THIS WORKS A LOTTO-LOGIX SYSTEM
Robert Perkis
by Robert Perkis Copyright 2001
The lottery system I'm about to describe appears to work. It
has been tested against several games and has done as well as
I will relate. It is not a cure all, it doesn't guarantee you
will win millions, but it is a solid first step. I hope you
will test the system against at least 20 past game draws
before using it and will offer suggestions either public or
private.
As rec.gambling.lottery regulars know, there has been only one
effective system to date for eliminating numbers. Royce Penny
of http://www.geocities.com/wallstreet/7746/ and I Robert
Perkis of http://www.lotto-logix.com/ have long espoused the
simple method of NOT using the numbers from the prior draw in
the next draw.
This elegant solution works roughly 45 times out of 100 and
when it does, in a 6/49 game sweeps 8 million combinations off
the table. Of course this is balanced by the 55 draws out of
100 where one or two numbers are found among those of the
prior draw making only lower tier prizes possible. The idea
however, is to have an advantage in some draws at a reasonable
exchange rate.
While I appreciated the ease and value of this simple method
of number elimination, I regretted the loss of the easiest
winning number pick up, the one or two winning numbers often
found in the last draw. I was also frustrated by the auto
imposed limit of only six numbers being eliminated at best.
When Florida expanded the game from 6/49 to 6/53 I realized we
needed something better, a system that would use the numbers
from the last draw, open the door to eliminating more than six
numbers and work as well as the old system.
This method is very easy to get used to, especially if you
store past drawings in a lotto software that allows you to
print out a limited range of past drawings. You will want to
print out the last 15 to 20 past drawings from the game
history. We will use only the last 15 but, maybe you will see
something useful further back that I missed.
Here we have the last 15 drawings of the Florida 6/53 broken
by a blank line between the groups we will be working with.
E-Group 3 Lines
783 07/28/01 07-31-36-42-43-50
784 08/01/01 15-24-28-32-50-51
785 08/04/01 04-05-20-38-40-49
D-Group 3 Lines
786 08/08/01 05-08-11-13-23-40
787 08/11/01 13-23-27-43-44-50
788 08/15/01 04-15-25-28-34-51
C-Group 4 Lines
789 08/18/01 01-04-17-32-52-53
790 08/22/01 09-11-15-29-31-47
791 08/25/01 01-09-37-42-44-46
792 08/29/01 18-28-31-32-48-49
B-Group 3 Lines
793 09/01/01 06-17-26-43-45-50
794 09/05/01 03-09-10-24-33-35
795 09/08/01 02-11-25-29-30-35
A-Group 3 Lines
796 09/12/01 07-16-28-46-47-49
797 09/15/01 17-19-21-24-49-53
798 09/19/01 09-25-27-38-43-52
Below these we write out the numbers 01 through 53 and next to
each number we put a check mark for each time a number appears
in the groups A = 796-798 and C = 789-792. Note that for the
moment we are skipping group B = 793-795. There should be a
total of 42 check marks for a pick six game when this part is
finished.
01 ** 16 * 31 ** 46 **
02 17 ** 32 ** 47 **
03 18 * 33 48 *
04 * 19 * 34 49 ***
05 20 35 50
06 21 * 36 51
07 * 22 37 * 52 **
08 23 38 * 53 **
09 *** 24 * 39
10 25 * 40
11 * 26 41
12 27 * 42 *
13 28 ** 43 *
14 29 * 44 *
15 * 30 45
Now here's where it gets a little tricky. We are going to use
the three lines B = 793-795. If a number from the three lines
793 through 795 has a check mark next to it we add another for
each time it appears in the three lines. However, if the
number has no check marks, (a blank space) we put an "X".
B-Group
793 09/01/01 06-17-26-43-45-50
794 09/05/01 03-09-10-24-33-35
795 09/08/01 02-11-25-29-30-35
01 ** 16 * 31 ** 46 **
02 X 17 *** 32 ** 47 **
03 X 18 * 33 X 48 *
04 * 19 * 34 49 ***
05 20 35 50 X
06 X 21 * 36 51
07 * 22 37 * 52 **
08 23 38 * 53 **
09 **** 24 ** 39
10 X 25 ** 40
11 ** 26 X 41
12 27 * 42 *
13 28 ** 43 **
14 29 ** 44 *
15 * 30 X 45 X
If we consider the numbers with an "X" as eliminated, we have
put 9 number aside and taken millions of combinations off the
table we no longer have to worry about matching. The trouble
is, this was a very good example.
