Sums and number boundaries...

thornc

Member
Re: Numbers between 10 and 35

GillesD said:
In the 1992 draws of Canadian Lotto 6/49, only 35 draws (1.76%) had all the 6 winning numbers between 10 and 35 (including those two values).

The distribution of the sum for these 35 draws is:

- 0 with the sum between 81 and 90
- 1 with the sum between 91 and 100
- 3 with the sum between 101 and 110
- 4 with the sum between 111 and 120
- 8 with the sum between 121 and 130
- 7 with the sum between 131 and 140
- 9 with the sum between 141 and 150
- 2 with the sum between 151 and 160
- 1 with the sum between 161 and 170

Just as I thought, thanks Gilles.
The events are independent... But they do seem to work!
 

PAB

Member
Use of Excel

Hi GillesD,

I read your post in regard to you keeping your Lottery Database in Excel. I am interested in the effect of last digits and decades within Lottery draws.
Would it be possible for you to let me have the formulas ( as opposed to the actual data ) for each of the last digit distribution for the following please :-

1-1-1-1-1-1
2-1-1-1-1-0
2-2-1-1-0-0
2-2-2-0-0-0
3-1-1-1-0-0
3-2-1-0-0-0
3-3-0-0-0-0
4-1-1-0-0-0
4-2-0-0-0-0
5-1-0-0-0-0

I would like the formulas for the TOTAL of each distribution as opposed to the actual totals for the draws to date ( adding all the totals would equal 13,983,816 Combinations ).

Would it also be possible to let me have the same information for the distribution of decades.

Thank you very much for your help

Regards
PAB
 

winhunter

Member
hmmm

GillesD,

In retrospect,

I think you should be able to utilize the .dll files from WINHunter inside Excel. Although, you would need to look at alot of the WINHunter code up close to get a better understanding of the architecture, but I believe it could be done.




ThornC,

I want to explore this idea further. As I said a few months back, this is an excellent candidate for a trigger. And since you seem to be experiencing some positive proof of your findings, this would be an excellent way to build files to support your ideas. Then other users could test the same ideas against their histories.



Andrew
 

winhunter

Member
Re: Use of Excel

PAB said:
Would it also be possible to let me have the same information for the distribution of decades.



WINHunter display's the Index value, Sum and Deltas for drawing histories. I suppose I could also add the Decade information. Might not be a bad idea, but WINHunter really does not process that information. Would only appear in the Scan window.



Andrew
 

GillesD

Member
Distribution - Last digit / Decades

To PAB

There might be a formula where you put in the proper information and you get the answer to your question but I am not that good. Maybe somebody (Nick K. ?) has one formula that works in all cases.

The problem with last digit and decades is that probabilities are not equal for in all cases. For example, for last digits, there are only 4 numbers ending in 0 but 5 numbers ending in 1 to 9.

I have all the formulas for each distribution for last digit and decades but I used logic in each case. Let's look at an example and develop the formula: the 2-2-2 distribution for last digit.

There are 2 possibilities:
A - all three last digits are from 1 to 9 (no 0 involved),
B - one last digit is 0 and the other two are from 1 to 9.

Let's look at the first one:
- there are COMBIN(9,3) = 84 combinations with three numbers taken between 1 and 9 (from 1-2-3 to .... 3-5-7 to ... 7-8-9);
- for each last digit, there are COMBIN(5,2) = 10 combinations (for last digit 2, they are from 2-12 to ... 2-42 to ... 32-42);
- for all three last digits are from 1 to 9, the number of combinations is then COMBIN(9,3)*COMBIN(5,2)*COMBIN(5,2)*COMBIN(5,2) or 84,000.

Now for the second possibility:
- there are COMBIN(4,2) = 6 combinations with the number 0 as last digit (10-20, 10-30, 10-40, 20-30, 20-40 and 30-40);
- there are COMBIN(9,2) = 36 combinations with two numbers taken in 1 to 9 (from 1-2 to .... 3-7 to ... 8-9);
- in this case, again, there are COMBIN(5,2) = 10 combinations for each last digit (for 7, they are from 7-17 to ... 7-47 to ... 37-47);
- so when one last digit is 0 and the other two are taken between 1 and 9, the number of combinations is then COMBIN(4,2)*COMBIN(9,2)*COMBIN(5,2)*COMBIN(5,2) or 21,600.

