Hi lottoboy
Sorry to say that either I did not understand your question or, you don’t understand the answer. The above postings of mine were an attempt to demonstrate for you the possible use of the median and standard deviation. It is not really a strategy, although I suppose it could be used as one with the understanding you are going to be wrong about 98% of the time – over time. Most of us in this forum are looking for, and working on, strategies much better than that.
You asked - “How to set the sum and tolerance level.†You use the WNH sums and difference filter for that, as they are in the program as it stands, complete with a help manual and instructions. There is nothing to add to the program to get those results.
You asked if Stan can check the approximations I have given. Off course he can, but he probably knows the answer already since they are so fundamental to his program. Besides, I think he has more important things to do than check something so simple for you. Why don’t you check the figures I have given against your own Lottery? Post your results and people will give you further guidance. That way you may learn something that may be useful in building your own strategies.
Bobijohn
Simulators
21 posts
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by lottoboy » Mon Jun 02, 2008 6:05 am
[quote=Bobijohn]
Hi lottoboy
Sorry to say that either I did not understand your question or, you don’t understand the answer. The above postings of mine were an attempt to demonstrate for you the possible use of the median and standard deviation. It is not really a strategy, although I suppose it could be used as one with the understanding you are going to be wrong about 98% of the time – over time. Most of us in this forum are looking for, and working on, strategies much better than that.
You asked - “How to set the sum and tolerance level.†You use the WNH sums and difference filter for that, as they are in the program as it stands, complete with a help manual and instructions. There is nothing to add to the program to get those results.
You asked if Stan can check the approximations I have given. Off course he can, but he probably knows the answer already since they are so fundamental to his program. Besides, I think he has more important things to do than check something so simple for you. Why don’t you check the figures I have given against your own Lottery? Post your results and people will give you further guidance. That way you may learn something that may be useful in building your own strategies.
Bobijohn
[/quote]
Hi, Bobijohn:
Thanks for your reply! I think that you should not understand my question totally. First, I'm interesting with your way which is new clue to use
Median and StdDev for estimating the ranges of Sum at least to me. Second, I realy not clear something in your reply post. For example, you said:"...you will have all 11 columns in tolerance about 1.5% of the time (67.5%^11=1.5%) taken over >1000 draws." I'm not sure if the 11 is maltiplicand, index or others? (67.5% x 11=7.425 or 67.5%^11=1.3254%.) Also, what's the meaning of "taken over >1000 draws" and "to have the winning ticket once per year assuming..."??
I'm so sorry to bother you again.
Best,
lb
Hi lottoboy
Sorry to say that either I did not understand your question or, you don’t understand the answer. The above postings of mine were an attempt to demonstrate for you the possible use of the median and standard deviation. It is not really a strategy, although I suppose it could be used as one with the understanding you are going to be wrong about 98% of the time – over time. Most of us in this forum are looking for, and working on, strategies much better than that.
You asked - “How to set the sum and tolerance level.†You use the WNH sums and difference filter for that, as they are in the program as it stands, complete with a help manual and instructions. There is nothing to add to the program to get those results.
You asked if Stan can check the approximations I have given. Off course he can, but he probably knows the answer already since they are so fundamental to his program. Besides, I think he has more important things to do than check something so simple for you. Why don’t you check the figures I have given against your own Lottery? Post your results and people will give you further guidance. That way you may learn something that may be useful in building your own strategies.
Bobijohn
[/quote]
Hi, Bobijohn:
Thanks for your reply! I think that you should not understand my question totally. First, I'm interesting with your way which is new clue to use
Median and StdDev for estimating the ranges of Sum at least to me. Second, I realy not clear something in your reply post. For example, you said:"...you will have all 11 columns in tolerance about 1.5% of the time (67.5%^11=1.5%) taken over >1000 draws." I'm not sure if the 11 is maltiplicand, index or others? (67.5% x 11=7.425 or 67.5%^11=1.3254%.) Also, what's the meaning of "taken over >1000 draws" and "to have the winning ticket once per year assuming..."??
I'm so sorry to bother you again.
Best,
lb
- lottoboy
- Posts: 738
- Joined: Sat Dec 08, 2007 8:37 pm
by Bobijohn » Mon Jun 02, 2008 5:46 pm
Hi lottoboy,
67.5%^11=1.3254% is correct. I just rounded to the next half percent for simplicity. The 11 refers to the 11 columns in the WNH table. Taken over >1000 draws means that I had more than 1000 draw results in my data base on which to do my analysis.
So, if a Lotto has two draws per week then there are 104 drawings per year – say 100 for simplicity. If we can expect a correct answer (winning ticket) 1.3254% (say 1.5% for simplicity) of the time, then we can expect to have a winning ticket once every 104/1.3254 = 78 draws. Speaking just generally I used 100/1.5 and said that approximates to 1 drawing per year. However, it is purely an expectation – could happen the next draw or, it may take many more than 104 draws before the winning ticket is drawn. It is purely simple statistics with no guarantee as to when the correct outcome will occur.
