Off-Shelf Software to Analyze Lottery Drawings
Ion Saliu
Somebody started a thread on using SQL or Access to analyze the
lottery. SQL is a database programming language. I pointed out that SQL
is not very capable of analyzing the lottery, since it does not have
built-in statistical functions to satisfy such a requirement. I
recommended instead the use of Excel, the spreadsheet that is bundled
with Access in Microsoft Office Suite. First of all, I am not a
software salesman. I only posted in the thread because I assumed the
initial writer already had Excel on his PC.
I showed how to use an Excel add-in: “Data Analysis”. Its “Summary
Statistics” generates very useful reports regarding lottery drawings
files. You can read my post again. I need to mention now that
my “findings” are, in fact, a consequence of the so-called “normal
probability rule”. The rule is in turn derived fro the “Gaussian
Distribution”. The normal probability rule establishes the probability
for an event in correlation with “the most probable result” in a
probability distribution. A file with lottery drawings represents a
probability distribution. The “median” is one of the most probable
results in drawings series.
Let’s take as an example the median of the first numbers in the
drawings. In the case of my lotto-5 drawings file, the median of the
first number is 5. The standard deviation was 6. The rule of normal
probability states that the probability of any first number in the
drawings is 95.4% less than 2 standard deviations away from the median,
and 99.7% less than 3 standard deviations away. Trying to make it more
clearly. My lotto-5 drawings file has over 1000 drawings. Over 97% of
the numbers drawn in the first position are between 1 and 18 (which
approximates 3 standard deviations from the median). Over 98% of the
numbers drawn in the 5th position are between 21 and 39 (again,
approximately within 3 standard deviations). This is valid for one
number position at a time. If I want the simultaneous probability of
the 1st and 5th numbers, it is .97 x .98 = .95. There is a 95%
probability that the 1st AND the 5th numbers will be between 1-18 AND
21-39. The simultaneous probability for ALL five numbers was around 60
% for my lotto-5 file. In over 60% of the drawings, the winning numbers
were as follows:
1st number: between 1 – 18
2nd number: between 3 – 21
3rd number: between 11 – 29
4th number: between 1 9 – 37
5th number: between 21– 39
Total combinations of 18 taken 5 at a time is: 8,568.
My lotto-5 game uses 39 balls. Therefore, total combinations of 39
taken 5 at a time is: 575,757.
Now, important to note that Excel does not have a function to generate
lottery numbers. The user needs to program such functions in Visual
Basic for Applications.
Or, do what I do: write specialized lottery software outside Excel or
Microsoft Office, using general-purpose programming languages.
But if you do not program at all, it is obvious that it is far better
to select numbers from the ranges determined by the normal probability
rule. If you are to choose 5 numbers from the most frequent 5 ranges,
the winning probability is 1 in 8,568. Choosing 5 numbers from a field
of 39 has a winning probability of 1 in 575,757. In 60% of the cases,
you will face odds of only 8,568 to 1 if you choose numbers from the
most frequent ranges.
Finally, the facts I have presented here are meaningful only if you
already have Excel on your PC.
The code I use in my software is entirely my business and I do not make
it available. My freeware software at http://saliu.com/LottoWin.htm
does not include the probability feature presented here. I do not
accept analyzing your lottery drawings files for any fees.
Keep in mind, however, that everything I have presented here represents
undeniable mathematics. You may hear some nay-sayers screaming foul
play. Your best reaction is to verify the facts yourself. Also, always
question the motivations of any opinions. Here are my motivations.
Number one, I am a soldier of the truth. I am searching for the truth,
no matter how hard it is to reach it. Second, I am trying to introduce
my software and my Web site to a larger audience. This second
motivation seeks both the financial profit and serving the public.
Ion Saliu
Sent via Deja.com http://www.deja.com/
Before you buy.
John Griffin
>you will face odds of only 8,568 to 1 if you choose numbers from the
If that's an advertisement, it's a lie - otherwise it's just
an ignorant assertion.
Any set of numbers you pick is exactly as likely to be
drawn as any other.
CDEX
No offense is intended, so please don't assume one. It's just a statement
of odds from choosing numbers from any subset.
The "subset" argument was put to rest in RGL about a year ago. It should
still be rattling around in Deja.com.
It had been pretty much worked over previously by Dick Adams, John Griffin,
and Mike Sharkey a couple of years earlier. I raised the discussion again
last year. Several contributors took the newsgroup through the analysis of
subsets. In particular, Mario showed the probabilities in detail.
The odds of winning from a combination in any subset are exactly balanced by
the odds of guessing that subset out of the general pool. It does not
matter how one divides the pool into various categories of subsets. The
player's actual odds for winning remain the same.
For example, one could separate from the pool all combinations of a given
Sum, or Even/Odd split, or Consecutives, or Repeats, or any mixture of these
attributes, or others. That includes variances from a lottery model or any
other statistics one extracts from the history. If one 'knows' in advance
of the draw that the winning combination will be from that subset, one's
chance of winning is of course enhanced by playing only the combinations in
that subset. But no one 'knows' that in advance. If one does not 'know'
that information in advance, and plays all of the combinations from that
selected subset, one's chance of winning is equal to the ratio of that
subset to the pool.
Best regards,
Joe Roberts
CDEX
Ion Saliu
To John Griffin and Joe Roberts:
Both of you are wrong. John Griffin goes even beyond being wrong. By
his nature probably he gets mad at any person who expresses something.
