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Topic closed. 35 replies. Last post 11 years ago by Thoth.

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The Quantum Master
West Concord, MN
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 Posted: November 12, 2006, 4:11 pm - IP Logged

The equation I posted was derived this way.

Normal Distribution -  y = (1 / s Ö2 p) e -((x - m)² / s²)

m = (6 / 1000)s = 0.006 s

s = Öm = Ö0.006 s

Plugging in the values gives this equation.

y = (1 / Ö0.012 p s) e -((x - 0.006 s)²/(0.012 s)

yet another finger check...  I was very tired when I typed it in.

Normal Distribution is actully -    y = (1 / s Ö2 p) e -((x - m)² / 2 s²)        missed the 2 in the part  s²  -->   2 s²

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Jehocifer

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 Posted: November 12, 2006, 11:19 pm - IP Logged

I certainly don't consider myself to be a mathematician, but I'm counting 1569 draws from October 11th through November 8th.

Not every state in the list has drawings every day of the week ...and some don't even have midday drawings.  979 should be the right amount of draws (I think).

~Probability=Odds in Motion~

Mid-Missouri
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 Posted: November 12, 2006, 11:38 pm - IP Logged

I sure am glad you all are trying to figure out the probability cause I'm as curious as can be. And I am definitely letting you all do the math... LOL. But I did want to tell you it hit again last night as 041, right on cue...It goes no more than 3 days without hitting. Cool!

Thanks Guys,

Bryan  :)

The Quantum Master
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 Posted: November 13, 2006, 12:04 am - IP Logged

This is really unusual, if the real count for the total number of draws is 979 for that sample, then there is really something seriously wrong. I would have a better chance of winning a straight pick 3 using a single number of any kind three times in a row. Something is wrong. The first thing I would suggest is separate the computer picked numbers from the natural picked numbers and see if there is some relative difference. Something has to be causing this, or a combination of things could be causing this.

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Jehocifer

The Quantum Master
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 Posted: November 13, 2006, 12:14 am - IP Logged

Also, on the opposite end of the scale. Look for deficient numbers that have a low hit rate.

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The Quantum Master
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 Posted: November 13, 2006, 1:01 am - IP Logged

I did a quick check on my home state WI for the entire run of numbers.

I did not find any 014 anomaly in either the natural or computer picks.

The contributor must be coming from some other state or combination of states.

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Jehocifer

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 Posted: November 13, 2006, 1:41 am - IP Logged

finger check on that last one, it's actually 1 in 24,939,240,465 for 979 draws.

Thanks for taking your time to help on this one JadeLottery!  There is brilliance in your logic and the way you apply those formulas.  However, the odds of 1 in 24,939,240,465 for 21 hits in 979 draws seem extraordinarily inflated to me.

I was expecting the odds/probability to be more around one in several hundred thousand or possibly as high as 1 in 2,000,000 or so.  From a statistacal perspective, the frequency of boxed combos that show 10 and 11 hits within 979 consecutive trials would seem to suggest a much better probability for a combo to actually hit 21 times within that time frame.  Since the average rate of occurrence for a boxed no match combo is about 1 in every 167 trials (.006), then you could expect the combo to be drawn .006 X 979 = 5.87 times within 979 games as a standard.  There are also plenty of combos that hit 8 and 9 times and plenty of combos that hit into the teens.  I can't imagin the odds would be that bad to see a combo hit 21 times.

As far as finding a formula to find the permutations, perhaps I did a poor job of describing the problem.  Just as you stated, it doesnt matter if the 979 trials are separate drawings spread across any combination of multiple states or if they are 979 consecutive trials within an individual state, the probability should be the same.  Here is a simple example to illustrate my question:

Imagine three states playing the Pick 1 game. Each state has a single hopper loaded with the ten digits 0 through 9.  Whats is the chance the that exactly two of the three states will pull the digit 7?  We know that answer is 27 in 1000 because we know that there are 10^3 total permutations, which is the total possibilities...and that there are 27 doubles.  The 27 doubles are calculated by knowing that the outcome fits the pattern of AAB, where A is the digit "7" and B is any digit other than 7.  Since the B could be any of 9 possible digits, we multiply 9 by the total ways the pattern can be arranged...9*(3!/2!)=27.  Now suppose we were playing the imaginary Pick 1 lottery in four different states.  Whats is the chance that exactly two of the states will each select a 7?  We know there are 10^4 permutations, but now there are two possible patterns that fit the outcome in which exactly two of the four digits are 7's:  AABB and AABC.