Most of the time, we will eliminate 6 to 7 numbers and 9 times
out of 20 none of those numbers will appear in the drawing.
Sometimes this method only eliminates four or five numbers,
how do we remove some more?
When numbers from the B-Group don't eliminate they add to the
check mark count. When a number has five check marks from the
last 10 drawings, it is almost sure not to hit again. I have
found when a number has 4 check marks there is a 50/50 chance
it will hit again, but if there are two numbers with 4 check
marks there is almost a 100% chance one will hit and the other
will not. Something of a gamble here.
I have also found we can skip or ignore the D-Group.
If we are desperate to eliminate at greater risk some more
numbers, use Group-E numbers to put an "-x" next to remaining
blanks, but we do not add them to the check mark count.
E-Group 3 Lines
783 07/28/01 07-31-36-42-43-50
784 08/01/01 15-24-28-32-50-51
785 08/04/01 04-05-20-38-40-49
01 ** 16 * 31 ** 46 **
02 X 17 *** 32 ** 47 **
03 X 18 * 33 X 48 *
04 * 19 * 34 49 ***
05 -x 20 -x 35 50 X
06 X 21 * 36 -x 51 -x
07 * 22 37 * 52 **
08 23 38 * 53 **
09 **** 24 ** 39
10 X 25 ** 40 -x
11 ** 26 X 41
12 27 * 42 *
13 28 ** 43 **
14 29 ** 44 *
15 * 30 X 45 X
So here we are with 14 numbers eliminated and number 9 as a
possible having a count of four checks. By now I'm sure you
are wondering what the winning numbers were . . .
799 09/22/01 01-17-23-39-44-53
A TOTAL SUCCESS: Your mileage may vary.
Additional Tips:
This method has broken the last 10 drawings down into three
groups. The ones with the X -x we don't play and those with
check marks and those without check marks. The ratio of
numbers with check marks being drawn to those without is
roughly 2 to 1. You would want your combinations to have 4 or
5 numbers with check marks and 1 or 2 numbers from those
without check marks.
Play all the remaining numbers on six or seven tickets or
wheel them all to about a 90% guarantee using a 3if5 or 3if6
open cover wheel.
If you eliminate in the range of 9 to 12 numbers, take
insurance and play those numbers in two combinations.
This method works equally as well with Pick-5 games, the only
difference is the lines have 5 numbers instead of six, but use
the same groups A - E and the same number of lines each.
This method has not yet been tested against very large games
like the PA 6/69 or the New York 6/59.
In some tests I would reach 20 draws and find I had only 8
where the method worked, an additional test or two would pick
up the difference. I did notice success often came in
clusters, so don't be discouraged if the first attempts don't
work because you will then have a cluster of draws where it
does. This is not to say there may be some games where this
doesn't work, hence you should test before risking your funds.
Please let me know, who, what, when and where if you test and
find whether this works for you or not. If you see a way to
improve on this system without increasing the risk too much,
please post or drop me an email.
Good luck to you. Robert Perkis / http://www.lotto-logix.com/
Thunder Child
I will try and test your strategy with the UK Lottery and keep you posted
with what I think the groups are (cos.. I had to read this through twice). I
may need some help here but if your right then this beats tossing.
Kind Regards
TC
Robert Perkis <rob...@icdus.com> wrote in message
news:3BB24227...@icdus.com...
Robert Perkis
09/26/2001 was 06-18-31-38-50-53 the ABC first part of
the system eliminated 7 numbers none of which hit, the
Part E eliminated 7 more one of which hit, so the price
of eliminating 14 out of 53 this time was one winning
number. RP
Nigel Hurneyman
beat the Sharkey Challenge with this system.
Good luck,
NigelH
Ratty
______________________________________________________________
Don't tolerate spam. Report to http://spamcop.net
Robert Perkis
the lottery and by doing so exceed standard deviation by something
like three times to rule out random good luck.
This system remains firmly within the boundaries of probability
just like the system of eliminating the previous six drawn numbers.
Try the system Nigel, see if it doesn't take a handful of numbers
off the list 9 times out of 20 and when doing so sweep millions of
combinations off the table. Don't even bother with the E part, just
try the A-B-C for the past couple of drawings. Remember, it doesn't
matter where you pull numbers out of the stack, the stack is shortened
just the same.
When a lottery wants to bump up the odds by millions, they only add
one or two numbers, people say what's an additional number more or
less, yet the difference between a 6/48 and a 6/49 lottery is a
million combinations. Successfully eliminate a single number for a
draw and you've taken a million combinations off the table.