You need a whole lot more study in mathematics and statistics. I am not
here for the purpose of teaching such studies. The subsets I presented
in my post are THE most representative samples in the series. They
always are. Their probability is at least 60% and one can increase it
by increasing the ranges (for example standard deviation+1 or +2). If
you apply the gambling formula you’ll discover that an event of 60%
probability only rarely skips 2 or 3 drawings in a row. You see two
skips of the subset, it is a good chance the subset will hit the next
drawing.
But if you are not too much into theory, why don’t you just check some
real lottery drawings? Do you know the results of other readers of this
forum? Not everybody is versed in mathematics. But everybody has the
intelligence to check the facts. Get your lotto-5, or lotto-6, or lotto-
7 ASCII drawings files and analyze them in Excel. Then try to be
truthful and publish your results in this thread. Fair enough?
Ion Saliu
Ion Saliu
--
••• I've had enough! If it doesn't have a formula, it is garbage.
http://www.saliu.com/
Verification
These are the latest results from Pennsylvania lotto-5 game (Cash 5):
1) Saturday 11/20/1999: 8, 11, 14, 35, 39
2) Friday 11/19/1999: 15, 17, 21, 23, 39
3) Thursday 11/18/1999: 1, 6, 11, 14, 35
4) Wednesday 11/17/1999: 4, 5, 9, 19, 29
5) Tuesday 11/16/1999: 1, 10, 22, 31, 39
6) Monday 11/15/1999: 10, 17, 27, 34, 37
7) Sunday 11/14/1999: 10, 13, 19, 27, 34
Drawings #1 and #2 have 5 winning numbers each from the subset
generated by the normal probability rule. Drawing #3 has 4 winning
numbers; drawing #4 has 4 winners; drawings #5 and #6 have 5 winners
each; drawing #7 has 5 winners from the most representative sample.
How about that, Bobby and Johnny? What is your real motivation of
contesting the facts? Did you already know about them and don't want
other people use the knowledge? Or you are plain ignorants in this
field, while pretending you are experts?
royce penny
>
(...snip for brevity only...please see
previous post...)
>
> Now, important to note that Excel does not have a function to generate
> lottery numbers. The user needs to program such functions in Visual
> Basic for Applications.
>
> Ion Saliu
>
------------------
Ion, just to clarify the statement above, Excel does have a
function to generate lottery numbers.
It is the RAND() function. Thus,
To generate a random real number between a and b, use:
RAND()*(b-a)+a
For a lotto 649 use the function "=RAND()*49+1" in six
adjacent cells.
It is important to note that the cells used should be set to
the number format with decimal set to 0 - otherwise you will
get a result that it not an integer.
Pressing the F9 key will generate a new set of six numbers
each time between 1 and 49.
If you have the Analysis ToolPak also installed, you can use
the RANDBETWEEN function which returns a random number
between the numbers you specify. A new random number is
returned every time the worksheet is calculated.
The syntax for this function is RANDBETWEEN(bottom,top),
where bottom is the smallest integer RANDBETWEEN will
return, and top is the largest integer RANDBETWEEN will
return.
--
Royce Penny
Royce Penny's Money Machine
http://www.geocities.com/wallstreet/7746
royce penny
> lottery numbers. The user needs to program such functions in Visual
> Basic for Applications.
> Microsoft Office, using general-purpose programming languages.
> But if you do not program at all, it is obvious that it is far better
> to select numbers from the ranges determined by the normal probability
> rule. If you are to choose 5 numbers from the most frequent 5 ranges,
> the winning probability is 1 in 8,568. Choosing 5 numbers from a field
> of 39 has a winning probability of 1 in 575,757. In 60% of the cases,
(...snip for brevity...)
>
> Ion Saliu
>
-------------------
Ion - I do not understand your logic above when you state
that 60% of the time the chance of a win is 1 in 8568.
For example, take only a sample subset of the ranges you
mention above, where the first number is between 1 and 9,
the second between 10 and 19, the third between 20 and 25,
the fourth between 26 and 32, and the fifth between 33 and
39.
All these combinations meet your criteria, and this small
subset of your criteria has 9*10*6*7*7 = 26,460 combinations
of the 575,757 available. This exceeds the 8,568 that you
calculate, and does not include all the possibilities.
For any lotto 5/39, for a subset to appear 60% of the time,
then it would require 60% (give or take a little) of the
combinations that are available.
--
Royce Penny
Royce Penny's Money Machine
http://www.geocities.com/wallstreet/7746
.
CDEX
> How about that, Bobby and Johnny? What is your real motivation of
> contesting the facts? Did you already know about them and don't want
> other people use the knowledge? Or you are plain ignorants in this
> field, while pretending you are experts?
> Ion Saliu
Please don't get personal or insulting. Please, just the numbers only.
They are universal.
> The subsets I presented in my post are THE most representative
> samples in the series. They always are. Their probability is at least
> 60% and one can increase it by increasing the ranges (for example
> standard deviation+1 or +2). If you apply the gambling formula you’ll
> discover that an event of 60% probability only rarely skips 2 or 3
> drawings in a row.
Yes. But you play 60% of the combinations for a win 60% of the time.
The extension of the subset logic as in:
> Their probability is at least
> 60% and one can increase it by increasing the ranges
... is that the probability of winning is directly proportional to the size
of the subset. And it is that.
I have a 100 percent chance of winning if I increase the size of the subset
to 100 percent of the nCr for the game. I have a 0 (zero) percent chance of
winning if I decrease the size of the subset to zero, i.e. if I do not play
anything at all.
And my chance of winning for any subset between 0...100 percent of nCr is
exactly in proportion to the size of that subset. No more, no less.
It does not matter by what method that subset was gained: by min-max
ranges, medians, and variances of each position, or any other means.
Size matters, as well as how one uses it.
Please, let's stay civil and go with numbers.
