Unbelievable ... how to hit a jackpot.

Re: Unbelievable ... how to hit a jackpot.

Postby D1Magnet » Wed Aug 14, 2019 4:19 pm

Hi edymurph, thank you, please see my reply
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Re: Unbelievable ... how to hit a jackpot.

Postby edymurph » Wed Aug 14, 2019 4:45 pm

Hi D1 ...
I wonder if you understand my logic here.

I will give you an example.
The current draw (1852) has sum (1,104,582)

I will make the forecast for the (1853)

It will be between ... (1,105,662-1,105,906)

I can make it narrower.

It will be between ... (1,105,662-1,105,846)

100% sure.
In the spreadsheet I sent you I think you realized where this final 3-digit average comes from.
But as I said ... I don't know how this can help, I don't remember the formulas in those numbers either.
I didn't test your idea ... of turning the final 3 digits into numbers (500) to use in match wining number, because I did a test a while ago with numbers (8,9,10) with the full packet inside the complex filter. and it took almost all day.
My pc is very slow.
But that's it...
Thank you
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Re: Unbelievable ... how to hit a jackpot.

Postby edymurph » Thu Aug 15, 2019 3:21 am

Hello D1 ...
I was thinking here ...
if we also did the reverse calculation for the total sum or single sum numbers.
I will explain using the 1640 contest.

We know that in the 1640 contest came the 543 and we also know that it could come between (476-589) an average of 113.

In total came 866,362, we also know that could come between (866,228-866,452) an average of 226.

Now ... if we divided either of these two values ​​(543 or 866,362) by (2)?
Thus we would have a smaller average.
Or maybe we divided by (4,6,8 ...) the average would be too narrow.
I don't know if you understand me, but it's a crazy idea.
Thank you...
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Re: Unbelievable ... how to hit a jackpot.

Postby D1Magnet » Thu Aug 15, 2019 8:13 pm

Hi edymurph

I understand, it would take a long while to filter them & it seems you might still be left with a lot of tickets

I think I understood what you meant by dividing the answer number by 2 to try and narrow it’s range. (You could also do this on the sum number 866,362 also) I tried this on the actual answer number. (I also tried this further down to 20,30 decimal places over the 3 columns but the answer didn’t seem to converge on a single number (although you could keep trying as they’re just simple excel calculations), even then you’d have to convert it back to it’s original answer number)

In the spreadsheet I’ve attached, if you look at the rows 31 & 37 that I’ve highlighted in pink, you can see that I transferred those decimal numbers across the sheets into the boxes with pink highlight in them.

For example, in line 37, you can see (from the decimal calculation) min is 55, answer was 63 and max was 69, 62 is in the middle, so perhaps you’d forecast 62 or 63’s answer

You'd also consider the direction estimate, if the previous answer is above or below it's calculated median.

In those boxes you see a decimal pattern series where I’ve recalculated the decimals back to their original answer prediction estimates (click in the cell to see the formula, and you should see what I mean)

Can you please also remind me how you determined for draw 1640 that 476 is the minimum and 589 is the maximum?

(another idea) You could also look at what the lowest & highest number was for the last say 100 draws (or whatever number you wanted)

Thanks
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Re: Unbelievable ... how to hit a jackpot.

Postby edymurph » Fri Aug 16, 2019 2:43 pm

Hello D1 ...
Thanks again for the calculations.
Then I'll take a quiet look at the spreadsheet.

(... Can you please also remind me how you determined to draw 1640 that 476 is the minimum and 589 is the maximum? ...)

To get this average you should only use as many draws as you want.
Example. To find the average of 1640 (analyze only the 1640 results in the software).
1641 (analyze only the 1641 results in the software) and so on.
Then use the excel filter to see the average of min and max. OK?

Remember this prediction?
(I will make the forecast for the (1853)

It will be between ... (1,105,662-1,105,906)

I can make it narrower.

It will be between ... (1,105,662-1,105,846)

It came exactly
(1,105,754)
It is within the forecast.
I can do this for any future draw based on the last draw, but as the average is high between them, I don't know how I can use it to my advantage.

I hope you understand, any questions you may ask.

But first have a question about another subject.
Where is the filter (Matrix Distribution)?
Or can I use Number Groups for this?