I don't want to rehash all the discussions of the past years. I
think this is something lottery players will find useful to give
them an advantage in a reasonable number of drawings, paid for
by those drawings where it doesn't. This is all most of us ask
for, a better shot some of the time. Robert Perkis
Martin Harrison
I've run your system through the entire history of the UK 6/49 lottery
results from draw 17 onwards (585 draws). The results follow. In each
case I've calculated on the basis of playing all combinations
prescribed by the system.
Basic, Groups ABC as prescribed:
Expected wins: 248
Actual wins: 234
Groups ABC as prescribed, + eliminating 5+ checks:
Expected wins: 241
Actual wins: 228
Groups ABCE as prescribed, + eliminating 5+ checks:
Expected wins: 126
Actual wins: 125
Groups ABCE as prescribed, + eliminating 5+ checks, playing 1 or 2
unchecked:
Expected wins: 85
Actual wins: 74
These figures would suggest the system doesn't work too well for the
UK lottery.
What is your premise here? Why do you think combinations resulting
from this system should beat expectation?
Mart
Robert Perkis <rob...@icdus.com> wrote in message news:<3BB24227...@icdus.com>...
Paracelsus
> To beat the Sharkey Challenge a system must first propose to beat
> the lottery and by doing so exceed standard deviation by something
> like three times to rule out random good luck.
>
> This system remains firmly within the boundaries of probability
> just like the system of eliminating the previous six drawn numbers.
>
> Try the system Nigel, see if it doesn't take a handful of numbers
> off the list 9 times out of 20 and when doing so sweep millions of
> combinations off the table. Don't even bother with the E part, just
> try the A-B-C for the past couple of drawings. Remember, it doesn't
> matter where you pull numbers out of the stack, the stack is shortened
> just the same.
>
> When a lottery wants to bump up the odds by millions, they only add
> one or two numbers, people say what's an additional number more or
> less, yet the difference between a 6/48 and a 6/49 lottery is a
> million combinations. Successfully eliminate a single number for a
> draw and you've taken a million combinations off the table.
>
> I don't want to rehash all the discussions of the past years. I
> think this is something lottery players will find useful to give
> them an advantage in a reasonable number of drawings, paid for
> by those drawings where it doesn't. This is all most of us ask
> for, a better shot some of the time. Robert Perkis
>
Sorry, Robert, I'm confused here. If this doesn't beat the lottery outwith
the boundaries of probability, why go to all this trouble? Why not just
eliminate a certain amount of numbers randomly? And in what way does "THIS
WORK"? (Sorry if I'm going over old ground.)
Paracelsus
Martin Harrison <galc...@yahoo.co.uk> wrote in message
news:5d249201.01092...@posting.google.com...
Scott Rudy
field of numbers. At 81% wouldn't you expect to match about 5.6 of the
numbers drawn about 81% of the time? If you choose 81% of the total number
field at random wouldn't you come out with about the same results?
Look to see which past drawings the numbers are drawn from the most often
and choose numbers from the 6 highest number producing past drawings. In the
Ohio Super Lotto the top 6 past drawings are the last 1, 2, 3, 4, 5 and 7th
past drawings 1 being the most recent.
These past drawings can yeild 25 to 34 numbers but the average is 29
numbers. This strategy matches 4 to 6 numbers of the numbers drawn about 73%
of the time and 5 to 6 numbers around 45% of the time, this is based on the
last 38 drawings.
Good Luck and may all of your numbers be winners.
Scott Rudy
Thunder Child
I tried the experiment a couple of times and realised there has to be an
easier way for number elimination.
In fact there are several.
I'm conducting an experiment which is to isolate a definitive number of
lines to trap the winning numbers. Much like the Szur-Can theme, I have
managed to wittle it down to just under 65,000. In the last 10 weeks the
lines have captured 2 x 5+ wins.
Question: How long should the experiment run, in order 'exceed standard
deviation by something like three times to rule out random good luck.'?
Cheers
TC
Scott Rudy <sa...@erinet.com> wrote in message
news:3bb3c3a7$0$1525$4c5e...@news.erinet.com...
Martin Harrison
> Christ, Mart.......you didn't even give me time to sharpen me pencil!
>
LOL!
Of course, I cheated by using a computer, but if you want to check
these figures by hand, feel free...
Robert Perkis
even your conclusion. ;-) I recall hearing Pablo Picasso in
a hardware store once exclaimed, "You mean people actually
pay money for this stuff?!"