Many thanks,
Joe Roberts
CDEX
Robert Perkis
subset to work with for different lottery methods,
whether that preferred subset be 12 numbers or
42 of 49 or even all 49. You see if the method
captures numbers within the subset often enough
to be useful and adjust the size accordingly.
Agreed everything rides the see-saw overall, it is
sometimes better to accept a known loss in return
for an expected advantage within a series of draws.
Too bad such advantages are not overwhelming
when they occur. Robert Perkis
John Griffin
>To John Griffin and Joe Roberts:
>Both of you are wrong. John Griffin goes even beyond being wrong. By
>his nature probably he gets mad at any person who expresses something.
>You need a whole lot more study in mathematics and statistics. I am not
>here for the purpose of teaching such studies. The subsets I presented
>in my post are THE most representative samples in the series. They
>always are. Their probability is at least 60% and one can increase it
>by increasing the ranges (for example standard deviation+1 or +2). If
>you apply the gambling formula you’ll discover that an event of 60%
>probability only rarely skips 2 or 3 drawings in a row. You see two
>skips of the subset, it is a good chance the subset will hit the next
>drawing.
>But if you are not too much into theory, why don’t you just check some
>real lottery drawings? Do you know the results of other readers of this
>forum? Not everybody is versed in mathematics. But everybody has the
>intelligence to check the facts. Get your lotto-5, or lotto-6, or lotto-
>7 ASCII drawings files and analyze them in Excel. Then try to be
>truthful and publish your results in this thread. Fair enough?
>Ion Saliu
I'll leave the fruitless but entertaining exercises to you. The "results"
you refer to have **absolutely nothing** to do with what numbers will be
drawn next. There is no cause and there is no effect between any pair or
even among all pairs of draws. Maybe it will help you if you consider the
fact that it's possible that the entire universe of draws could exactly
fit everything you say without you ever winning as much as a free ticket.
No set of numbers is more likely to be drawn than any other.
If you think you have provided any information that shows
otherwise, take up a different field of study, because you
have no hope of understanding this one.
Your error is so fundamental that I think it might be useless to try to
explain it to you. In general, if your numbers appear to offer better
odds given some particular circumstance, and you ignore (or fail to
understand) that the probability of that circumstance is a factor in the
probability of your numbers popping up, you're on the wrong track (if
you're on any track).
Bottom line: You spend hours playing with a lotto history, and then
buy a set of numbers that you've selected thusly. I buy a quick-pick.
We have something in common: precisely the same chance of winning.
I hope you do win a few million bucks - but if you do, it will be dumb
luck.
royce penny
>
> Royce Penny wrote:
> >
> > All these combinations meet your criteria, and this small
> > subset of your criteria has 9*10*6*7*7 = 26,460 combinations
> > of the 575,757 available. This exceeds the 8,568 that you
> > calculate, and does not include all the possibilities.
> >
> ...AND divided by 1x2x3x4x5.
> Only in the digit games your calculation is valid. The drawing can have
> the first digit equal with the 2nd digit and the 3rd digit less than
> the first two, etc.
>
> Royce Penny wrote:
> >
> > For any lotto 5/39, for a subset to appear 60% of the time,
> > then it would require 60% (give or take a little) of the
> > combinations that are available.
> > --
>
> The calculations for combinations do not follow that logic. For
> example "24 taken 6 at a time" does not represent 50% of "48 taken 6 at
> a time". It is 134,596 in the first case and 12,271,512 in the latter.
> The ratio is just 1.1%.
> In the case I presented, 18 taken 5 at a time is 8,568. 39 taken 5 at a
> time is 575,757. The ratio is 1.5%. Yet, the subset had 5 out of 7
> winning drawings (5 winning numbers)! See below.
>
--------------------
Ion - I am *trying* to follow your logic - but am having
some difficulty. Where I get myself lost is where you get
the 18 number subset from the 39 to make up to
C(18,5)=8,568. I still believe that you have much more than
the 8,568 combinations because I see that you have more than
18 numbers in play for each draw.
Please explain.
Thanks,
Ion Saliu
In over 60% of the drawings, the winning numbers
> > were as follows:
> > 1st number: between 1 – 18
> > 2nd number: between 3 – 21
> > 3rd number: between 11 – 29
> > 4th number: between 1 9 – 37
> > 5th number: between 21– 39
> > Total combinations of 18 taken 5 at a time is: 8,568.
> > My lotto-5 game uses 39 balls. Therefore, total combinations of 39
> > taken 5 at a time is: 575,757.
> All these combinations meet your criteria, and this small
> subset of your criteria has 9*10*6*7*7 = 26,460 combinations
> of the 575,757 available. This exceeds the 8,568 that you
> calculate, and does not include all the possibilities.
>
Ion Saliu's response:
...AND divided by 1x2x3x4x5.
Only in the digit games your calculation is valid. The drawing can have
the first digit equal with the 2nd digit and the 3rd digit less than
the first two, etc.
> For any lotto 5/39, for a subset to appear 60% of the time,
> then it would require 60% (give or take a little) of the
> combinations that are available.
> --
The calculations for combinations do not follow that logic. For
example "24 taken 6 at a time" does not represent 50% of "48 taken 6 at
a time". It is 134,596 in the first case and 12,271,512 in the latter.
The ratio is just 1.1%.
In the case I presented, 18 taken 5 at a time is 8,568. 39 taken 5 at a
time is 575,757. The ratio is 1.5%. Yet, the subset had 5 out of 7
winning drawings (5 winning numbers)! See below.
By the way, thank you for your Excel tip. I had only known of the RAND
function, but I still don't find it usable for lottery. The second
function you mention seems to be more capable. As a matter of fact, I
always write my own computer programs to deal with lottery or gambling.