I would like you to look at my print and tell me if there is something I am doing wrong with this distribution and filtering.
Thank you...
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Re: Unbelievable ... how to hit a jackpot.

Postby D1Magnet » Sat Aug 17, 2019 8:48 pm

Hi edymurph, thank you I see easily now how you set the min, max and average on a draw by draw basis :D I sent you a spreadsheet of my calculations.

Matrix Distributions Filtering

Go to:
Winning numbers, analyze, select the number of draws, then select “matrix distribution” under the basic folder. From here you can press the filter icon after you have selected 1 or more rows (which are then highlighted in blue)
(press CTRL+left click to highlight more than 1 row)

Also in the drop down box where it says “5 horizontal” you can see the other 3 options available

See also 4. matrix distribution under
Help, help contents, search, type in “matrix distribution”

This explains how on one row 3 numbers were found (from your screenshot, 10,12,18). Then 1 number was found on each row, 5, then 25, then 33. Therefore it is 3,1,1,1

Mega Sena Filtering

From what I understand, Mega Sena draws individual digits (rather than whole numbers) from separate ball cages, but basically it is a pick 6 from 60 game, so the results display with 6 winning numbers between 1-60 https://en.wikipedia.org/wiki/Mega-Sena

From my screenshot “matrix distributions filter more than 1 row” (which would work the same for Mega Sena), perhaps it may be easier to filter this way, from the selected blue rows in my screenshot.

Running this matrix distributions filter (from the screenshot) on Demo 6/49 lotto reduces the package just under 10%, (say 9% reduction only)

As when you were using number groups from your screenshot, the 3 numbers (10,12,18) hit horizontally in the number range of 10-18, not vertically like your 9 custom number groups are setup. (I see the logic of how you setup the 9 custom discrete groups though) Also you have 9 custom discrete groups setup & you are trying to combine all 9 into a pattern that ends at 6 eg 3,2,1,1,1,1 but there are still nine groups having to be condensed into this 6 pattern.

Hope this helps.

Thanks
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Re: Unbelievable ... how to hit a jackpot.

Postby edymurph » Sun Aug 18, 2019 3:22 am

Hi D1 ... Yes ... I understood your concept for the matrix distribution filter. I realize that I chose the numbers wrong, but being chosen in lines of 9, would be 7 lines. I went to the help section and the explanation was for 6/49. Although, in the mega sena the 9 horizontal distribution form is 7 lines, where we have up to 6 draw positions left me a doubt. If there are 7 lines of 9, how is the configuration? min 1-1-1-1-0-0 max 3-2-1-1-1-1 If I used the group number filter I don't think it would work because there is still one line left without setting in the filter. In both horizontal and vertical lotofacil this works exactly. But don't worry about it. I'll change the subject. I saw in your spreadsheet that you mentioned where I subtracted 30 from 662. I think you'll understand where I got this value from the spreadsheet I sent you. But I found it interesting that this value is similar in result 1640 and 1852. If this is a pattern, we may be on the way. Any questions, I'm here. Thank you.
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Re: Unbelievable ... how to hit a jackpot.

Postby D1Magnet » Sun Aug 18, 2019 8:03 pm

Hi edymurph,

Thank you, yes I understand now how you’re using actual sum (1-632) & amount summed (540-663) to come up with the difference of 30 or 31. And yes it would be nice to see if there is a further pattern in this on more draws.

Filter Screenshot

The matrix distributions filter seems to be the filter to use, as in your screenshot, the 9 horizontal was a matrix distributions statistic, so it would be easier to filter with this corresponding filter.

If there are 7 lines of 9, how is the configuration? min 1-1-1-1-0-0 max 3-2-1-1-1-1


From the help file for matrix distributions:

9 Horizontal
This matrix has 9 columns, the numbers are arranged from the left to the right. The count of rows depends on the pool size.
two numbers are on separate rows (23 and 47), the rest of the numbers are evenly distributed into two rows. So the 9 Horizontal matrix distribution value is 2-2-1-1
________________

2-2-1-1 equals 6

In Mega Sena 9 horizontal, you still have 9 columns (so having 7 rows is ok also)
Because it is a pick 6 lottery, only 6 numbers are drawn, so even with only 1 per row drawn, the result is 1-1-1-1-1-1

max 3-2-1-1-1-1 in your example above, would be what you counted over the 10 draws in your screenshot (counting downwards over the 6 columns) & it can only be 6 columns as 6 numbers are drawn in this game.

