Lotto Tips & Strategies Methods and Systems for Winning the Jackpot. 
01162011, 08:08 PM

Registered User


Join Date: Jan 2011
Posts: 18


How many 3 number combinations are there for 649
Trying out some new strategies for lotto 649,
and was wondering how many 3 number combinations
there are for a 649 lotto ?

01172011, 01:15 AM


Registered User


Join Date: Oct 2004
Location: South Africa
Posts: 869


Hi yamaha74
There are 18424 combinations for a 3 number combo in 649 lotto
BlouBul

01172011, 08:57 PM

Registered User


Join Date: Jan 2011
Posts: 18


Cool, thanks for the info !
How's the math work for that ?
I've been playing "current drawn number
eliminations" for 649, where i take the
last 2 or 3 draws, and delete / or circle
the numbers that have been recently drawn for the
last 2 or 3 draws, and eliminate them from
this table, so that you get about 30 to 36
numbers to start making combinations out of.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 24 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49
...then start playing combinations from there.
From what i've seen, out of every 2 to 4 draws,
there is always a fresh set / 6 new numbers, that
have' nt been drawn in the last 1 or 2 draws.
... closest i've "ever" come to winning millions,
was when i pulled 12 numbers out of all the
numbers i eliminated, and only bought
two dollars worth of tickets
the xxxxxx were the winning numbers,
but i picked
xxxyyy
yyyxxx
...won $20, ...but the amount i could
have won with playing different combinations
could have been big $$$$...

01182011, 06:35 AM


Registered User


Join Date: Aug 2009
Location: UK
Posts: 358


Quote:
Originally Posted by yamaha74
Cool, thanks for the info !
How's the math work for that ?
...

Well to think it through in simple terms, lets start with pairs.
if we were to list all the partners for number 1 in pairs we'd have 1,2...1,3,,,, right up to 1,49 and there would be 48 of them.
The same appies to all the other number pairs . So we have 49 numbers each with 48 possible partners giving us 49 x48 = 2352. However there are duplications in the list since pair 1,2 is the same as pair 2,1 etc. So the number of pairs is 2352 divided by 2 =1176.
Now lets look at triples. we have 49 numbers each with a possible 48 second partners and a possible 47 third partners giving us 49 x48 x47 =110544 triples. Again we have duplications in the list since
1,2,3
1,3,2
2,1,3
2,3,1
3,1,2
3,2,1
are all the same combinations. So we have to divide 110544 by 6 . This gives us 18424 unique combinations.
The quick way if you have Excel is to use the formula =COMBIN(49,3) to get the same answer.

01182011, 08:45 PM

Registered User


Join Date: Jan 2011
Posts: 18


thanks for the info.
...ever get that feeling someday
you might actually win ?
...been really close more than a few times,
playing number eliminations, etc

01192011, 12:04 AM


Registered User


Join Date: Jul 2006
Location: Encinitas, CA U$A
Posts: 4,499


y'all can also use the WolframAlpha site to calculate C(49,3)
[url]http://www.wolframalpha.com/input/?i=C%2849%2C3%29[/url]
or get your 1176 with this one: C (49,3)(49*48)/2
[url]http://www.wolframalpha.com/input/?i=C++%2849%2C3%29%2849*48%29%2F2[/url]
cya,
blitzed

01192011, 01:01 AM


Registered User


Join Date: Sep 2007
Location: Nairobi
Posts: 82


Quote:
Originally Posted by yamaha74
thanks for the info.
...ever get that feeling someday
you might actually win ?
...been really close more than a few times,
playing number eliminations, etc

Hi yamaha74,
I have as well missed twice the main one as a result of elimination. What elimination method do you use?
Share if you want
Thank you