If you add the two values (84,000 and 21,600). you get 105,600 the number of combinations with a 2-2-2-0-0-0 distribution for the last digit (or 0.755% of the 13,983,816 combinations).

Using this method, you can determine the number of combintations for each distribution
 

PAB

Member
Hi GillesD,

Thanks for the reply.
I actually have all the last digits information for every draw, they are automatically calculated whenever I enter new draw numbers.

I am working on the distribution 4-1-1-0-0-0 but without much success at the moment. I just seem to have a mental block trying to work this out.

Anyway, thanks for your time.

Best Regards
PAB
 

doc

Member
Addendum

Just to add to this so far. I will be so bold as to remind everyone that there have been 2011 draws to date in the CN649. Total combos that fit the original parameters defined by thornc are precisely 230,230---or 1.646% of the total available (for any 649). This equates to about 1 in every 60.7 draws (theoretical expected average). Actual results so far (to draw 2011) is 36---with the last being at draw 1997. This is about 1.79% or somewhat slightly better than the expected at this point. I wouldn't be prepared to bet the farm on such an outcome. There are much bigger fish to fry. Research shows that (theoretically) over 86% of all random combinations have a sum between 100-199. I'll take 86 out of a hundred every time.:cool:
 

PAB

Member
Hi GillesD,

I have done some work on distribution of last digits but am unsure if I am on the right lines. There are 3 I cannot work out.
The ones I have done are below.

Distribution 1-1-1-1-1-1
=COMBIN(9,6)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)

=COMBIN(4,1)*COMBIN(9,5)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)

Total = 2,887,500

Distribution 2-1-1-1-1-0
Distribution 2-2-1-1-0-0

Distribution 2-2-2-0-0-0
=COMBIN(9,3)*COMBIN(5,2)*COMBIN(5,2)*COMBIN(5,2)

=COMBIN(4,2)*COMBIN(9,2)*COMBIN(5,2)*COMBIN(5,2)

Total = 6,930,000

Distribution 3-1-1-1-0-0

Distribution 3-2-1-0-0-0
=COMBIN(9,1)*COMBIN(5,3)*COMBIN(8,1)*COMBIN(5,2)*COMBIN(7,1)*COMBIN(5,1)

=COMBIN(4,1)*COMBIN(9,1)*COMBIN(5,3)*COMBIN(8,1)*COMBIN(5,2)

=COMBIN(4,2)*COMBIN(9,1)*COMBIN(5,2)*COMBIN(8,1)*COMBIN(5,1)

=COMBIN(4,3)*COMBIN(9,1)*COMBIN(5,2)*COMBIN(8,1)*COMBIN(5,1)

Total = 924,000

Distribution 3-3-0-0-0-0
=COMBIN(9,2)*COMBIN(5,3)*COMBIN(5,3)

=COMBIN(4,3)*COMBIN(9,1)*COMBIN(5,3)

Total = 3,960

Distribution 4-1-1-0-0-0
=COMBIN(9,2)*COMBIN(5,2)*COMBIN(5,2)*COMBIN(5,2)

=COMBIN(4,4)*COMBIN(9,1)*COMBIN(5,2)*COMBIN(8,1)*COMBIN(5,1)

Total = 39,600

Distribution 4-2-0-0-0-0
=COMBIN(9,1)*COMBIN(5,4)*COMBIN(8,1)*COMBIN(5,2)

=COMBIN(4,4)*COMBIN(9,1)*COMBIN(5,2)

=COMBIN(4,2)*COMBIN(9,1)*COMBIN(5,4)

Total = 3,960

Distribution 5-1-0-0-0-0
=COMBIN(9,2)*COMBIN(5,5)*COMBIN(5,1)

=COMBIN(4,0)*COMBIN(9,2)*COMBIN(5,5)

=COMBIN(4,1)*COMBIN(9,1)*COMBIN(5,1)

