 12 14 20 21 22 25 27 28 31 34 38 40 41 43
45 47 48 49

some numbers (like 31) althought they are very popular,
spread their occurance around between columns, so they are not
as popular in this analysis as say a 01 or a 49 which are only
found in the outside columns....you will find it is quite easy to
predict a smaller range of outside numbers...it's the ones in the
middle which create the problems ! using excel you can plot the
columns on a graph...the outside columns are almost identical
but reverse and 1/x, the inner columns form normal distributions
with a skew either left of right, but they behave very well for
basic statisical analysis.

 

[W Kaenzig]

gaps

A gap of 1 occurs more often than any other. A gap of 2 a little less, a gap of 3 a little less than a gap of 2. And, so on... Gaps of more than 26 should never be used as they occur very infrequently.
There are many of considerations other than just gaps.
W.

 

[mirage]

Re: gaps

A gap of 1 occurs more often than any other. A gap of 2 a little less, a gap of 3 a little less than a gap of 2. And, so on... Gaps of more than 26 should never be used as they occur very infrequently.
There are many of considerations other than just gaps.
W.

It is logical that a "gap" of "0", ie., no gap, in any 6/49 draw, whether it is just 2 consecutively, or any combination(s) of consecutives or no gap(s), does occur more than a gap (space) of 1, a gap of 2, a gap of 3, a gap of 4, a gap of 5, etc, etc, etc. and you are right about the decreasing frequency. There is a slight variation between statistical expectation and historical data.

Other than the obvious reason for gaps of 1, to use your way of stating it, (or "0"-no gap or space to use mine) being more frequent, can you explain why the decreasing frequency for gaps of increasing size? Would be appreciated!

 

[kosteczki]

Wow this has turned into an interesting conversation. Sorry I have not posted anything I have just been too bussy and have not had time for anything, I haven't even bought my lotto tickets in 3 draws and Saturday my 4 numbers hit.

But anyways I should have time this weekend to work on some stuff and post my findings.

 

[W Kaenzig]

gaps (gasp!)

A gap of zero does not exist. As with this one (13,983,816) any lotto game is merely sequential possibilities. As the sequentials go up so does the gap between nrs.
Only in pick 3 or pick 4 where each ball is taken from a different container would one have a zero gap.
The way to win at lotto is to eliminate (or reduce significantly) all possible repetitions. And, all possible repetitions can be calculated using simple math (none of that algebra stuff which can get one lost in the forest while tracking probabilities). W

 

[W Kaenzig]

data

Here are some tables which I developed some time ago which may help those who want to use (or abuse) :

3 odd 3 even 4,655,200 33.2899%
4 odd 2 even 3,491,400 24.9674%
4 even 2 odd 3,187,800 22.7964%
5 odd 1 even 1,275,120 9.1185%
5 even 1 odd 1,062,600 7.5988%
6 odd 177,100 1.2665%
6 even 134,596 .9625%
---------------- ---------
13,983,816 100.00%

ea single nr occurs 1,712,304 times (49 different)
ea pair occurs 178,365 (1,176 different)
ea tripl occurs 15,180 (18,424 different)
eq quad occurs 990 (211,876 different)
eq quint occurs 44 (1,906,884 different)

enjoy !! W

 

[kosteczki]

ea single nr occurs 1,712,304 times (49 different)
ea pair occurs 178,365 (1,176 different)
ea tripl occurs 15,180 (18,424 different)
eq quad occurs 990 (211,876 different)
eq quint occurs 44 (1,906,884 different)
QUOTE]

What do you mean by EA and EQ

 

[W Kaenzig]

typos

Sorry, my fingers hit the wrong keys, and my eyes didn't catch the mistakes. (ea) means each. (eq) was meant to be each but I slipped. W

 

[millsy]

9/29/05
to get a gap of zero would mean say 38-38 =0 , since 38 cannot
be drawn twice in lotto 649/canada - there will never be a zero gap. to consider gaps of 1 ie #38-#37=1, would be the same as
looking at your analysis for consecutive numbers...consecutive numbers is a better way to analyse and filter data for ONE gap.
...go back to my first post on gaps..you will see i mention "gap patterns"..ie..you must consider all the the gaps created for a
given outcome...for instance, last night 4, 11, 21, 27, 38, 41
the gap pattern is 7, 10, 6, 11, 3
check this against historical "gap patterns" lots of them are duplicated...implication...start with any number, say 01 thru 05,
then calculate what the winning ticket might be, using the most
popular gap patterns....if you started with 01 for instance and
transposed last night's gap pattern, the ticket would be..
01 , 08 , 18 , 24 , 35 , 38
remember to eliminate tickets with duplicate combo's of 6 or 5
which have already ocurred.
recall in last message i mentioned using the top 4 per column..
any 3 or more...(out of 24)..last night 5/6 hit !
i've had 5/6 on 7/16, 7/20, 8/24, 9/7, 9/28 - pretty good results!
(although 3x used the bonus so only won 4/6 prize!) too bad