01222011, 02:39 PM

Registered User


Join Date: Jan 2011
Posts: 18


WESTERN 649 RECENT WINNING NUMBERS
Draw Date Winning Numbers Bonus EXTRA
Wed. Jan 19, 2011 7 20 22 26 38 46
bonus 21
Sat. Jan 15, 2011 8 15 19 31 32 34
bonus 12
Wed. Jan 12, 2011 16 18 19 22 35 46
bonus 28
Sat. Jan 08, 2011 6 19 25 28 31 32
bonus 41
Wed. Jan 05, 2011 1 9 21 39 40 47
bonus 23
Sat. Jan 01, 2011 11 22 25 27 28 46
bonus 21
Wed. Dec 29, 2010 4 14 26 35 38 41
bonus 39
Sat. Dec 25, 2010 6 10 12 15 16 44
bonus 13
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48 49
Example...
...i start by eliminating jan 19, and jan 15 numbers
7,20,22,26,38,46, and 8,15,19,31,32,34
which would leave 37 numbers left to choose
combinations from. Jan 19 bonus #21 is a repeater,
so that will give me 36 numbers for combo's,
If you want to eliminate others, look for more repeaters,
such as 6,12,25,28,35,39,41, which can give you as few as 29
numbers to pick combo's from.
Once you've picked all the combo's you want to play,
google "numbers ever won", ...and you can check
to see how many times the combo's you picked have
ever won before, and make adjustments to you're combo's
from there. I try and make combo's that have the least
amount of recently drawn winning numbers/combinations
...sometimes this system works well, ...sometimes the repeating
numbers keep popping up, and i dont win a dime.
But then i just adjust my combo's a little,
and play them again next draw...
i figure the odds of using "cold" numbers seems to be more practical
than using numbers that have recently been drawn

01232011, 02:33 AM


Registered User


Join Date: May 2006
Location: SouthAfrica
Posts: 37


Lotto Hit Tracker (4Drawn Numbers)
The following E.A.L plugin seems to be doing the same thing you are trying to accomplish.
Unfortunately it is only for registered users, but keep this info for future reference.
http://www.slx.za.net/rtfm/eal/plugins/lotto_hit_tracker_4dn.html

02052011, 07:23 PM

Registered User


Join Date: Jan 2011
Posts: 18


So lets say i eliminate the recently drawn numbers
and end up with 36, or 30 numbers to start picking
combo's with.
If i use the same math as 49x48x47/6 = 18424,
is it correct that 36x35x34/6 = 7140 unique combo's
to match 3 numbers, or 30x29x28/6 = 4060 combo's ?
Then if i start comparing those 7140, or 4060, sets of
3 number combos that have recently been drawn and
eliminate them, the odds of matching 3 numbers could
be reduced even more drastically,
as well as the possibility of matching 4, 5, or 6 numbers.
I figure the odds of winning the lottery could be figured out
mathematically, or at least the odds could be greatly reduced
by using some type of system.
Any similar / or other idea's to try ?

02082011, 07:18 AM


Registered User


Join Date: Aug 2009
Location: UK
Posts: 358


Your arithmetic is correct, but thats where your argument comes to a stop. If we use the example of 30 numbers making 4060 triples we could ask ourselves how many lines we could eliminate by just pairing triple A with triple B from your set. The initial thoughtless answer would be 4060 x 4059 possible partner triples from your set =16,479,540 ! Clearly nonsense a) because thats more than there are lines available and b) because it includes invalid lines which are really not lines of 6 different numbers but have duplicate numbers taken from both triple A and triple B. So you've got a lot of hard work just sorting out which of the above lines are valid. Add to that the fact that each triple in your shortlist has a possible 15,180 parent lines of 6 it might appear in and you've got 4060 of them making 61,630,800 lines of 6  obviously containing overlaps and duplicates from each triple, but probably using all the other 49 numbers mixed in with your triple, if your original 4060 triples used all 49 numbers. I think you can see you've got your work cut out.
All to little avail, since the whole thing assumes that a recent triple has less chance of appearing in the next draw than any other, which is a fallacy, so a lot of hard work for nothing.

02082011, 09:15 PM

Registered User


Join Date: Jan 2011
Posts: 18


I was more thinking, that if winning the western 649
is 1/6,991,908, its possible to start figuring out combinations
that might be drawn, instead of what has just been drawn.
Lets say the last 2 draws were 5, 12, 17, 28, 33, 41, 47,
and 1, 14, 23, 30, 38, 44
What are the chances that "any" of those 3 number
combinations will show up in the next draw,
compared to a new set of 3 number combinations.
Granted it might happen, but the possibility's of it
happening consecutivly draw after draw, is astronomical.
From what i've seen, fresh sets / non repeating numbers
happen every 3 or 4 draws.
Just trying to narrow down the odds of numbers that "might"
be drawn !
Any better ideas, im all ears !

02092011, 06:04 AM


Registered User


Join Date: Aug 2009
Location: UK
Posts: 358


Quote:
Originally Posted by yamaha74
I was more thinking, that if winning the western 649
is 1/6,991,908, its possible to start figuring out combinations
that might be drawn, instead of what has just been drawn.
Lets say the last 2 draws were 5, 12, 17, 28, 33, 41, 47,
and 1, 14, 23, 30, 38, 44
What are the chances that "any" of those 3 number
combinations will show up in the next draw,
compared to a new set of 3 number combinations.
Granted it might happen, but the possibility's of it
happening consecutivly draw after draw, is astronomical.
From what i've seen, fresh sets / non repeating numbers
happen every 3 or 4 draws.
Just trying to narrow down the odds of numbers that "might"
be drawn !
Any better ideas, im all ears !