Total = 396

I would be grateful if you could let me know if there are any mistakes, and maybe help me with the 3 I am stuck on.
Thanks in advance.
PAB
 

thornc

Member
Re: Addendum

doc said:
Just to add to this so far. I will be so bold as to remind everyone that there have been 2011 draws to date in the CN649. Total combos that fit the original parameters defined by thornc are precisely 230,230---or 1.646% of the total available (for any 649). This equates to about 1 in every 60.7 draws (theoretical expected average). Actual results so far (to draw 2011) is 36---with the last being at draw 1997. This is about 1.79% or somewhat slightly better than the expected at this point. I wouldn't be prepared to bet the farm on such an outcome. There are much bigger fish to fry. Research shows that (theoretically) over 86% of all random combinations have a sum between 100-199. I'll take 86 out of a hundred every time.:cool:

Doc case you hadn't noticed Gilles had already checked that, the two statement I make are exclusive of each other, meaning that they won't happen togheter!
And yes you're right about going for a sum in the 100-190 range, but my observation tends to note that you can use numbers from a 35(39) pool size to make your picks rather than from the full 49; which is a good thing on my book!!
 

PAB

Member
Sums and Number Bounderies

Hi GillesD,

I wonder if you have had time to look at my post for the 14th of May 2003 ( above ).

Thanks in advance.

PAB
 

barge

Member
trigger!

Thanks to Thornc, there was a trigger of sum 68 in the UK649 three draws ago. The highest number in last nights draw was 32!
Unfortunately I only managed to get three of them...............but the trigger worked.

Well done

Barge

:)
 

PAB

Member
Sums and Number Boundaries...

Hi GillesD,

I was clearing out some of my Excel files and came across the unfinished one for "COMBIN" Formulas for Distribution of Last Digits.
I wonder if you have had time to look at my post for the 14th of May 2003 ( above ). I think that I have made it far more complicated than it actually is.

Thanks very much in advance.

PAB
 

PAB

Member
Excel Formulas for Last Digits

Hi GillesD,

Not to worry, I think I have managed to construct the Excel Formulas for Last Digits. For some reason I seemed to have a mental block, a bit like not seeing the wood for trees.
Anyway, I think they are O.K. and I have posted them below :-

Distribution 1-1-1-1-1-1
=COMBIN(9,6)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)+COMBIN(4,1)*COMBIN(9,5)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)

Distribution 2-1-1-1-1-0
=COMBIN(9,1)*COMBIN(5,2)*COMBIN(8,4)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)+COMBIN(4,1)*COMBIN(9,1)*COMBIN(5,2)*COMBIN(8,3)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)+COMBIN(4,2)*COMBIN(9,4)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)

Distribution 2-2-1-1-0-0
=COMBIN(9,2)*COMBIN(5,2)*COMBIN(5,2)*COMBIN(7,2)*COMBIN(5,1)*COMBIN(5,1)+COMBIN(4,1)*COMBIN(9,2)*COMBIN(5,2)*COMBIN(5,2)*COMBIN(7,1)*COMBIN(5,1)+COMBIN(4,2)*COMBIN(9,1)*COMBIN(5,2)*COMBIN(8,2)*COMBIN(5,1)*COMBIN(5,1)

Distribution 2-2-2-0-0-0
=COMBIN(9,3)*COMBIN(5,2)*COMBIN(5,2)*COMBIN(5,2)+COMBIN(4,2)*COMBIN(9,2)*COMBIN(5,2)*COMBIN(5,2)

Distribution 3-1-1-1-0-0
=COMBIN(9,1)*COMBIN(5,3)*COMBIN(8,3)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)+COMBIN(4,1)*COMBIN(9,1)*COMBIN(5,3)*COMBIN(8,2)*COMBIN(5,1)*COMBIN(5,1)+COMBIN(4,3)*COMBIN(9,3)*COMBIN(5,1)*COMBIN(5,1)*COMBIN(5,1)