Comments embedded below.
Martin Harrison wrote:
>
> Hi Robert,
>
> I've run your system through the entire history of the UK 6/49
> lottery results from draw 17 onwards (585 draws). The results
> follow. In each case I've calculated on the basis of playing all
> combinations prescribed by the system.
>
> Basic, Groups ABC as prescribed:
> Expected wins: 248
> Actual wins: 234
I set out to find a method equivalent to eliminating the six numbers
from the prior draw yet not restricted to six numbers, apparently I
was successful. I am trusting your figures as averages, I have no
idea how you dealt with draws with more or less numbers eliminated.
> Groups ABC as prescribed, + eliminating 5+ checks:
> Expected wins: 241
> Actual wins: 228
Again, I don't know how many cases there were with 5+ checks next
to a number, only six failures of this portion of the theory may
or may not be excessive if it is out of a hundred+ such draws.
> Groups ABCE as prescribed, + eliminating 5+ checks:
> Expected wins: 126
> Actual wins: 125
Ah, the added risk of eliminating more numbers, of course this is
only shooting for jackpots, if you can eliminate 10+ numbers and
have 5 among those left in most of the remaining half that's not
so bad.
>
> Groups ABCE as prescribed, + eliminating 5+ checks, playing 1 or 2
> unchecked:
> Expected wins: 85
> Actual wins: 74
One person wrote and said, "...why go through all this fuss and
bother, just play the first 36 numbers and forget the rest."
That "theory" would eliminate 13 numbers in every draw, I ran
them against the 600 U.K. draws 11/19/94 - 09/22/01 and came
up with 64 times all six would have been within the 36. We are
all within the ballpark here.
The same report told me my friend would have had (not using the
bonus number here) 235 draws with five numbers and 189 draws with
four numbers among his 36. I expect my method would prove equal
to these figures.
>
> These figures would suggest the system doesn't work too well
> for the UK lottery.
On the contrary, any lottery player would be happy with these
results in contrast to the usual success rate of pulling the
numbers out of thin air to eliminate. It sounds great, just
don't play 14 numbers, but try picking them with nothing to
go on or just an rng. I don't want to die saying I should of
played black.
>
> What is your premise here? Why do you think combinations
> resulting from this system should beat expectation?
>
> Mart
You can't beat expectation, but you can play at it. Most of
the time lottery players had no standardized method of play
resulting in flailing around with no idea at all how to take
numbers off the table. The usual result is wheeling 18 - 24
numbers chosen by some magic method and getting two or three
on their wheel.
I don't want to rehash all the arguments of the past over
why lottery players prefer a strategy to tossing everything
up in the air and playing what comes down heads.
Thank you for the test, the results are as expected, hurray!
Robert Perkis / http://www.lotto-logix.con/
Robert Perkis
with it's automatic consequential resulting combinations containing
those numbers being eliminated, the other is combination rejection
for unexpected criteria.
What you would want to do is both. Eliminate some numbers to take
more than half the combinations off the table that way, then you
can whittle down what's left by removing aspects you don't expect
to see in the next drawing.
The experiment should be an extenuation of the system being applied
against the history run to see if it holds true. Some systems that
look good against past history, like the 24 numbers that were found
in the most jackpots and five number wins, show no likelihood to
predict the next draw.
You test these things until you feel it good enough to put your
own hard earned cash down on the counter to play it. We're doing
this to win, not write a research paper on applied lottery techniques.
Go to my http://www.lotto-logix.com/information.html and try to get
yourself a demo of Lottohat and the 16bit free version of Lotwin you
will have to hunt around a bit on their site to find it from the
other versions and register the free version, it is worth it if you
don't already have it. I can think of no software better made to
work these systems combined or independently described here. This is
not to say other software aren't good for other theories, but these
are doing exactly what we would need.
Good luck to you, keep us informed.
Robert Perkis / http://www.lotto-logix.com/
Martin Harrison
Robert Perkis <rob...@icdus.com> wrote in message news:<3BB4C4A4...@icdus.com>...
> This is super Martin, everything is well within expectation
> even your conclusion. ;-) I recall hearing Pablo Picasso in
> a hardware store once exclaimed, "You mean people actually
> pay money for this stuff?!"
Glad you like it. I've got some figures on elimination counts for you
below. I forgot to mention originally that I've ignored the bonus ball
in all of these tests.
>
> Comments embedded below.