 

[W Kaenzig]

math calculations

The data is correct but I slipped up on some of the letters as I previously confessed.
Study the game and ask yourself some questions:

Say the following draw occurs:

3 8 10 25 31 47


How many pairs are represented here? How many triples? How many quads? How many quints ?

One gets in all cases 15 pairs, 20 triples, 15 quads, and 6 quints.
There are 13,983,816 possible drawings.
How many different pairs are there? Multiply the highest possible pair and divide by the lowest possible -- (48x49) divided by (1x2). This = 1, 176
So: (13,983,816x15) divided by 1,176 = 178,365 occurances of all different pairs.

How many possible quints, and how often do they occur?

(49x48x47x46x45) divided by (1x2x3x4x5) = 1,906,884

(13,983,816x6) divided by 1,906,884 = 44

You can continue this on your own.....

Later, I'll talk about the phantom numbers..... W

 

[W Kaenzig]

0 gaps

Thank you, Millsy --- You explained it better than I. W

 

[W Kaenzig]

total picture

In order to understand the game of lotto, one has develop a single sheet that shows quantities for each of the 6 integers by position. I call it my alphabet soup.
Set up within excel (or pencil & paper) 8 columns.
col 1 represents each of the nrs from 1 to 49. col 8 represents how each nr = 1,712,304.
The 6 columns between 1 and 8 represent quantities by position.
I label these columns a,b,c,d,e,f.. You can use whatever letters you like, but I like these.
the integer 1 is easy. All of it's quantities go in column a.
But interger 2 in column a? It = 1,533,939. Then, subtract all of it's possibilities from 1,712,304. This difference of 178,365 is the quantity for integer 2 in column b.
How does one arrive at 1,533,939 ? Pretrend you are calculating the odds for a game of 6 of 48. (Indiana's lotto). (48x47x46x45x44x43) divided by 720 (1x2x3x4x5x6) x 6 divided by 48 = 1,533,939.
Not convinced? Let's do it a different way.

 

[W Kaenzig]

continued

To continue: Sorry, I'm not doin a good job of editing my typing.

If integer 2 is in col b, there are 4 higher nrs in front of it -- 3-49.
How many is that ? 47 Thus, (47x46x45x44) divided by (1x2x3x4) = 178,365. How about a 3rd way to check it? I said that all pairs occur 178,365!!
Your brain is turning to mush, isn't it? Mine has done so many times.

Some of you can continue this, but some may not. Ask some questions, as I'm not here to insult, only to help.

It took me a considerable amount of time to learn and understand all of this. W

 

[mirage]

Yes, but what is the logical or mathematical reason for the decreasing frequency of gaps or spaces of increasing size? I have noticed this occurence but I don't know the logical explanation-it just makes "intuitive" sense that I personally can't put into words, or by using math language. The best way that I can understand it is geometrically-somehow it makes sense in a draw of finite numbers. There is an underlying principle here that I can't put into words.

Btw, another term for consecutives is Gail Howards old term, "neigboring (US) or neighbouring (Canadian/English) pairs".
(Should I ask this question in Q & A section-maybe Johnph77 can answer it.)

 

[W Kaenzig]

gap frequency

I'm familiar with Gail Howard. I have 1 of her books --- "Lottery Master Guide". I learned a lot from her, and she confirmed some of my own studies.