I always set filters to eliminate combinations. That would be very hard
to implement in Excel. Besides, I want stand-alone applications, which
perform far better.
As of the frequency of the most representative sample in my lotto-5
drawings, please read my previous posting. The method hit 5 times in
the past week. In the two losing drawings, the method had four winners
in each one. Today, Nov.21, 1999 (7PM, after my post), the method had
five winning numbers again: 1, 10, 19, 22, 37. One download link for PA
lottery is: http://www.palottery.com/pastres.cfm
The method is even more effective when used in conjunction with the
gambling formula.
Ion Saliu
> Royce Penny
> Royce Penny's Money Machine
> http://www.geocities.com/wallstreet/7746
>
> .
>
--
••• I've had enough! If it doesn't have a formula, it is garbage, or bul
http://www.saliu.com/
Nick Koutras
1 NB from 1-18
1 NB from 3-21
1 NB from 11-29
1 NB from 19-37
1 NB from 21-39
Sets satisfying the hypothesis 507,762
Sets NOT satisfying hypothesis 67,995
-------------------------------------------
Total sets 575,757
I hope this helps,
Nick Koutras
royce penny wrote:
> Ion Saliu wrote:
> >
> > Royce Penny wrote:
> > >
> > > All these combinations meet your criteria, and this small
> > > subset of your criteria has 9*10*6*7*7 = 26,460 combinations
> > > of the 575,757 available. This exceeds the 8,568 that you
> > > calculate, and does not include all the possibilities.
> > >
> > Ion Saliu's response:
> > ...AND divided by 1x2x3x4x5.
> > Only in the digit games your calculation is valid. The drawing can have
> > the first digit equal with the 2nd digit and the 3rd digit less than
> > the first two, etc.
> >
> > Royce Penny wrote:
> > >
> > > For any lotto 5/39, for a subset to appear 60% of the time,
> > > then it would require 60% (give or take a little) of the
> > > combinations that are available.
> > > --
> > Ion Saliu's response:
> >
> > The calculations for combinations do not follow that logic. For
> > example "24 taken 6 at a time" does not represent 50% of "48 taken 6 at
> > a time". It is 134,596 in the first case and 12,271,512 in the latter.
> > The ratio is just 1.1%.
> > In the case I presented, 18 taken 5 at a time is 8,568. 39 taken 5 at a
> > time is 575,757. The ratio is 1.5%. Yet, the subset had 5 out of 7
> > winning drawings (5 winning numbers)! See below.
> >
> --------------------
> Ion - I am *trying* to follow your logic - but am having
> some difficulty. Where I get myself lost is where you get
> the 18 number subset from the 39 to make up to
> C(18,5)=8,568. I still believe that you have much more than
> the 8,568 combinations because I see that you have more than
> 18 numbers in play for each draw.
>
> Please explain.
>
> Thanks,
> --
Sean B
---> snip
>
> The simultaneous probability for ALL five numbers was around 60
>% for my lotto-5 file. In over 60% of the drawings, the winning numbers
>were as follows:
>1st number: between 1 – 18
>2nd number: between 3 – 21
>3rd number: between 11 – 29
>4th number: between 1 9 – 37
>5th number: between 21– 39
>Total combinations of 18 taken 5 at a time is: 8,568.
>My lotto-5 game uses 39 balls. Therefore, total combinations of 39
>taken 5 at a time is: 575,757.
>-
->
Interesting post Ion. We did cover a thread over a year ago
on the probability of each ball. Its a good strategy to play
say 4x to 49 knowing that it will win on 50% of draws, even
perm all the numbers from say 42-49.
I actually calculated the probability of each ball and plotted
graphs in Quattro Pro. The 1st and last ball graphs are a
different shape peaking at 1 and 49.
Playing the expected range for each ball provides an almost
continuous cover of all 49 numbers, from memory 1-15, 16-33 etc.
The balls do not fall within these ranges that often, but it is
a strategy and is as good as any other.
I gave up on this because of the overlapping probabilities.
If you put on a spreadsheet the probability for each ball at
each location 1 to 49 and add them up you get the expected
probability of 1.00000000 for each number 1 to 49. In other
words each number has the same probability of being drawn.
So the strategy is as good as playing say 3 odd + 3 even.
Keep up the good work, I'm sure you will find something
to tip the scales in your advantage eventually.
Sean B
royce penny
>
> I used LotWin32 to calculate the quantity of sets satisfying the hypothesis:
> 1 NB from 1-18
> 1 NB from 3-21
> 1 NB from 11-29
> 1 NB from 19-37
> 1 NB from 21-39
> Sets satisfying the hypothesis 507,762
> Sets NOT satisfying hypothesis 67,995
> -------------------------------------------
> Total sets 575,757
>
> I hope this helps,
>
> Nick Koutras
>
Thanks Nick, I was just going to do the same iteration to
see what the exact number of combinations are that meet
Ion's criteria.
From your calculations, Ion's results should show that he
meets the criteria 88% of the time, since 88% of the
combinations for the 5/39 are within the five number ranges
that he specifies.
A reverse hypothesis using the 67,995 sets that do *not*
meet the criteria might be a thought to pursue, since they
should show once every 8 or so draws, and probably have an
equal chance as any other sets at the lower tier payouts.
Ion Saliu
--
••• http://www.saliu.com/
Ion Saliu's response:
Yes, I realize I miscalculated. Actually, I used my software to
generate the combinations. The software already has built-in filters (I
presented them with my optimized number generator at
http://saliu.com/LottoWin.htm). Using the built-in filters and other
filters I got under 8,000 combinations. It was very close to C(18,5).