>
> Martin Harrison wrote:
> >
> > Hi Robert,
> >
> > I've run your system through the entire history of the UK 6/49
> > lottery results from draw 17 onwards (585 draws). The results
> > follow. In each case I've calculated on the basis of playing all
> > combinations prescribed by the system.
> >
> > Basic, Groups ABC as prescribed:
> > Expected wins: 248
> > Actual wins: 234
>
>
> I set out to find a method equivalent to eliminating the six numbers
> from the prior draw yet not restricted to six numbers, apparently I
> was successful. I am trusting your figures as averages, I have no
> idea how you dealt with draws with more or less numbers eliminated.
The elimination count in this case varied from 2 to 12, with an
average of 6.38.
To get the expected wins, I sum the expectations from each individual
draw. For example if 6 numbers are eliminated, then for that draw the
expectation is C(43,6)/C(49,6) = 0.436. The sum of these fractions for
all draws gives the total expectation.
>
> > Groups ABC as prescribed, + eliminating 5+ checks:
> > Expected wins: 241
> > Actual wins: 228
>
>
> Again, I don't know how many cases there were with 5+ checks next
> to a number, only six failures of this portion of the theory may
> or may not be excessive if it is out of a hundred+ such draws.
112 more numbers were eliminated here (in 585 draws), including 2 more
in each of 4 individual draws. The range was again 2 to 12, with the
average up to 6.57.
>
> > Groups ABCE as prescribed, + eliminating 5+ checks:
> > Expected wins: 126
> > Actual wins: 125
>
>
> Ah, the added risk of eliminating more numbers, of course this is
> only shooting for jackpots, if you can eliminate 10+ numbers and
> have 5 among those left in most of the remaining half that's not
> so bad.
The elimination average is up to 10.81 at this point, with the range 4
to 19.
You would have got (exactly) 5 correct in 223 draws playing all these
combinations.
>
> >
> > Groups ABCE as prescribed, + eliminating 5+ checks, playing 1 or 2
> > unchecked:
> > Expected wins: 85
> > Actual wins: 74
The elimination average and range is unchanged at this point, though
obviously much fewer combinations come into play.
>
> One person wrote and said, "...why go through all this fuss and
> bother, just play the first 36 numbers and forget the rest."
I was kinda thinking along those lines myself.
>
> That "theory" would eliminate 13 numbers in every draw, I ran
> them against the 600 U.K. draws 11/19/94 - 09/22/01 and came
> up with 64 times all six would have been within the 36. We are
> all within the ballpark here.
>
> The same report told me my friend would have had (not using the
> bonus number here) 235 draws with five numbers and 189 draws with
> four numbers among his 36. I expect my method would prove equal
> to these figures.
That would be expected.
> >
> > These figures would suggest the system doesn't work too well
> > for the UK lottery.
>
>
> On the contrary, any lottery player would be happy with these
> results in contrast to the usual success rate of pulling the
> numbers out of thin air to eliminate. It sounds great, just
> don't play 14 numbers, but try picking them with nothing to
> go on or just an rng. I don't want to die saying I should of
> played black.
I guess 'work' is subjective in this conext!
I see what you mean though. Having a method saves you from making any
nasty 'random' decisions. I'm finally beginning to see what you're
about, Robert ;).
Worth pointing out though, that playing just the first 36 numbers is
also methodical (and less time-consuming).
>
> >
> > What is your premise here? Why do you think combinations
> > resulting from this system should beat expectation?
> >
> > Mart
>
>
> You can't beat expectation, but you can play at it. Most of
> the time lottery players had no standardized method of play
> resulting in flailing around with no idea at all how to take
> numbers off the table. The usual result is wheeling 18 - 24
> numbers chosen by some magic method and getting two or three
> on their wheel.
>
> I don't want to rehash all the arguments of the past over
> why lottery players prefer a strategy to tossing everything
> up in the air and playing what comes down heads.
>
> Thank you for the test, the results are as expected, hurray!
Glad to help.
Mart
Geoff Bache
This left me 34 numbers to play wheeled to 17 lines.
1 of the previous weeks numbers came out and 2 from yours, leaving 3.
I'm pleased to say that the 3 lined up...£10.
02 38 44
So, for me. On the UK 6/49. Week 1. Hits 1.
I'll try again next week
Geoff.
Frank
Mart,
you answered to Robert:
>
>I guess 'work' is subjective in this conext!
>
>I see what you mean though. Having a method saves you from making any
>nasty 'random' decisions. I'm finally beginning to see what you're
>about, Robert ;).
>
"Random" decisions lead to a method too, but sometimes we forget that. The problem is that with that method we cannot test it in past draws like a "persistent" way of selecting (forecasting, predicting) numbers.
