(UK) Lotto Skips

[PAB]

This particular topic was touched on in another thread but because it was off topic I thought I would start a new thread and see what happens.

I have calculated all the UK Main 649 Lotto Skips from it's inception back in 1994 and come up with the following data analysis:-

Ball 01, Minimum = 0, Maximum = 52, Average = 6.4291, Median = 4
Ball 02, Minimum = 0, Maximum = 44, Average = 5.6751, Median = 4
Ball 03, Minimum = 0, Maximum = 38, Average = 5.8461, Median = 4
Ball 04, Minimum = 0, Maximum = 54, Average = 6.7772, Median = 4
Ball 05, Minimum = 0, Maximum = 47, Average = 6.4010, Median = 4
Ball 06, Minimum = 0, Maximum = 26, Average = 5.1054, Median = 4
Ball 07, Minimum = 0, Maximum = 45, Average = 6.4815, Median = 4
Ball 08, Minimum = 0, Maximum = 40, Average = 5.7242, Median = 4
Ball 09, Minimum = 0, Maximum = 27, Average = 5.4286, Median = 4
Ball 10, Minimum = 0, Maximum = 36, Average = 6.2890, Median = 4
Ball 11, Minimum = 0, Maximum = 23, Average = 4.8770, Median = 3
Ball 12, Minimum = 0, Maximum = 47, Average = 6.3574, Median = 4
Ball 13, Minimum = 0, Maximum = 49, Average = 7.7104, Median = 5
Ball 14, Minimum = 0, Maximum = 46, Average = 6.7033, Median = 4
Ball 15, Minimum = 0, Maximum = 37, Average = 6.2763, Median = 4
Ball 16, Minimum = 0, Maximum = 47, Average = 7.0943, Median = 5
Ball 17, Minimum = 0, Maximum = 72, Average = 7.0871, Median = 4
Ball 18, Minimum = 0, Maximum = 41, Average = 6.2466, Median = 4
Ball 19, Minimum = 0, Maximum = 42, Average = 6.2063, Median = 4
Ball 20, Minimum = 0, Maximum = 33, Average = 6.7110, Median = 5
Ball 21, Minimum = 0, Maximum = 46, Average = 6.3673, Median = 4
Ball 22, Minimum = 0, Maximum = 43, Average = 6.1125, Median = 4
Ball 23, Minimum = 0, Maximum = 36, Average = 5.2576, Median = 3
Ball 24, Minimum = 0, Maximum = 36, Average = 5.7286, Median = 4
Ball 25, Minimum = 0, Maximum = 35, Average = 5.3767, Median = 4
Ball 26, Minimum = 0, Maximum = 36, Average = 5.9658, Median = 4
Ball 27, Minimum = 0, Maximum = 35, Average = 5.3811, Median = 4
Ball 28, Minimum = 0, Maximum = 45, Average = 5.7297, Median = 4
Ball 29, Minimum = 0, Maximum = 39, Average = 5.8213, Median = 4
Ball 30, Minimum = 0, Maximum = 32, Average = 5.7408, Median = 4
Ball 31, Minimum = 0, Maximum = 32, Average = 4.9482, Median = 3
Ball 32, Minimum = 0, Maximum = 44, Average = 6.3177, Median = 4
Ball 33, Minimum = 0, Maximum = 28, Average = 5.1423, Median = 4
Ball 34, Minimum = 0, Maximum = 38, Average = 6.1892, Median = 4
Ball 35, Minimum = 0, Maximum = 32, Average = 5.5814, Median = 4
Ball 36, Minimum = 0, Maximum = 38, Average = 6.1390, Median = 4
Ball 37, Minimum = 0, Maximum = 36, Average = 6.1087, Median = 4
Ball 38, Minimum = 0, Maximum = 28, Average = 5.1418, Median = 3
Ball 39, Minimum = 0, Maximum = 52, Average = 6.5179, Median = 4
Ball 40, Minimum = 0, Maximum = 34, Average = 5.7110, Median = 4
Ball 41, Minimum = 0, Maximum = 34, Average = 6.6200, Median = 5
Ball 42, Minimum = 0, Maximum = 34, Average = 6.3878, Median = 4
Ball 43, Minimum = 0, Maximum = 47, Average = 5.9057, Median = 4
Ball 44, Minimum = 0, Maximum = 34, Average = 5.3497, Median = 4
Ball 45, Minimum = 0, Maximum = 40, Average = 5.5008, Median = 4
Ball 46, Minimum = 0, Maximum = 44, Average = 6.2521, Median = 4
Ball 47, Minimum = 0, Maximum = 48, Average = 5.9923, Median = 4
Ball 48, Minimum = 0, Maximum = 37, Average = 5.2625, Median = 4
Ball 49, Minimum = 0, Maximum = 28, Average = 5.5538, Median = 4