If you accept that fact that any lotto game where all numbers are pulled from the same container can not repeat, then you also accept the fact that games such as this are in sequential order.

assume the following:
a b c d e f

6 7 8 9 10 11
6 7 8 9 10 12
6 7 8 9 10 13
6 7 8 9 10 14 etc, etc

as the possibilities change, gaps between b and a, c and b, d and c, e and d remain the same. However, the gap between f and e steps up by gap of only 1 : 1,2,3,4

Perhaps this may it clarify it for you. W

 

[millsy]

9/29/05
ok..now we're getting interesting ! so here is one for you to check out....in col a ..only occurances of 01 thru 32 +38 to date
col b..all except 01, 40 and 44 thru 49, col c, all except for 44 &
01, 02 & 45 thru 49, col d, all except for 01 thru 05, & 48 & 49
col e - all except for 07 thru 10 +12 & 49, col f..use 18 and greater..so do a nested loop to create your starting file...
should be 3million or so..but it will contain ALL winners to date
and likely MOST winners for MOST of the time...the occasional
outlier happens.

then remove all combo 6's already occurred ( incl bonus )
and remove all combo 5's already occurred....remove any #'s
which have the combo 3's which have NOT yet occurred....

now..start to create "sets" of tickets which have the following
traits.....no duplicate combos of 5 (within a set)..then filter this
set further for no duplicate combos of 4 and again for no duplicate combos of 3 and finally, no duplicate pairs....

the max you will come up with will be around 78-82 tickets...
in this set of tickets, you should ALWAYS survive any first two
balls which fall..because you should have a set with say 78
tickets..each containing 15 pairs..no duplicates so you've
covered approx 78 x 15 = 1170 of the possible 1176 unique pairs....ideally, you would sort your file looking for tickets with 2 or more from the last 21 and 3 or more from the top 4 x column
(24 numbers )....

what does this mean ??..you'll survive every draw for the first two
balls that fall...now you're playing a 4 ball lottery on 15 tickets

see if you can program this one...i have been having trouble with it....i know how it should work, but i can't get it to happen...
ideally you should end up with a "stacked in your favour" set of
78 tickets..( i use 80 above for example just to capture the 6
extra pairs with get eliminated by the programming rules...there is a law for this..i forget what it is called...some guy in sweden or norway or something like that....anyway,, see what you can
do !!

 

[mirage]

Re: gap frequency

I'm familiar with Gail Howard. I have 1 of her books --- "Lottery Master Guide". I learned a lot from her, and she confirmed some of my own studies.

If you accept that fact that any lotto game where all numbers are pulled from the same container can not repeat, then you also accept the fact that games such as this are in sequential order.

assume the following:
a b c d e f

6 7 8 9 10 11
6 7 8 9 10 12
6 7 8 9 10 13
6 7 8 9 10 14 etc, etc

as the possibilities change, gaps between b and a, c and b, d and c, e and d remain the same. However, the gap between f and e steps up by gap of only 1 : 1,2,3,4

Perhaps this may it clarify it for you. W

I'm trying but not really First of all I'm having logical problems with your statement, "if you accept the fact", etc...
Same container, no repeats=games, must be sequential order. (Huh?)

However, that's OK, one should be able to work with the rest, but then somehow the analog of the process is too...prosaic (simple). No, I don't want fancy, it's just that it's missing something - either a logical step or dimension - is missing. There must be something more. What I can say that it is a step toward explaining it..... but it doesn't explain it completely. However, thanks for your input though! Maybe I can formulate it myself after some thought. (If I can manage it, it will be in very common sense logic-nothing fancy at all. )

Meanwhile back to looking at all the possibilities and bits and pieces--did you know that the more filters you use, the more jackpots you lose? It's true, unfortunately.

 

[W Kaenzig]

simple

Mirage, you are right. Your trying to over-complicate it. It is very simple as far as math is concerned. Even though the following drawing may occur, 35 2 7 9 12 40 -- one justs puts these numbers in numerical order, 2 7 9 12 35 40. Hence, the sequence of all possibilities range from 1 2 3 4 5 6 through 44 45 46 47 48 49. W

 

[W Kaenzig]

Millsy, I followed you with the first part. However, you got into programing vs what has happened, and I don't give a hoot what has happened. Wheels or abbreviated wheels taken by themselves are short-sighted as the program that generates these various wheels do not consider other factors that are very relevant.
I, too, propose eliminating repetition but not by just using wheels. I spent well over a year trying to come up abbreviated wheels using libertybasic, which I am throughly familiar with. The problem which I was never able to overcome was the "bunching" of too many low numbers vs bunching of too many high numbers.
I finally gave up and quit being a lotto addict. But, it never left my mind. Finally, about 2 months ago, I tried a different and self-learning approach. I wanted to keep each entity as closely balanced as possible. So, I'm doing this manually in excel. So far, it works, and meets my expectations, but I'm only 51% complete. W