Indeed, without filters, there are more combinations in the most
representative sample.
However, there are NOT NEARLY as many as your calculations indicate.
Keep in mind that many numbers are OVERLAPPED. Therefore a combination
with the first two numbers 5, 16 ... is valid; but 16, 5 ... is not
valid, since the strategy deals with drawings in ASCENDING ORDER only.
Combinations such as 6, 6 ... are not valid, either. I will come back
with more on this...
Ion Saliu
--
••• http://www.saliu.com/
Sean B Saint
Here's the number of times that each ball appears in out of
all 13,983,816 draws in each position 1 to 49.
If you add each column up it comes to 13,983,816
Added horizontally each line comes to 1,712,304
so all numbers 1-49 stand an equal chance.
The graphs for 1 and 6 peak at 1 and 49.
The graphs for 2,3,4 and 5 are a skewed bell.
When converted to percentages of 13,983,816 its
possible to determine the percentage expectation
for a ball to fall within a certain range.
i.e. the 1st ball will be 1,2,3,4 or 5 on 49.519956%
of draws. Other balls could also be in this range.
A good strategy would be to leave out 1-5 and play
6-49 in a 44 Ball Lottery on about 50% of draws.
Hope this helps.
Sean B
=================================================
Import into spreadsheet and plot graphs.
"All Balls 1-6 6/49 Lottery"
1 ,2 3, 4, 5, 6
1, 1712304, 0 , 0 , 0 , 0 , 0
2, 1533939, 178365 , 0 , 0 , 0 , 0
3, 1370754, 326370 , 15180 , 0 , 0 , 0
4, 1221759, 446985 , 42570 , 990 , 0 , 0
5, 1086008, 543004 , 79464 , 3784 , 44 , 0
6, 962598 , 617050 , 123410 , 9030 , 215 , 1
7, 850668 , 671580 , 172200 , 17220 , 630 , 6
8, 749398 , 708890 , 223860 , 28700 , 1435 , 21
9, 658008 , 731120 , 276640 , 43680 , 2800 , 56
10, 575757 , 740259 , 329004 , 62244 , 4914 , 126
11, 501942 , 738150 , 379620 , 84360 , 7980 , 252
12, 435897 , 726495 , 427350 , 109890, 12210 , 462
13, 376992 , 706860 , 471240 , 138600, 17820 , 792
14, 324632 , 680680 , 510510 , 170170, 25025 , 1287
15, 278256 , 649264 , 544544 , 204204, 34034 , 2002
16, 237336 , 613800 , 572880 , 240240, 45045 , 3003
17, 201376 , 575360 , 595200 , 277760, 58240 , 4368
18, 169911 , 534905 , 611320 , 316200, 73780 , 6188
19, 142506 , 493290 , 621180 , 354960, 91800 , 8568
20, 118755 , 451269 , 624834 , 393414, 112404 , 11628
21, 98280 , 409500 , 622440 , 430920, 135660 , 15504
22, 80730 , 368550 , 614250 , 466830, 161595 , 20349
23, 65780 , 328900 , 600600 , 500500, 190190 , 26334
24, 53130 , 290950 , 581900 , 531300, 221375 , 33649
25, 42504 , 255024 , 558624 , 558624, 255024 , 42504
26, 33649 , 221375 , 531300 , 581900, 290950 , 53130
27, 26334 , 190190 , 500500 , 600600, 328900 , 65780
28, 20349 , 161595 , 466830 , 614250, 368550 , 80730
29, 15504 , 135660 , 430920 , 622440, 409500 , 98280
30, 11628 , 112404 , 393414 , 624834, 451269 , 118755
31, 8568 , 91800 , 354960 , 621180, 493290 , 142506
32, 6188 , 73780 , 316200 , 611320, 534905 , 169911
33, 4368 , 58240 , 277760 , 595200, 575360 , 201376
34, 3003 , 45045 , 240240 , 572880, 613800 , 237336
35, 2002 , 34034 , 204204 , 544544, 649264 , 278256
36, 1287 , 25025 , 170170 , 510510, 680680 , 324632
37, 792 , 17820 , 138600 , 471240, 706860 , 376992
38, 462 , 12210 , 109890 , 427350, 726495 , 435897
39, 252 , 7980 , 84360 , 379620, 738150 , 501942
40, 126 , 4914 , 62244 , 329004, 740259 , 575757
41, 56 , 2800 , 43680 , 276640, 731120 , 658008
42, 21 , 1435 , 28700 , 223860, 708890 , 749398
43, 6 , 630 , 17220 , 172200, 671580 , 850668
44, 1 , 215 , 9030 , 123410, 617050 , 962598
45, 0 , 44 , 3784 , 79464 , 543004 , 1086008
46, 0 , 0 , 990 , 42570 , 446985 , 1221759
47, 0 , 0 , 0 , 15180 , 326370 , 1370754
48, 0 , 0 , 0 , 0 , 178365 , 1533939
49, 0 , 0 , 0 , 0 , 0 , 1712304
Nick Koutras
What you came up to, with your Table below, is the Pascal's Triangle &
the
relation of it with the Binomial Coefficient!
The way you have constructed your Table is of interest!
I will recommend two web pages for anyone how interests to read more
about it
http://forum.swarthmore.edu/workshops/usi/pascal/pascal_binomial.html
http://www.treasure-troves.com/math/BinomialCoefficient.html
Below is a table, which you can expand more easily for up to 10 Lotto
and ANY size.