Well unless theres something in the drawing process or the lottery machine, that leaves a "memory trail"  each draw is an independent event. So what happenned in the last 2 draws is gone forever and the new draw can choose any combination, even the same ones again. Its just as likely as any other combination.
What you you are suggesting would reduce your personal percieved odds, but not the real odds. The odds against any triple matching any triple in another draw (ignoring bonus), be it a past draw or a future one is 1 in 57. Because you get two tickets for the price of one that is 1 in 29 to quote the Western649 lottery website.
What you are doing is making a calculated 'side bet' that in the next draw such a triple repeat is unlikely. You wish to construct a system based on that side bet. You are not alone, all systems are based on an assumption or a side bet that "this won't happen in the next draw".
Theres nothing wrong with that as long as you realise that the assumption won't hold forever, and sometimes you will lose that side bet.
I calculate that you would eliminate 607,180 combinations leaving 6,384,728 still in play  so that would be your new odds against winning a jackpot.
What you have to consider, is  is all the hard work and number crunching worth that percieved impovement ? Don't get me wrong, Theres nothing wrong with your or any other system, provided you accept that you are making (in your case) a 1 in 29 bet upfront that your system will work on this occasion.

02092011, 11:18 AM


Registered User


Join Date: May 2006
Location: SouthAfrica
Posts: 37


Double Matrix Play
Using the E.A.L Lotto Hit Tracker (4Drawn Numbers) plugin as mentioned above, you will be amazed
at how many times a number from a previous Lotto draw comes up in the next one.
Here is some data based on the SouthAfrican Lotto (6/49) using this E.A.L plugin.
You have set your range at 30 or 36 numbers, but with this plugin the range limit is at 25 numbers.
007 Times  6 Numbers + Bonus
012 Times  6 Numbers
039 Times  5 Numbers + Bonus
045 Times  5 Numbers
134 Times  4 Numbers + Bonus
118 Times  4 Numbers
173 Times  3 Numbers + Bonus
151 Times  3 Numbers
119 Times  2 Numbers + Bonus
110 Times  2 Numbers
049 Times  1 Number + Bonus
032 Times  1 Number
010 Times  0 Numbers + Bonus
002 Times  0 Numbers
This above mentioned list basically informs you of how many times Winning Lotto numbers have come up
in the range limit (1  25)
Example: Taking the 6 Numbers + Bonus, the 7 draws are as follow:
Draw  Lotto Numbers  Hit Tracker Location (4Drawn Numbers)
039  101415204647 B21  010203071022 B@04
105  101319303242 B06  050718202123 B@12
176  131820303337 B34  020410131518 B@19
232  081618303248 B45  071115161822 B@06
257  192130404648 B15  030608131719 B@12
576  020811214748 B23  060810131421 B@18
941  060824324547 B21  050811141525 B@21
But I have developed my own system where you select Lotto numbers, E.A.L then uses your numbers as a
Lotto wheel rule and looks up numbers as to where they are on a stats list.
This being mentioned under the "Double Matrix Play" menu items:
http://www.slx.za.net/rtfm/eal/addons/user_dbase_01.html
So instead of focusing on only one aspect of using past Lotto numbers, the "Double Matrix Play" system
can be used to increase you chances of winning by treating the "E.A.L Lotto Hit Tracker Data" as
additional Lotto draws.

02102011, 10:31 AM


Registered User


Join Date: Aug 2009
Location: UK
Posts: 358


It is a statistical fact that any 6/49 result has a 0.564035 probability of having at least one ball matching ANY other result, not just the one before it ! So a better than 50/50 chance that pick any draw in lotto history, at least one ball from it will appear in the next future draw. I have done extensive research on this for the UK lotto and have built a system around it. To use it, I'm making a side bet that 56% of a selection of (up to 25) chosen past lotto draws will have at least one ball in the future draw. This creates a table of most common possible repeat balls from which I select the highest counts of possible matches to create a selection pool. I've never managed to capture even a match 5 at the upper end of my shortlist, mainly because 40% of the draws I select do not have a match with the draw I'm predicting, so dilute the pool with wrong numbers. Like I said, its a calculated side bet, sometimes I'm lucky and get more draws that will repeat in my list than ones that don't.

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