I hope someone finds this interesting!

Regards,
PAB


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12:45, restate my assumptions.
(1) Mathematics is the language of nature.
(2) Everything around us can be represented and understood through numbers.
(3) If you graph the numbers of any system, patterns emerge. Therefore, there are patterns, everywhere in nature.

 

[jack]

Hello, pab, here the formula corrected objective is to find a midpoint
Ie after calculating all 49 numbers put the average value noa
In ascending order the 49 numbers in a row, and aim to see the 3 numbers along
Along the line, it is a matter of physics and logic ok
Example is the mean of 17 is 10.23, ... and the average of 32 is 9, 44, in ascending order
The average value (formula below) the numerto 32, comes in front of 17 and so
With the other numbers, put on the line in order of increasing average output
looking error of the formula:

F = (A - (B + C)) / 3 Formula No. 6

the variable F can not be two sides of the equation.

Making the isolation of F:

F = (A - B + F) / 3

F / 3 = A - B + F

B - F A = - (F / 3)

B - A = 2F / 3

F = (3 * (B - A)) / 2


Here is how to get the value of F

 

[jack]

Fazendo o isolamento de F:

F = ( A - B + F ) / 3

F / 3 = A - B + F

B - A = F - ( F / 3 )

B - A = 2F / 3

F = ( 3 * ( B - A)) / 2

 

[Icewynd]

Jack,

What is F? What are you trying to calculate?

And what values do we use for A, B, C?

 

[jack]

legend
01 = number of lottery
= Last 02 draw
03 = number of output per cycle / Average r / d
04 = (A) LARGE (>) SPACE BETWEEN CYCLES
05 = (B) SMALL (<) SPACE BETWEEN CYCLES
06 = (F) SUPPLEMENT, (A-(B + F)) / 3
= 07 repetitions FREQUENCIES OF CYCLES
08 = MORE SPACE INSIDE THE CYCLE
09 FINAL = (A-F)
10 = FREQUENCY OF REPEATED CYCLE
11 = (C) SUM OF FREQUENCIES
12 = (D) AMOUNT OF FREQUENCIES BETWEEN CYCLES
13 = MEDIA FREQUENCY BETWEEN CYCLES = (C / D)
PAB FIRST YOU HAVE TO DO THE STATISTICS in each cycle, and the cycles together
You can improve the formula, is the desire, goal is to find an average number of 49/6 are all together in a sector of the line
Example = 12,45,45,36,01,25,35,08,09 ...... up to 49

 

[Icewynd]

This particular topic was touched on in another thread but because it was off topic I thought I would start a new thread and see what happens.

I have calculated all the UK Main 649 Lotto Skips from it's inception back in 1994 and come up with the following data analysis:-

So 50% of the hits on most numbers come in at less than 4. That's worth knowing. Of course, the other 50% is the pesky side -- the ones that skip up to 72!

 

[jack]

F = average frequency value of each digit in the cycle, attention, this formula can be enhanced,
Goal is to see the average of all the numbers and put a line
To see where the numbers are grouped in the next draw, we have to improve some paramentros or pab or someone from the forum will be (give permission) to move the formula ok

 

[PAB]

Hi Icewynd,

So 50% of the hits on most numbers come in at less than 4. That's worth knowing. Of course, the other 50% is the pesky side -- the ones that skip up to 72!