1 2 3 4 5 6 7 8 9 10 <== Lotto set size
1 1
2 2 1
3 3 3 1
4 4 6 4 1
5 5 10 10 5 1
6 6 15 20 15 6 1
7 7 21 35 35 21 7 1
8 8 28 56 70 56 28 8 1
9 9 36 84 126 126 84 36 9 1
10 10 45 120 210 252 210 120 45 10 1
11 11 55 165 330 462 462 330 165 55 11
12 12 66 220 495 792 924 792 495 220 66
13 13 78 286 715 1287 1716 1716 1287 715 286
14 14 91 364 1001 2002 3003 3432 3003 2002 1001
15 15 105 455 1365 3003 5005 6435 6435 5005 3003
16 16, =(15+105)
=120
I hope the above is self explanatory!
Nick Koutras
CDEX
Ion,
I hope we can establish a dialogue on this.
Going with your 11/21 post: <819548$arb$1...@nnrp1.deja.com>...
> The subsets I presented in my post are THE most representative
> samples in the series. They always are. Their probability is at least
> 60% and one can increase it by increasing the ranges (for example
> standard deviation+1 or +2). If you apply the gambling formula you’ll
> discover that an event of 60% probability only rarely skips 2 or 3
> drawings in a row.
That's true. But it does not affect your chances of winning.
Your chance of winning is directly proportional to the size of your subset.
You are playing 60% of the combinations for a winning match 60% of the time.
We can start with the full range of combinations, and we can define some
subset of those combinations. We can define it based on any attributes and
criteria we like: e.g., min-max ranges and variances of numbers in each
digit position. Their distribution is a simple Gaussian curve.
We can produce the curve in two ways: By calculation, or by looking
backward at the game's history. As the game's history progresses, the
historical curve will increasingly resemble the calculated one.
If one looks at the historical curve, and snips off the lower 20% and upper
20% 'tails' of the curve, one sees what has happened in 60% of the draws.
If one does the same with the calculated model, one sees what is likely to
happen in 60% of the future draws, within some stated degree of confidence.
That degree of confidence is based on the *amount* of future draws one
considers. For a long series (3..4 X nCr) it is high. For the *next* draw,
it is (1/nCr).
To put it differently:
Let's consider two players. Two persons divide the distribution of total
combinations between them, such that one person retains 90% of the
combinations and the other retains 10%. In order to have an assured win,
each player must play, and must pay for, all of the combinations contained
in his portion.
... The two players have a cost ratio of 9:1.
Let's also assume that the game's next 10 draws will exactly conform to the
calculated model -- i.e., they will fit the distribution exactly according
to the player's portions.
... The two players have a win ratio of 9:1.
Note: We have assumed a "given" that the game's "next 10 draws" will
conform to the 90%/10% distribution model. That, in itself, is a huge given
for a short series of only 10 draws. But let's accept it for the
discussion.
Let's position ourselves at the start of that series of 10 draws. We have a
"given" that 9 of the draws will conform to the first player's distribution
of plays, and that 1 of those draws will conform to the second player's
distribution of plays.
It is nine times as likely that the "next" draw will match the first
player's distribution, than that it will match the second player's. We have
that assumption for the 10-draw series.
But we do not have it for any individual, isolated, single "next" draw.
Remember our assumption: One draw of the ten will fit the second player's
distribution. That 'one of ten' event could occur on the first draw, or
second draw, or on any draw including the tenth draw. We simply do not know
the distribution at the "next" draw level.
Or in other terms: If we "knew" that the next draw as an individual,
isolated and single event were to fit the second player's distribution, then
our "knowledge" of that "1-draw" event is something unrelated to the
"10-draw series" distribution model we started with.
Difficult language. Let us see if we can pin this down.
If we started with information whose level of precision consisted of an
assumed "1 of 10" distribution -- and nothing more than that information --
then our expectation must be based on that information. Our expectation of
a win must be "1 of 10". We cannot automatically transform it to a "1 of
5", or "1 of 4", or anything else, *unless* we are given new information
at that new level. And we cannot transform it to a "1 of 1" expectation
(meaning, an expectation for the next draw) unless we are given new
information at that level. We do not have that information at the beginning
of the series.
Now let us move ahead 9 draws. Let us assume that the first 9 draws
happened to fit the first player's distribution. The 10th draw (the next
draw) will then conform to the second player's -- **again**, going with
the huge assumption we began with, that the set of 10 draws will exactly
conform to the calculated distribution. That was a wild assumption for 10
draws -- but let's roll along with it.
It's a given now that the second player will win in the next draw, with his
10% outlay of cost for that draw. The first player will lose in the next
draw, with his 90% outlay of cost. In fact, the first player (if he has any
sense) will want to shift his play to that 10% also.
But we acquired that information one draw at a time. We did not know that
information in advance of *any* one of the preceding 9 draws. We knew it
only after those draws had taken place. Again visualizing it at the start,
both of the players would have had to outlay their respective costs for each
of those draws. The net result is exactly the same: (9 X Cost, 9 X Wins),
(1 X Cost, 1 X Win).
Once more, just to avoid misunderstanding: We started with a huge
assumption that we would "know" the outcome of the next 10 draws, as an
aggregate series, and not at any level more detailed than for that aggregate
series. But even that is a huge assumption.
The result is: He who plays a subset of 90% of the total possible
combinations matches 90% of the wins; he who plays a subset 10% of the
total possible combinations matches 10% of the wins.
It does not matter by what attributes those total possible combinations have
been divided into their respective subsets.
Back to the 60% subset based on positional ranges. It's true that
combinations which have 60% probability of occuring will indeed occur in 60%
of an extended series of drawings, within a degree of confidence that is
*solely* based on the length of that series. When you reduce that 'series'
to one draw, the probability for the combination in that draw is 1/nCr.