Distribution 3-2-1-0-0-0
=COMBIN(9,1)*COMBIN(5,3)*COMBIN(8,1)*COMBIN(5,2)*COMBIN(7,1)*COMBIN(5,1)+COMBIN(4,1)*COMBIN(9,1)*COMBIN(5,3)*COMBIN(8,1)*COMBIN(5,2)+COMBIN(4,2)*COMBIN(9,1)*COMBIN(5,3)*COMBIN(8,1)*COMBIN(5,1)+COMBIN(4,3)*COMBIN(9,1)*COMBIN(5,2)*COMBIN(8,1)*COMBIN(5,1)

Distribution 3-3-0-0-0-0
=COMBIN(9,2)*COMBIN(5,3)*COMBIN(5,3)+COMBIN(4,3)*COMBIN(9,1)*COMBIN(5,3)

Distribution 4-1-1-0-0-0
=COMBIN(9,1)*COMBIN(5,4)*COMBIN(8,2)*COMBIN(5,1)*COMBIN(5,1)+COMBIN(4,1)*COMBIN(9,1)*COMBIN(5,4)*COMBIN(8,1)*COMBIN(5,1)+COMBIN(4,4)*COMBIN(9,2)*COMBIN(5,1)*COMBIN(5,1)

Distribution 4-2-0-0-0-0
=COMBIN(9,1)*COMBIN(5,4)*COMBIN(8,1)*COMBIN(5,2)+COMBIN(4,2)*COMBIN(9,1)*COMBIN(5,4)+COMBIN(4,4)*COMBIN(9,1)*COMBIN(5,2)

Distribution 5-1-0-0-0-0
=COMBIN(9,1)*COMBIN(5,5)*COMBIN(8,1)*COMBIN(5,1)+COMBIN(4,1)*COMBIN(9,1)*COMBIN(5,5)

Regards
PAB
 

GillesD

Member
Formulas

Yes this is the formulas I use except in one case.

In the 3-3-0-0-0-0, you use COMBIN(5,3) while I use COMBIN(5,2).

But since both equal 10, this does not affect the end result,
 

PAB

Member
Excel Formulas

Thanks very much GillesD for taking the time to look over and confirm the Excel Formulas.

All the best
PAB
 
Re: Use of Excel

GillesD said:
My databases are in Excel files (.xls) and other files refer to these files to get the proper data.

I have one file that calculates automatically:
- ratio of odd/even numbers (from 6/0 to 0/6)
- distribution of the sum of numbers into 9 sub-groups
- distribution of range (high - low number) into 7 sub-groups
- ratio of high/low numbers (from 6/0 to 0/6)
- the time the 6 numbers of a draw repeat in the next
- distribution of consecutives in a draw (from 1-1-1-1-1-1 to 6-0-0-0-0-0
- distribution of last digit (0 to 9) and the spread of LD (from 1-1-1-1-1-1 to 5-1-0-0-0-0)
- distribution of numbers by decades (1-9, 10-19, ...) and the spread in decades (from 2-1-1-1-1 to 6-0-0-0-0)
- distribution of numbers by dozens (1-12, 13-24, ...) and the spread in decades (from 2-1-1-1-1 to 6-0-0-0-0)

All calculations are automatically updated as I opened the file. In this case, I use only available functions in Excel. Such a large file is unusual, normally I keep a file to keep track of one statistic (whether the Delta distribution, the distribution on numbers in the 6 positions, etc.). In some cases, I calculate without and with the bonus number.

I use VBA as last resort to facilitate data entry or calcultate large numbers of possibilities (pairs, triple, quads, ...). I would rather use formulas as results are almost instantneous even with a large database.

My files are usually set for 2000 draws which mean that I will have to update quite a few in the near future.

Is it possible to have spreadsheet template.
Thank you.
Michael168
 
Request from GillesD

Hi GillesD,

I am interest in your spreadsheet. If possible can I have a copy of your spreadsheet template which contains the formula so that I can use it for keeping the draw results.

Thanks & Regards
Michael168
 

GillesD

Member
Getting my template

Sure, most likely it coulf be forwarded to you by LT.

One problem: the headings are in French, does that create a problem to you? The text in the file is fairly simple, just words like impair/pair for odd/even or somme for sum.

I could easily translate the text on the summary page but would take longer on calculation pages.
 

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