And as a method it must be tested in the future draws too, but we don't know which idea will assist us when selecting numbers in the future.
Sometimes we read in RGL some criticism to predictions that forgets this way of thinking.
So, before we forecast some numbers, we may be sure of the limitations of our method and how to test it in the future.
I remember your help about the "math question" thread, that I put here to know how to test a method.
There is a group of RGL readers that is pursueing this way. That's not an easy way ... may be we'll met something ...
Frank
>Mart
Royce Penny
By random expectation, to eliminate nine numbers, you should
be successful 30.75% of the time for a 6/53, and 27.45% of
the time for a 6/49.
I can propose a modification to your system as a cross
reference that may prove to give more positive results and a
cross reference (at least for a 6/49 - but try it for the
6/53). This has historically worked for the 6/49 where 7
numbers are drawn, and has about a 50% success ratio - which
so far does exceed random expectation - so maybe it will not
work in the future.
The method will complement yours and may give more
definitive results. This method is that 2 of the numbers
will come from the previous three draws, two for the 4th to
the 8th previous draws, and two from the remaining numbers.
(A "321" vs the "222" also works very well) If you mark
these numbers with a "Y1", "Y2", and "Y3" on your sheet, you
can try to match only those that meet your criteria, but
also give you no more that three of the individual Y1, Y2,
or Y3's.
I will try this for the Canada Lotto649 to see how it has
historically worked.
--
Royce Penny
Royce Penny's Money Machine
http://www.geocities.com/lottoking.geo
-------------------------------------------
Royce Penny
The "Y1" numbers are those that *occur* in previous draws 1
to 3. The "Y2" values are those numbers that are *unique*
to the 4th to 8th previous draws (ie: they did not appear in
the "Y1" group). The "Y3" numbers are those that have not
shown at all in the previous 8 draws. Thus, there is only
one instance of any number in any of the three groups.
In the next week or two I will run Robert's method through
the Canada Lotto 649 history file, and the combined method,
and post the results to the group.
--
Royce Penny
Royce Penny's Money Machine
http://www.geocities.com/lottoking.geo
--------------------------
Royce Penny
previously posted to this group. (The method is listed in
Robert's post at the bottom of this post.)
The data below applies to the Canada Lotto 649 for a draw
history of 1835 draws to October 15, 2001 and includes the
bonus number for a total of 7 numbers drawn.
The third chart shows that by using all 3 steps of
elimination, an average of 9 numbers can be eliminated with
a 56.57% success ratio. In other words, for the Canada
Lotto 649 history file, you can eliminate 9 numbers, and be
correct in selecting 6 or 7 numbers over 50% of the time.
The maths indicate that when 9 numbers are eliminated (40
selected), you should correctly select 6 numbers only 27.45%
of the time. Thus, this method (STEP 3) appears to be
nearly twice as effective as random expectation when applied
to the Canada Lotto649 history file.
STEP1
-----
This is the first step in elimination described by Robert.
Overall, it catches 6 or 7 numbers 83% of the time and
averages 44 numbers selected from the 49.
To read the chart below, in 121 draws, 7 numbers were
successfully selected, and in 128 draws, 6 numbers were
successfully selected when 6 numbers were eliminated (43
numbers remaining) by STEP 1:
Select7 Select6 Select5 Select4 Select3 Select2
------- ------- ------- ------- ------- -------
30 0 0 0 0 0 0
31 0 0 0 0 0 0
32 0 0 0 0 0 0
33 0 0 0 0 0 0
34 0 0 0 0 0 0
35 0 0 0 0 0 0
36 0 0 0 0 0 0
37 0 1 1 0 0 0
38 0 1 0 1 0 0
39 2 5 6 0 0 0
40 8 24 16 5 0 0
41 40 61 25 12 1 0
42 73 91 49 9 4 0
43 121 128 66 15 2 0
44 176 157 53 8 0 0
45 188 133 23 4 0 0
46 132 72 10 2 0 0
47 56 19 0 0 0 0
48 30 2 0 0 0 0
49 3 0 0 0 0 0
------- ------- ------- ------- ------- -------
# draws 829 694 249 56 7 0
45.18% 37.82% 13.57% 3.05% 0.38% 0.00%
------- ------- ------- ------- ------- -------
STEP 2
------
This is the second step in elimination described by Robert.