- - - - -
Please note that there is no 'moral' implication. There's none in a
Gaussian curve. It's just a curve.
If a player wishes to play some subset, he's welcome to do so. There is no
"right" or "wrong" associated with slicing apart the total combinations and
parking them into various categories of trays.
One 'attribute' of combinations that makes a subset is no better or worse
than some other 'attribute' that makes some other subset. For example, a
subset that is comprised of positional number ranges yielding 60%
probability in extended play is no better or worse than a different subset
that is comprised of Sums, Even/Odd, Repeat numbers, Consecutive numbers,
and so on yielding 60% probability. Different reasons for associating
combinations, same probability.
The game is a little like a birthday cake being sliced at a party. No one
can eat the whole cake. There is a diminishing return for doing so, so we
go for a slice. Every slice is a little different. We pick the slice we
like, and we're happy with it. Other folks do their slices with equal
enjoyment.
- - - - -
Hoping this makes sense. I respect your viewpoint, and an extended trial
would be welcome to show any benefit to the player.
CDEX
I am sorry, but I cannot see your numbers.
I'm really struggling to see the 8,568 combinations. Maybe I missed
something, so please explain it.
You wrote in Message-ID:
<816l66$nbj$1...@nnrp1.deja.com> ...
> In over 60% of the drawings, the winning numbers
> were as follows:
>
> 1st number: between 1 - 18
> 2nd number: between 3 - 21
> 3rd number: between 11 - 29
> 4th number: between 19 - 37
> 5th number: between 21 - 39.
>
> Total combinations of 18 taken 5 at a time is: 8,568.
> My lotto-5 game uses 39 balls. Therefore, total
> combinations of 39 taken 5 at a time is: 575,757.
>
... (snip references to Excel and Visual Basic)
>
> But if you do not program at all, it is obvious
> that it is far better to select numbers from the
> ranges determined by the normal probability rule.
> If you are to choose 5 numbers from the most frequent
> 5 ranges, the winning probability is 1 in 8,568.
> Choosing 5 numbers from a field of 39 has a winning
> probability of 1 in 575,757. In 60% of the cases,
> you will face odds of only 8,568 to 1 if you choose
> numbers from the most frequent ranges.
>
- - - - -
I do not see the odds of "8,568 to 1" in "60% of the cases".
That is because I do not see where the 8,568 combinations are made from "18
numbers taken 5 at a time" in the positional limits given.
Are you really saying that 1.5% of the game's combinations win in 60% of the
drawings?
These are the positional limits:
> 1st number: between 1 - 18
> 2nd number: between 3 - 21
> 3rd number: between 11 - 29
> 4th number: between 19 - 37
> 5th number: between 21 - 39.
The results you have posted for the past eight drawings are:
> 11/21/1999: 1, 10, 19, 22, 37
> 11/20/1999: 8, 11, 14, 35, 39
> 11/19/1999: 15, 17, 21, 23, 39
> 11/18/1999: 1, 6, 11, 14, 35
> 11/17/1999: 4, 5, 9, 19, 29
> 11/16/1999: 1, 10, 22, 31, 39
> 11/15/1999: 10, 17, 27, 34, 37
> 11/14/1999: 10, 13, 19, 27, 34
Indeed in these eight drawings, there five drawings with 5-matches. That
is, all five numbers are within the positional limits you gave above.
But in those eight drawings, the player would have needed to play the
following *minimum* ranges of numbers. That is, these are the tightest,
narrowest, smallest ranges. They are based on the narrowest limits, either
of the actual winning numbers in those draws, or of your given ranges
(whichever is narrower):
1st position: 1...15
2nd position: 5...17
3rd position: 11...27
4th position: 19...35
5th position: 29...39.
These ranges are smaller than the limits you gave above. But even with
them, there are still 39 numbers in play.
I just do not see the "18 numbers taken 5 at a time" in the five positional
ranges you gave.
- - - - -
Considering the five positional ranges:
There are 39 numbers in play, separated into minimum-maximum ranges for each
position.
Using those five ranges, the lowest possible combination played will be
1-3-11-19-21. The highest will be 18-21-29-37-39. All of the combinations
below or above this range will be excluded.
Within that range, some of the combinations will be excluded also, if each
number in them does not fit inside that number's specified positional range.
For example, all combinations in which the first number is higher than 18
will be excluded.
This does reduce the amount of combinations. But it does not reduce it down
to the level of "18 numbers taken 5 at a time", or 8,568.
With all 39 numbers restricted to the five ranges you show above, there are
371,036 combinations.
That's 64.44% of the game's total of 575,757 combinations.
Of course that subset of 371,036 combinations will have some winner in about
64% of the draws.
Help me out, please. Where do the 8,568 combinations come from?
CDEX
In your run of the numbers: <3838EF9F...@hotmail.com>...
> I used LotWin32 to calculate the quantity of
> sets satisfying the hypothesis:
> 1 NB from 1-18
> 1 NB from 3-21
> 1 NB from 11-29
> 1 NB from 19-37
> 1 NB from 21-39
> Sets satisfying the hypothesis 507,762
> Sets NOT satisfying hypothesis 67,995
> -------------------------------------------
> Total sets 575,757
... Did that maybe not include the 'positional limits' on the placement of
the numbers?
I ran it with the 39 numbers restricted to their five discrete positional
limits (in LD Free Wheeling) and got: 371,036 sets. That equals 64.44% of
the total sets.
Running it with the five ranges of numbers without regard to their
positional placement does indeed give: 507,762 sets.
Robert Perkis
if so, how are we dealing with column overlapping?