Bobijohn
67.5%^11=1.3254% is correct. I just rounded to the next half percent for simplicity. The 11 refers to the 11 columns in the WNH table. Taken over >1000 draws means that I had more than 1000 draw results in my data base on which to do my analysis.
So, if a Lotto has two draws per week then there are 104 drawings per year – say 100 for simplicity. If we can expect a correct answer (winning ticket) 1.3254% (say 1.5% for simplicity) of the time, then we can expect to have a winning ticket once every 104/1.3254 = 78 draws. Speaking just generally I used 100/1.5 and said that approximates to 1 drawing per year. However, it is purely an expectation – could happen the next draw or, it may take many more than 104 draws before the winning ticket is drawn. It is purely simple statistics with no guarantee as to when the correct outcome will occur.
Bobijohn
- Bobijohn
- Posts: 212
- Joined: Tue Feb 10, 2009 3:27 pm
by lottoboy » Mon Jun 02, 2008 9:18 pm
[quote=Bobijohn]
Hi lottoboy,
67.5%^11=1.3254% is correct. I just rounded to the next half percent for simplicity. The 11 refers to the 11 columns in the WNH table. Taken over >1000 draws means that I had more than 1000 draw results in my data base on which to do my analysis.
So, if a Lotto has two draws per week then there are 104 drawings per year – say 100 for simplicity. If we can expect a correct answer (winning ticket) 1.3254% (say 1.5% for simplicity) of the time, then we can expect to have a winning ticket once every 104/1.3254 = 78 draws. Speaking just generally I used 100/1.5 and said that approximates to 1 drawing per year. However, it is purely an expectation – could happen the next draw or, it may take many more than 104 draws before the winning ticket is drawn. It is purely simple statistics with no guarantee as to when the correct outcome will occur.
Bobijohn
[/quote]
Hi, Bobijohn:
I'm clear now. Thanks for your sincere and detail reply!
Good luck,
lb
Hi lottoboy,
67.5%^11=1.3254% is correct. I just rounded to the next half percent for simplicity. The 11 refers to the 11 columns in the WNH table. Taken over >1000 draws means that I had more than 1000 draw results in my data base on which to do my analysis.
So, if a Lotto has two draws per week then there are 104 drawings per year – say 100 for simplicity. If we can expect a correct answer (winning ticket) 1.3254% (say 1.5% for simplicity) of the time, then we can expect to have a winning ticket once every 104/1.3254 = 78 draws. Speaking just generally I used 100/1.5 and said that approximates to 1 drawing per year. However, it is purely an expectation – could happen the next draw or, it may take many more than 104 draws before the winning ticket is drawn. It is purely simple statistics with no guarantee as to when the correct outcome will occur.
Bobijohn
[/quote]
Hi, Bobijohn:
I'm clear now. Thanks for your sincere and detail reply!
Good luck,
lb
- lottoboy
- Posts: 738
- Joined: Sat Dec 08, 2007 8:37 pm
by Falcon » Sun Jul 06, 2008 2:20 pm
Hi Stan,
With the introduction of the Median column into the WN History Simulator panel the Simulation Direction advice ie Increase or Decrease has become a bit confusing. The Difference values can now be showing a negative value whereas the Direction advice can be showing an increase - this is because these direction advices continue to be derived from the Average value.
I suggest the Direction advices should be derived now from the Median values and their relationship to the previous SUM value.
This matter has been raised previously in this thread but appears to have been overlooked.
regards
falcon
With the introduction of the Median column into the WN History Simulator panel the Simulation Direction advice ie Increase or Decrease has become a bit confusing. The Difference values can now be showing a negative value whereas the Direction advice can be showing an increase - this is because these direction advices continue to be derived from the Average value.
I suggest the Direction advices should be derived now from the Median values and their relationship to the previous SUM value.
This matter has been raised previously in this thread but appears to have been overlooked.
regards
falcon
- Falcon
- Posts: 442
- Joined: Fri Feb 23, 2007 3:13 am
by Falcon » Sun Jul 06, 2008 5:10 pm
Hmmm,
Not quite correct what I have said immediately above. After a re read of the Help File I find that the Increase/Decrease advice is independent of both the Average and the Median value columns. This is really to be preferred as one can use the percentage value median value difference value and average value all as independent pointers as what the column is likely to do particularly if they all in agreement. Thus Stan quite happy to leave it as it is.
falcon
Not quite correct what I have said immediately above. After a re read of the Help File I find that the Increase/Decrease advice is independent of both the Average and the Median value columns. This is really to be preferred as one can use the percentage value median value difference value and average value all as independent pointers as what the column is likely to do particularly if they all in agreement. Thus Stan quite happy to leave it as it is.
falcon
- Falcon
- Posts: 442
- Joined: Fri Feb 23, 2007 3:13 am
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