AABB:  9*(4!/2!)/2!) = 54

AABC:  (9!)/((9-2)!)/(2!)*(4!/2!/1!/1!) = 432

There are 54 + 432 = 486 possible ways that exactly two 7's could be drawn from the hoppers for all four states.  The odds are thus 486 in 10,000 or .0486.

Without too much more boring detail (I know this is elementary stuff for you), imagine finding the odds of drawing exactly ten 7's in 22 states (still playing Pick 1).  You would have to list all possible outcome patterns that contain exactly ten "A"s  to find all the possibilities: here are just three of the many that are possible:

 A A A A A A A A A A B B B B B B B B B C C C

 A A A A A A A A A A B B B B B B B B C C C C

 A A A A A A A A A A B B B B B B B B C C D D

Is there a formula that can be used to calculate all pattern possiblities?  How about the occurrence of a particular digit in specific amounts...so you dont have to find and list all of these patterns from scratch??  This is the solution I had in mind for finding the true probability of the combo hitting 21 times in 979 trials—but writing the patterns out and running the formulas would probably take a super computer and several years of my time lol!

~Probability=Odds in Motion~

The Quantum Master
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 Posted: November 13, 2006, 2:30 am - IP Logged

Thanks for taking your time to help on this one JadeLottery!  There is brilliance in your logic and the way you apply those formulas.  However, the odds of 1 in 24,939,240,465 for 21 hits in 979 draws seem extraordinarily inflated to me.

I was expecting the odds/probability to be more around one in several hundred thousand or possibly as high as 1 in 2,000,000 or so.  From a statistacal perspective, the frequency of boxed combos that show 10 and 11 hits within 979 consecutive trials would seem to suggest a much better probability for a combo to actually hit 21 times within that time frame.  Since the average rate of occurrence for a boxed no match combo is about 1 in every 167 trials (.006), then you could expect the combo to be drawn .006 X 979 = 5.87 times within 979 games as a standard.  There are also plenty of combos that hit 8 and 9 times and plenty of combos that hit into the teens.  I can't imagin the odds would be that bad to see a combo hit 21 times.

As far as finding a formula to find the permutations, perhaps I did a poor job of describing the problem.  Just as you stated, it doesnt matter if the 979 trials are separate drawings spread across any combination of multiple states or if they are 979 consecutive trials within an individual state, the probability should be the same.  Here is a simple example to illustrate my question:

Imagine three states playing the Pick 1 game. Each state has a single hopper loaded with the ten digits 0 through 9.  Whats is the chance the that exactly two of the three states will pull the digit 7?  We know that answer is 27 in 1000 because we know that there are 10^3 total permutations, which is the total possibilities...and that there are 27 doubles.  The 27 doubles are calculated by knowing that the outcome fits the pattern of AAB, where A is the digit "7" and B is any digit other than 7.  Since the B could be any of 9 possible digits, we multiply 9 by the total ways the pattern can be arranged...9*(3!/2!)=27.  Now suppose we were playing the imaginary Pick 1 lottery in four different states.  Whats is the chance that exactly two of the states will each select a 7?  We know there are 10^4 permutations, but now there are two possible patterns that fit the outcome in which exactly two of the four digits are 7's:  AABB and AABC.

AABB:  9*(4!/2!)/2!) = 54

AABC:  (9!)/((9-2)!)/(2!)*(4!/2!/1!/1!) = 432

There are 54 + 432 = 486 possible ways that exactly two 7's could be drawn from the hoppers for all four states.  The odds are thus 486 in 10,000 or .0486.

Without too much more boring detail (I know this is elementary stuff for you), imagine finding the odds of drawing exactly ten 7's in 22 states (still playing Pick 1).  You would have to list all possible outcome patterns that contain exactly ten "A"s  to find all the possibilities: here are just three of the many that are possible:

 A A A A A A A A A A B B B B B B B B B C C C

 A A A A A A A A A A B B B B B B B B C C C C

 A A A A A A A A A A B B B B B B B B C C D D

Is there a formula that can be used to calculate all pattern possiblities?  How about the occurrence of a particular digit in specific amounts...so you dont have to find and list all of these patterns from scratch??  This is the solution I had in mind for finding the true probability of the combo hitting 21 times in 979 trials—but writing the patterns out and running the formulas would probably take a super computer and several years of my time lol!