Thanks for the reply.
Yes, they are quite interesting statistics so far to date.
I didn't say it was going to be easy to get the rest though .

Regards,
PAB


-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-
12:45, restate my assumptions.
(1) Mathematics is the language of nature.
(2) Everything around us can be represented and understood through numbers.
(3) If you graph the numbers of any system, patterns emerge. Therefore, there are patterns, everywhere in nature.

 

[jack]

Hello, will be hard work, statistical, to find (F) midpoint frequency)
A, b, c = demoninaçoes intervals are, again this formaula can be improved or changed to adapt to reference some cycles complement when a number is not finished the cycle of calculations is then placed in a row in ascending order by the average value the formula, ok goal is to see where the numbers are grouped next draw,
34 45 36 33 04 06 12 20 23 15 48 08 01 07 24 03 41 21 28 46 17 25 13 29 14 19 42 39 47 05 35 18 30 40 10 27 31 32 09 22 02 11 37 38 44 16 49 26 50
The next was = 12 28 46 17 25 13 29
Ice and pab, saw that starting from position 17 on line 5 of the 6 numbers were gathered,

 

[Icewynd]

Hello, will be hard work, statistical, to find (F) midpoint frequency)

But, PAB has already posted the Median for each digit which is the midpoint of the distribution of skips. So, what further measure of midpoint frequency is necessary?

 

[jack]

Hello, you have to do in each cycle and the cycles after all together and put on line
In ascending order of average value, for each closed loop curve has a different shape and each of the other cycles,

 

[PAB]

Hi Icewynd,

...PAB has already posted the Median for each ball which is the midpoint of the distribution of SKIPS.

Thanks Icewynd, I thought it was me for a minute.

I would like to just point out to everyone that this thread is for the discussion of SKIPS, and NOT, as what is being discussed in another thread about Numbers Elimination with regard to Cycles Completion.

Regards,
PAB


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12:45, restate my assumptions.
(1) Mathematics is the language of nature.
(2) Everything around us can be represented and understood through numbers.
(3) If you graph the numbers of any system, patterns emerge. Therefore, there are patterns, everywhere in nature.

 

[Frank]

Just to clarify something I posted earlier on skips, you may have noticed that I mentioned in the UK lotto regarding number 17 having a historic maximum of 73. You will have noticed if you did your own analysis that a figure of 72 comes up for that number - for those who use the conventional definition of a skip, e.g if its a direct repeat then its skip is zero.

I have never used this convention. I use for a direct repeat, skip =1. Two reasons really,

1. Its easy to work out a skip, its just drawnumber 2 - drawnumber1 for the skip value.

2. I don't like statistics where the dominant value is zero, and skips are a good example of zeroes being 15% of the total and the rest being progressively less.
So all my skips are one more than anyone elses.

Having said that, what follows will be amended to match everyone elses convention of direct repeat = skip of zero.

Its already been said that 50% of the hits on most numbers come in at less than 4. A more complete list for 6/49 is as follows:-

skip ..cumulative hits
0...... 15%
1...... 27%
2...... 37%
3...... 46%
4...... 54%
5...... 60%
6...... 66%
7 ...... 71%
8...... 75%
9...... 79%
10...... 82%
11...... 84%
12...... 87%
13...... 88%
14...... 90%

So the figure in the second column is the percentage total of all the preceding skip hits so far. So skips of 0,1,2,3,4,5 and 6 occupy 66% of the total number of skips. 90% of the skips are in the range 0 to 14.

If one were to plot the individual skip frequencies against skip values for any 6/49 lottery the result would be an exponential decay curve,
which tended towards zero as the skip values go higher. The more draws on the lottery, the more counts of skips per skip value, but the shape is always the same. See ..

http://www.lotterygen.co.uk/test/images/ukskips.png

Actually it is possible to work out the equation for this decay curve and I have deduced that the curve is:-

Count of skips= 1.032 x N x e^(-0.0154 x s) within the limits of accuracy of my graph. I used Excel charts to help me do this.