Thanks
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Re: Unbelievable ... how to hit a jackpot.

Postby edymurph » Tue Aug 20, 2019 3:28 pm

Hi D1 ...
I had done almost the same process to update the spreadsheet (Actuals and Changed Values- EDITED)
but for one small detail and my carelessness that went wrong.
I removed the values ​​(-2,6,20,32, -2,0, -16, -20,2, -2)
of the place, and I think that got in the way of updating.
But anyway, I appreciate the explanation for the problem.
Taking advantage I will put here the forecast for the 1856
(1,109,208-1,109,454)
More closed
(1,109,208-1,109,392)
If the value does not go above 632, but I find it unlikely.
Thank you
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Re: Unbelievable ... how to hit a jackpot.

Postby edymurph » Thu Aug 22, 2019 3:13 pm

Hi D1.
Once again the forecast is within the analysis (1,109,316)
This is because the 632 was not exceeded.
But these predictions are still not helping anything, as in the 1853 contest there was a difference from the predicted value of (92) using the narrow side.
In the 1855 contest there was a difference of the predicted value of (58) using the narrow side as well.
Already in 1856 had a difference of (76) using the narrow side.
These values ​​fluctuate, making it hard to find any pattern.
Let's see how it behaves in 1857.
Thank you...
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Re: Unbelievable ... how to hit a jackpot.

Postby D1Magnet » Fri Aug 23, 2019 7:06 pm

Hi edymurph, thank you, yes I see your forecast for 1856 was within the range again. If you calculate the average of 1,109,208-1,109,454 (602) and then the average of 1,109,208-1,109,392. (586) The answer was exactly the average of these two at 594 (or 586 +8 is 594, 602-8 is 594).

To do a few variations of average for the next contest 1857, where it is due to go up (as latest answer is under 597), it could be 1110506, but perhaps 1110520 or 1110550. These calculations seem to be finding these answers for the latest few contests (or perhaps near them eg, contest 1854 & 1855 they were 1 or 2 off one of these average answers).

Also for 1854, the narrow side average is 586 and answer was 584.
And 1853, it came exactly in the middle of your forecast at 1,105,754 (586)

As far as a pattern goes, I've just been noticing that the answers are being found around these average calculations over the last 5 contests.

Thanks
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Re: Unbelievable ... how to hit a jackpot.

Postby edymurph » Fri Aug 23, 2019 10:56 pm

Hello D1 ...
I clearly understood your prediction for the 1857 contest based on the 1856 contest and looking at the 1856 I realized that the prediction was exactly correct.
(1,109,300-1,109,331)
average 1,109,316
It would be nice if everyone were like that.
( :P :P :D :D :lol: :lol: :wink: :wink: )
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Re: Unbelievable ... how to hit a jackpot.

Postby D1Magnet » Sat Aug 24, 2019 7:08 pm

Hi edymurph, yes it would be great if each contest hit that precisely! :D :D :D
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Re: Unbelievable ... how to hit a jackpot.

Postby edymurph » Sun Dec 29, 2019 6:47 pm

Hello everyone.
Maybe here in the forum already have the answer to this question, but not yet found.
I would like to know something.
If I am using EL paid version with only EL working with only complex filter in single filtering with a high number of tickets ... increasing the number of threads to 6 or 7 on a PC of 8 RAM, system operational and java 64-bit and J-xmx4g file ...
the question is...
With only EL open with only the complex filter running with a single filter, will the number of threads increase the speed of this filtering?
This is my doubt.
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Re: Unbelievable ... how to hit a jackpot.

Postby stan » Mon Dec 30, 2019 1:54 am

edymurph wrote:Hello everyone.
Maybe here in the forum already have the answer to this question, but not yet found.
I would like to know something.
If I am using EL paid version with only EL working with only complex filter in single filtering with a high number of tickets ... increasing the number of threads to 6 or 7 on a PC of 8 RAM, system operational and java 64-bit and J-xmx4g file ...
the question is...
With only EL open with only the complex filter running with a single filter, will the number of threads increase the speed of this filtering?
This is my doubt.

Yes, it should increase the filtering speed.
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