I see a possible problem in you logic. I can't explain right now, I have to think about it for a while to format it in a way you can understand.

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Jehocifer

The Quantum Master
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 Posted: November 13, 2006, 6:37 am - IP Logged

Thoth,

Just writing to let you know, I'm still working this one out in my mind. Also, I will be gone for a few days on other projects. I'll check now and then to see what's going on, but probably won't have much time to post. Anyway, I want to pass on a few thoughts about the problem you restated. That helped me understand it more. I actually worked on something similar to this a while ago. It had to do with finding the basic permutational forms for Pick 1 through Pick N, N being any quantity of selections from the set of numbers 0 to 9. I found that when I gone past Pick 10, an interesting problem arose and it's closely relates to the problem you're trying to solve. The problem has to do with the assigning of variables to the problem. In any other math problem, you can add as many variables as you'd like, because typically there are an infinite set of numbers to use; mainly the set real numbers. In this case, we can't do that. We are dealing with a finite set of numbers and are restricted as to how many variables we can assign. The set of finite numbers are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The most variables we can assign are {A, B, C, D, E, F, G, H, I, J}.

When I did the work out for Pick 10 I had no problem. I was able to get all 512 basic permutational forms. However, when trying to do the work out for Pick 11 there was a computational problem. It had to do with taking 10 numbers 11 at a time. If you think about this, you can see this is impossible. After a few hours, a few beers, and a few nights rest, I realized the solution. The problem has to do with repetition. A question came to mind, "If all the numbers are different for the first 10 selections, what happens on the 11th selection?" I realized the no matter what number is selected, it will be the same number of at least 1 of the other 10. That's why the computation fail, I had used up all the available numbers in the set. So, assigning another variable would not work. I'd just be repeating 1 of the numbers. Solution, assign the 11th variable to be the same as 1 of the previous 10. I this case for Pick 11, the permutation is now a permutation of a permutation and would expand to a permutation of a permutation of a permutation in fractals of 10 for higher values Pick N.

Well, I hope that's some insight to what I'm thinking about. I'll get back with you later.

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Jehocifer

Harbinger
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 Posted: November 13, 2006, 10:54 pm - IP Logged

y = (1 / Ö0.012 p s) e -((x - 0.006 s)²/(0.012 s)

x - number of hits based on a given sample size
s - number of samples

y - probability that x number of hits will have happen by s number of samples

note: to get percent probability multiply y by 100%  Ok jade. lets do a little dimensional analysis.

x= 10

s =10

Pi= 3.14

e=2.73 e E1

y= (1/sqrootof .012 x 3.14 x 10) x e exp ( 10-.006 x10)sqd./ (.012x10)

y = 1/sqrtof .3768 x e exp(99.88/ .12)

y= 1/.613 x e exp 832

y=1.63 x e exp 832

y= 1.63 x e exp 832

the answer is equal to a number in a different dimension.

The Quantum Master
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 Posted: November 14, 2006, 2:09 am - IP Logged

y = (1 / Ö0.012 p s) e -((x - 0.006 s)²/(0.012 s)

x - number of hits based on a given sample size
s - number of samples

y - probability that x number of hits will have happen by s number of samples

note: to get percent probability multiply y by 100%  Ok jade. lets do a little dimensional analysis.

x= 10

s =10

Pi= 3.14

e=2.73 e E1

y= (1/sqrootof .012 x 3.14 x 10) x e exp ( 10-.006 x10)sqd./ (.012x10)

y = 1/sqrtof .3768 x e exp(99.88/ .12)

y= 1/.613 x e exp 832

y=1.63 x e exp 832

y= 1.63 x e exp 832

the answer is equal to a number in a different dimension.