Or are we talking about playing a number from each
group of Ten's? Thanks, Robert Perkis
Ion Saliu wrote:
> In article <3839567A...@yahoo.com>,
> lottoking....@yahoo.com wrote:
> > Nick Koutras wrote:
> > >
> > > I used LotWin32 to calculate the quantity of sets satisfying the
> hypothesis:
> > > 1 NB from 1-18
> > > 1 NB from 3-21
> > > 1 NB from 11-29
> > > 1 NB from 19-37
> > > 1 NB from 21-39
> > > Sets satisfying the hypothesis 507,762
> > > Sets NOT satisfying hypothesis 67,995
> > > -------------------------------------------
> > > Total sets 575,757
> > >
Nick Koutras
Yes I agree with your results.
I have to admit I did not enter the position of each Number on my original Post.
Your result of 371,036 sets is correct!!!
or 0.6444% of total sets are included into these hypotheses.
Regards,
Nick Koutras.
CDEX wrote:
> Nick,
>
> In your run of the numbers: <3838EF9F...@hotmail.com>...
>
> > I used LotWin32 to calculate the quantity of
> > sets satisfying the hypothesis:
> > 1 NB from 1-18
> > 1 NB from 3-21
> > 1 NB from 11-29
> > 1 NB from 19-37
> > 1 NB from 21-39
> > Sets satisfying the hypothesis 507,762
> > Sets NOT satisfying hypothesis 67,995
> > -------------------------------------------
> > Total sets 575,757
>
Sean B Saint
>Hi Sean,
>
>What you came up to, with your Table below, is the Pascal's Triangle &
>the relation of it with the Binomial Coefficient!
Thanks Nik, I didn't know I had Pascal's triangle, maybe
he's got mine ;-)
>The way you have constructed your Table is of interest!
>
See below.
>I will recommend two web pages for anyone how interests to read more
>about it
>http://forum.swarthmore.edu/workshops/usi/pascal/pascal_binomial.html
>http://www.treasure-troves.com/math/BinomialCoefficient.html
I'll have a look its years since I read anything relating
to Pascal.
The method I used was quite simple, just loop through
all 14M combinations and set up an array for each ball
of dimension 49 then as each ball passes through the
loop count the number of times each value 1 to 49
appears for balls 1 to 6. Its easier for a one off to do
just 1 ball at a time, save to disk and then paste as
columns into a spreadsheet or editor.
q = value of 1st ball
Increment FirstBallArray(q)
for v = 1 to 49
Print to file FirstBallArray(v)
etc for each ball
I did this in March 98 to find the expected 50% range for
each ball on a 6/49 it came out as:
Ball 1 : 1-5 inc
Ball 2 : 6-15 inc
Ball 3 : 17-28 inc
Ball 4 : 24-35 inc
Ball 5 : 35-44 inc
Ball 6 : 45-49 inc
As you can see this almost covers all the balls in the
Lottery so it looked fairly good. The bad news was
that when I checked the number of combinations
covered it was only 319,500 which is only just
over 2% of draws, so not really much use.
I did some further tests to find 60% coverage and
came up with:
Ball 1 : 1-14 inc
Ball 2 : 3-25 inc
Ball 3 : 8-34 inc
Ball 4 : 16-42 inc
Ball 5 : 25-47 inc
Ball 6 : 36-49 inc
Covers 60.73% of combinations.
This might be of use to someone as a playing strategy.
No overall increase in chance though.
Cheers
Sean B
Ion Saliu
about! Sorry, it happened at a very busy time for me. I will finish a
computer program that generates all possible combinations within ranges
and without any filters. For the lotto-5 first. I'll post it as a free
download. Again, keep in mind that my strategy works with sorted files
only. The drawings must be sorted first. Then, you check the winners
against sorted drawings. Thus a combination such as 3,17,20,29,38 is
valid. A combination like 17,3,20,29,38 is invalid!
Just give me a couple of days. Thanks!
Ion Saliu
Ion Saliu
ranges. The program uses no filtering at all. Indeed, I was wrong
regarding the total combinations to play! I mean, wrong, although not
quite as wrong as some other calculations indicated. I used first the
following ranges:
1 to 18
3 to 21
11 to 29
19 to 37
21 to 39
The program generated 371,036 combinations. That represents 64.4% of
all possible combinations for a lotto-5-39 game. The positive side is
that the ranges eliminated 204,721 combinations. That’s a very powerful
filter, as far as I am concerned.
I used next the following ranges:
1 to 9
8 to 16
16 to 24
24 to 32
31 to 39
The program generated 53,352 combinations. That represents 9.3% of all
possible combinations for a lotto-5-39 game. You should expect to win 1
every 10+ drawings using those ranges. The best situation to play the
ranges is after 4 to 8 skips (losses). In other words, if the ranges
don’t come up after 4 drawings, play them until they skipped 4 more
drawings.
Evidently, the ranges by themselves are not sufficient. They need to be
used in conjunction with other filters. Using other filters, I obtained
from 0 (zero) to 8,000 combinations to play. Of course, the winning
frequency is lower.
In my book, the best way to play the lotto games is to set filters that
eliminate the largest possible number of combinations. Of course, you
will have the chance to win more rarely, but a jackpot win
compensates.
If you are interested in the program, I uploaded it to my download site
(http://www.bwnow.com/lottowin/nss-folder/publicfolder). The name of
the program is RANGES5.EXE. You need to download also the file
RANGES.5, which consists of the ranges for the 5 numbers. The file is
in ASCII format, so you can use any text editor to change it. Right
now, the RANGES.5 file looks like this:
1 9
8 16
16 24
24 32
31 39