Where s is the skip value and N is the draw number. e is the mathematical constant and the quantity in the brackets is the power to which e is raised. After having deduced the formula I worked backwards and drew the theoretical curve over the top of the real one and its a pretty good fit. I also tried it using another set of results from four years ago and it still worked.

So you can work out approximately at a particular draw number, how many of a particular skip value there should be.

I was going to post how to work it out for yourself for your own 6/49 lottery but it might be boring for those who have forgotten what they learned at college!
If you have more than 1815 results, you'll probably get a much more accurate result!

 

[Icewynd]

Very interesting data Frank.

Perhaps you can use your data to answer a question I have often wondered about?

You demonstrate that 90% of the numbers come out within 14 skips. However, the numbers with skips of 15 or greater at 10% of the total have the potential to land in just over 1 in 2 drawn combinations. Do you have the data to show how many combinations have all their skips under a certain cuttoff?

I'm sure this could be demonstrated mathematically, but I am, alas, one of those who has forgotten much that I learned in college.

 

[Frank]

Hi Icewynd,

If I understand you correctly weve moved away from individual skips on a per ball basis and moved on to combinations as in actual results. I believe this would be a data mining excercise involving substituting each ball in a result by its skip value as it was on the date drawn. This would produce a table of skips at each draw number, replacing the actual result. I daresay it would be possible to then (for each result ) find the maximum skip value for the balls in that result. From this information I believe your question could be answered. I don't know if I'll find the time to do it soon, my wife will kill me if I don't get the decorating finished, but I'll look into it sometime, possibly PAB will beat me to it.

 

[PAB]

Excellent data and statistics as usual Frank .

...I believe this would be a data mining excercise involving substituting each ball in a result by its skip value as it was on the date drawn. This would produce a table of skips at each draw number, replacing the actual result. I daresay it would be possible to then (for each result ) find the maximum skip value for the balls in that result. From this information I believe your question could be answered.
Possibly PAB will beat me to it.

Icewynd

I have put this together from what I believe Frank has described above.
It replaces the Actual Drawn Numbers with their corresponding Skip values.
You could probably now calculate the statistics you are after.
Please let me know how you get on!

http://www.mediafire.com/view/?nst2arou58sht1j

Regards,
PAB


-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-
12:45, restate my assumptions.
(1) Mathematics is the language of nature.
(2) Everything around us can be represented and understood through numbers.
(3) If you graph the numbers of any system, patterns emerge. Therefore, there are patterns, everywhere in nature.

 

[PAB]

PREVIOUS FILE DELETED

Here is the new updated file.

http://www.mediafire.com/view/?74g4zl0uko5ymdw

Regards,
PAB


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12:45, restate my assumptions.
(1) Mathematics is the language of nature.
(2) Everything around us can be represented and understood through numbers.
(3) If you graph the numbers of any system, patterns emerge. Therefore, there are patterns, everywhere in nature.

 

[Icewynd]

Thanks, PAB.

I'll have a look and report back.

 

[Frank]

Thanks PAB for coming to the rescue so quickly. Mrs F sends her thanks.
I've downloaded your sheet and now I know the answer to Icewynd's question. It plots an interesting curve

Also I tested my line equation on the Canadian Lottery set you provided. It did reveal a typo in my original revelation of the formula for the skip distribution curve.

It should read 1.032 xN x e^(-0.154 x s) where s is the skip and N is the draw number.

It is spot on with the Canadian lotto curve. Now back to the decorating...

 

[PAB]

Hi Frank, you and Mrs. F are welcome!

...I tested my line equation on the Canadian Lottery set you provided. It did reveal a typo in my original revelation of the formula for the skip distribution curve.

It should read 1.032 x N x e^(-0.154 x s) where s is the skip and N is the draw number.

Yes, those pesky ZERO's are a pain at times .

Regards,
PAB


-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-∏-
12:45, restate my assumptions.
(1) Mathematics is the language of nature.
(2) Everything around us can be represented and understood through numbers.
(3) If you graph the numbers of any system, patterns emerge. Therefore, there are patterns, everywhere in nature.