In this case, 81.47% of the time, 6 or 7 numbers are
correctly chosen. On average, however, 5 numbers were
eliminated, or 44 numbers chosen from the field of 49. The
results here, on average, are very similar to the STEP1
results, although there are some instances where the field
of numbers selected are lower:
Select7 Select6 Select5 Select4 Select3 Select 2
------- ------- ------- ------- ------- -------
30 0 0 0 0 0 0
31 0 0 0 0 0 0
32 0 0 0 0 0 0
33 0 0 0 0 0 0
34 0 0 0 0 0 0
35 0 0 0 0 0 0
36 0 0 0 0 0 0
37 0 3 1 1 0 0
38 0 3 3 1 0 0
39 5 8 9 2 1 0
40 23 42 23 8 0 0
41 38 80 30 14 3 0
42 78 95 57 15 4 0
43 128 128 68 12 1 0
44 178 159 49 7 0 0
45 156 128 21 4 0 0
46 110 47 5 1 0 0
47 48 14 0 0 0 0
48 20 2 0 0 0 0
49 2 0 0 0 0 0
------- ------- ------- ------- ------- -------
# draws 786 709 266 65 9 0
42.83% 38.64% 14.50% 3.54% 0.49% 0.00%
------- ------- ------- ------- ------- -------
STEP 3
------
This is the third step in elimination by Robert. In this
case, in 56.57% of the draws, 6 or 7 numbers were
successfully selected. Here the average field of numbers is
reduced downward to 40 (calculated = 39.74) for an average
elimination of 9 numbers:
Select7 Select6 Select5 Select4 Select3 Select2
------- ------- ------- ------- ------- -------
30 0 0 0 0 0 0
31 0 0 1 0 0 0
32 0 1 1 1 2 0
33 0 3 2 2 1 0
34 0 4 8 5 2 1
35 1 18 24 8 3 0
36 18 35 36 25 13 2
37 24 54 56 41 3 1
38 32 103 86 45 10 2
39 62 109 94 34 10 1
40 73 116 104 26 5 0
41 53 112 58 16 1 0
42 48 75 42 3 1 0
43 24 31 15 3 1 0
44 19 10 2 0 0 0
45 7 3 0 0 0 0
46 1 1 0 0 0 0
47 1 0 0 0 0 0
48 0 0 0 0 0 0
49 0 0 0 0 0 0
------- ------- ------- ------- ------- -------
# draws 363 675 529 209 52 7
19.78% 36.78% 28.83% 11.39% 2.83% 0.38%
------- ------- ------- ------- ------- -------
--
Royce Penny
Royce Penny's Money Machine
http://www.geocities.com/lottoking.geo
----------------------------------------
Robert Perkis
the people who were all over me for suggesting this
method. I guess we have to live with them because we
don't have a moderated group with a killfile, but we
are driving with the parking brake on. Robert
Martin Harrison
> I have now completed my analysis of the Robert Perkis Method
> previously posted to this group. (The method is listed in
> Robert's post at the bottom of this post.)
>
> The data below applies to the Canada Lotto 649 for a draw
> history of 1835 draws to October 15, 2001 and includes the
> bonus number for a total of 7 numbers drawn.
>
> The third chart shows that by using all 3 steps of
> elimination, an average of 9 numbers can be eliminated with
> a 56.57% success ratio. In other words, for the Canada
> Lotto 649 history file, you can eliminate 9 numbers, and be
> correct in selecting 6 or 7 numbers over 50% of the time.
> The maths indicate that when 9 numbers are eliminated (40
> selected), you should correctly select 6 numbers only 27.45%
> of the time. Thus, this method (STEP 3) appears to be
> nearly twice as effective as random expectation when applied
> to the Canada Lotto649 history file.
There's probably a problem with your maths here, Royce.
Playing 40 numbers, I would expect to get 6 or 7 correct about 62% of
the time:
(C(40,7) + C(9,1)*C(40,6))/C(49,7) = 0.6192
So, at 57%, the RP system comes out slightly less effective than
random expectation, rather than nearly twice as effective. Alas, the
latter would have been more interesting...
Mart
<remainder of post snipped>
Royce Penny
Martin Harrison wrote:
>
(...see previous post for snip...)
>
> There's probably a problem with your maths here, Royce.
>
> Playing 40 numbers, I would expect to get 6 or 7 correct about 62% of
> the time:
>
> (C(40,7) + C(9,1)*C(40,6))/C(49,7) = 0.6192
>
> So, at 57%, the RP system comes out slightly less effective than
> random expectation, rather than nearly twice as effective. Alas, the
> latter would have been more interesting...