As x gets larger when s = 10, the value of y drops to near 0. Most average computers don't deal with numbers that have 1E-800's as exponents. It just converts it to the nearest value of 0. Besides, you really want to use this equation for higher values of s in the 100's, 1000's, or larger.

 s = 10 x y as a % 0 1.580540418 158.0540418 1 0.001032712 0.103271243 2 3.89863E-014 3.89863E-012 3 8.5036E-032 8.5036E-030 4 1.07165E-056 1.07165E-054 5 7.80299E-089 7.80299E-087 6 3.2827E-128 3.2827E-126 7 7.9791E-175 7.9791E-173 8 1.1206E-228 1.1206E-226 9 9.0924E-290 9.0924E-288 10 0 0

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Jehocifer

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 Posted: November 14, 2006, 8:42 pm - IP Logged

Select the Daily Numbers Games option from the Results drop-down menu on this site. Count the draws. I think you'll see that my count of 1569 draws is a lot closer to the correct amount than 979.

The Quantum Master
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 Posted: November 15, 2006, 1:31 am - IP Logged

Select the Daily Numbers Games option from the Results drop-down menu on this site. Count the draws. I think you'll see that my count of 1569 draws is a lot closer to the correct amount than 979.

Did a count by one week, it's 376 draws per week, removing one of the Iowa-Illinois repeats. That's 1504 draws for 4 weeks. The draw count of 1596 is correct. Also, I did a year to date check out for all states up to 45 weeks and the 014 hit count is within acceptable limits. There were a total of 16,920 draws; 014 had a .0161764 probabilty of hitting 115 times, that's about 1.61794 % or 1 in 61 chance of hitting 115 times. There's no 014 anomaly, it's probably in a correction/catch-up cycle for these last 30, 60, or 90 days; that's a reasonable expectation and explanation.

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Order is a Subset of Chaos
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Jehocifer

Mid-Missouri
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 Posted: November 15, 2006, 2:40 am - IP Logged

Thanks for checking it out guys. I'm going to continue to play it every 3rd day it misses until it shows the trend is over. Today is the 3rd day. It still hasn't when over the 3rd day. So, it really doesn't have to hit until tomorrow (16th) for it to break the pattern it has been following.

Wish me Luck,

Bryan  :)

Thanks Again...

The Quantum Master
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 Posted: November 15, 2006, 11:41 am - IP Logged

Thoth,

Just writing to let you know, I'm still working this one out in my mind. Also, I will be gone for a few days on other projects. I'll check now and then to see what's going on, but probably won't have much time to post. Anyway, I want to pass on a few thoughts about the problem you restated. That helped me understand it more. I actually worked on something similar to this a while ago. It had to do with finding the basic permutational forms for Pick 1 through Pick N, N being any quantity of selections from the set of numbers 0 to 9. I found that when I gone past Pick 10, an interesting problem arose and it's closely relates to the problem you're trying to solve. The problem has to do with the assigning of variables to the problem. In any other math problem, you can add as many variables as you'd like, because typically there are an infinite set of numbers to use; mainly the set real numbers. In this case, we can't do that. We are dealing with a finite set of numbers and are restricted as to how many variables we can assign. The set of finite numbers are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The most variables we can assign are {A, B, C, D, E, F, G, H, I, J}.

When I did the work out for Pick 10 I had no problem. I was able to get all 512 basic permutational forms. However, when trying to do the work out for Pick 11 there was a computational problem. It had to do with taking 10 numbers 11 at a time. If you think about this, you can see this is impossible. After a few hours, a few beers, and a few nights rest, I realized the solution. The problem has to do with repetition. A question came to mind, "If all the numbers are different for the first 10 selections, what happens on the 11th selection?" I realized the no matter what number is selected, it will be the same number of at least 1 of the other 10. That's why the computation fail, I had used up all the available numbers in the set. So, assigning another variable would not work. I'd just be repeating 1 of the numbers. Solution, assign the 11th variable to be the same as 1 of the previous 10. I this case for Pick 11, the permutation is now a permutation of a permutation and would expand to a permutation of a permutation of a permutation in fractals of 10 for higher values Pick N.

Well, I hope that's some insight to what I'm thinking about. I'll get back with you later.

Thoth,

I'm still working on this. However, I think it may have led me to something more interesting, this may become a new topic on it's own. I can not explain too much now. Thank you for sparking the idea. This may have already been doen by someone else, but I'm not sure. Most of the research I gone through tends to suggest it may not have been tried before. I'll let you know or look for a new topic in the Mathematics Forum. Thanks again.

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