WNH - reloaded

WNH - reloaded

Postby laurontario » Thu Dec 20, 2007 5:42 pm

Hello, EL users,

I started playing around with EL and trying to understand more about this.
I understand that WNH is a very powerful tool, which, if the estimates are correct, would lead to the jackpot.

I understand how the numbers are calculated in the 0 : -10 columns. It is also good that the charts show the min and max values for each column.

I did a little bit of research on this one and realized that the smallest value for column 0 is if all numbers drawn are from the bottom of the list (cold numbers, with the highest numbers of games out). On the other side, the highest number would appear if all numbers from previous draw are being drawn again.

Both these scenarios are possible in theory but actually never happened in the history, so that's why we need to eliminate them.

What I would like to see is the difference for each column from one game to another and what is the max, min and average difference for these throughout the entire history.

I am not sure if this is clear, but, let's say that if the game before we had Column 0 = 355 and at the last draw this changed to 383, then the difference would be +28.
I guess this can be put in a table, which could be sorted by occurence
I guess this way we could make our estimates much accurate, knowing that what happened the least (or never) in the past, is very unlikely to happen in the future.

Is there such a thing now in EL? If not, do you think it would be useful? (I know that it can definitively be implemented, from the programming point of view).

Any comments/suggestions on this?

Laurentiu
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Postby stan » Thu Dec 20, 2007 10:27 pm

[quote=laurontario]
Is there such a thing now in EL? If not, do you think it would be useful? (I know that it can definitively be implemented, from the programming point of view).
[/quote]

sure, go to 'summary table' in 'wn history' page and select 'differences only' in the 'show' combo-box
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Postby laurontario » Fri Dec 21, 2007 6:38 pm

Thank you Stan for your answer.

One more thing... I noticed that even though the History chart shows the Min and Max values for each column, the possible values which can be obtained are always different. I mean, for column zero, let's say, there is a min=326 and a max=486 (historically) . However, at this point, the minimum value obtainable is 280 and the max = 398, while the current value is 355 (values taken from the Canadian 649). Estimating that the chart would go up, the max value cannot exceed 398 despite the historical 486!

I was wondering if this cannot be calculated somehow and shown on the chart, either as a line (like the average) or as a value on the bottom, where the min/max are.

For the 0 column this can be easily calculated, but I think that for the rest, best way would be to run all 1.4 Million possible combinations and see which are the min/max possible values for each column (0 to -10).

This would give the user a better idea about where the ranges should be. Otherwise, if one estimates the value for a column outside of possible range, there will be no tickets left in the package - Stan, please correct me if I'm wrong.

Is this something which can be implemented easily? I think especially for novice users, it would give some kind of idea about where the range should be.

As a strategy, I think that if we could have this, then we could put it together the wn differences and make more accurate estimates.
Example:
current value = 355
max possible = 398
min possible = 280
Hist diff minim = -53
Hist diff max = 33
Min obtainable applying the min diff = 355-53 = 320
Max obt.... = 355+33 = 388
Historical min for the chart = 326
Hist. max = 486

Based on these, the range should fall somewhere between 355 - 388 (still, a very large interval).

I wonder, using this strategy could we filter down more effectively? Am I wrong? Am I missing something here?
I used this strategy on the random generated package and here are the results, after filtering for column 0 only:
before:
No of tickets: 160,000
Winnings at next draw:
3rd Prize = 2
4th = 162
5th = 2749
6th = 1958

After filtering:
No of tickets left = 106,431
Winnings:
3rd Prize = 0
4th = 153
5th = 2450
6th = 490

Perhaps, if we would have more information we could filter down more effectively.
Any opinions?

Laurentiu
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Postby laurontario » Sun Dec 23, 2007 7:48 am

[quote=laurontario:1198255101]

I was wondering if this cannot be calculated somehow and shown on the chart...
[/quote]

Forget about it - I just found it under Simulation :-D

Time to go to bed now :zzz:
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