>
> Mart
---------------------
Mart - I do not know where your maths come from...
If I eliminate 9 numbers, and select 6, then I have only
C(40,6) combinations to choose from. C(40,6) = 3,838,380
available combinations of the C(49,6), or 13,987,816
available. The fact is, that I have eliminated 10,145,436
of the combinations and am right over 50% of the time...
The fact that for 19.78% of the draws 7 numbers are
eliminated just enhances the results, since on these draws,
I have 7 chances to win the bonus and jackpot draw, rather
than one chance of either. Overall, I am still eliminating
10,145,436 combinations, and having a success ration
exceeding 50%.
Please explain where your maths come from.
Martin Harrison
Hi Royce.
I may have misunderstood what you were saying.
I interpreted what you said to mean that 57% of the time, either all 7
numbers drawn, or 6 of the 7 drawn, were among the numbers you hadn't
eliminated. I calculated the expectation for this (62%).
If this isn't what you meant, you'll need to clarify - sorry.
You may have seen my posts containing an analysis of applying RP's
system to the UK 6/49 past history. The results - at or about
expectation - were nowhere near as encouraging as you are claiming for
the Canada 6/49.
Would it be possible for you to let me have a copy of the draw history
you're working with? (Your web site only has the past 400 draws).
Private email would be fine. I'd really like to have a go at these
myself. If the system is beating expectation as you claim, then this
is potentially very interesting.
Mart
Franc Jose
your analysis only took in attention 6 or 7 hits in a pool of 40 numbers?
What about the other hits? 5in40, 4in40, 3in40,...0in40? Is that 5in40 has a
frequency more than expected?
It's important too, because you can have a system better then expected
relating to a XinY hits but worse in others hits.
Frank
"Royce Penny" <lottokin...@hotmail.com> wrote in message
news:3BCC3A80...@hotmail.com...
Royce Penny
6 as per Roberts post, I have recalculated the results for
STEP3, and now agree that the method meets random
expectation as noted below.
From the data for STEP3 from my previous post:
Total wins(6 or 5+) = 363*7+675 = 3216
Maximum wins = 1835*7 = 12,845
% wins(6 or 5+) = 3216/12,845 = 25.04%
Random expectation is C(40,6)/C(49,6) = 27.45% with 40
numbers
Random expectation is C(39,6)/C(49,6) = 23.33% with 39
numbers
(Actual average numbers = 39.74)
Thus Robert's method meets random expectation.
It should be pointed out that Robert's method *does* produce
a 6 or 5+bonus win in 56.57% of the actual draws, or, at
least once every two draws from the Canada Lotto649 history
file.
Also, if you wish, you can download a copy of the Canada
Lotto649 history file from my website as an unrelated link
at http://www.geocities.com/wallstreet/7746/can649.zip . It
is a space delimited ASCII text file that can be imported to
any spreadsheet or database file.
--
Royce Penny
Royce Penny's Money Machine
http://www.geocities.com/lottoking.geo
-----------------------------------
Royce Penny
>
> Royce,
>
> your analysis only took in attention 6 or 7 hits in a pool of 40 numbers?
>
> What about the other hits? 5in40, 4in40, 3in40,...0in40? Is that 5in40 has a
> frequency more than expected?
>
> It's important too, because you can have a system better then expected
> relating to a XinY hits but worse in others hits.
>
> Frank
Hi Frank - I did not calculate these values, but one can
readily do the calculation for each line from the data I
presented.
Also, see my other post on this thread that recalculated
STEP3 as meeting random expectation.
Franc Jose
"Royce Penny" <lottokin...@hotmail.com> wrote in message
news:3BCD883...@hotmail.com...
> Hi Frank - I did not calculate these values, but one can
> readily do the calculation for each line from the data I
> presented.
>
OK! I'll find.
> Also, see my other post on this thread that recalculated
> STEP3 as meeting random expectation.
> --
> Royce Penny
... about 6in40. But, if the frequency of 5in40 hits, and of 4in40 hits are
better than expected, may be RP system is worth to play although not, when
thinking about 6in40 hits.
Some time ago, Duncan and Martin told me (here) how to test the 7
frequencies (0inX to 6inX hits) with Chi-Squared test. I think you can do
the same (if you want) to really test this system.
Frank
Martin Harrison
Shame. For a moment there I thought you were onto something!
Thanks for the link to the history file. I'll run it through my
program when I get time.
Mart
Royce Penny <lottokin...@hotmail.com> wrote in message news:<3BCD86F2...@hotmail.com>...