Standard deviation of lotto sets

Reply #17

Nick Koutras — Thu, 30 Dec 2004 05:10:22 GMT

Quote: Originally posted by Hyperdimension on December 26, 2004 Hi, Ion Saliu has a program call FORMULA.exe and Superformula, both programs calculate the Standard deviation for an dvent of probability p in N of binomial dvents,I'll use Superformula for the next example,The program calculates p as a fraction of 2 values, 6 in 49 in this case,1st element of the fraction p = 62nd element of the fraction p = 49Enter the number of trials, N =2000Results: The standard deviation for an dvent of probab... [ More ]

Reply #16

Hyperdimension — Mon, 27 Dec 2004 15:25:14 GMT

Hi Johnph77, I understand your first explanation very well (excellent information), also the second one.. The reason I put continuing with the problem was not refering to your explanation Johnph77, it was because Bertil said in a previous post Hi, your comment is unrelated to the problem we were trying to solve , The problem of Bertil is Standard deviation of lotto sets, but when I start analyzing the numbers my computer crash.. If someone could help with this problem..Regards

Reply #15

Fenix — Mon, 27 Dec 2004 08:12:51 GMT

Really can't find any importance in this matter. If points around any curve represent draws, for the very next draw all 14 million points in the field hold the same mathematical probability 1 : 14 millions.

Reply #14

johnph77 — Mon, 27 Dec 2004 07:39:47 GMT

This is starting to bug me for some reason. I'm going to try this one more time as I feel my previous explanation was inadequate.I have a random number generator generating tickets for the lottery example given above. Defying all the odds in the known universe, the RNG has given me 13,983,815 different sets of numbers in the same amount of draws. What are the odds of drawing that last possibility in the 13,983,816th draw?You got it - the same odds of predicting any given set of numbers in the lo... [ More ]

Reply #13

Hyperdimension — Mon, 27 Dec 2004 06:51:11 GMT

Hi, Thank you for the explanation johnph77, Continuing with the problem, first I created the full wheel 6/49, in total 13,983,816 tickets, obtaining the next results.. Valid N Mean Minimum Maximum Std.Dev.1var 13983816 7.14286 1.000000 44.00000 5.7365642var 13983816 14.28571 2.000000 45.00000 7.4058723var 13983816 21.42857 3.000000 46.00000 8.1127264var 13983816 28.57143 4.000000 47.00000 8.1... [ More ]

Reply #12

johnph77 — Mon, 27 Dec 2004 05:55:13 GMT

Bertil -hyperdimension's math is based on probability, not possibility. If a lottery is completely random in its generation of tickets, and if there are 13,983,816 possibilities and the same number of tickets are sold, it is almost a mathematical certainty that all the tickets sold will not cover all the possibilities in the lottery. The odds of that happening far exceed the chances of winning.When the first ticket is sold, it is unique - there is no other ticket currently in that lottery with t... [ More ]

Reply #11

Bertil — Mon, 27 Dec 2004 02:39:31 GMT

Quote: Originally posted by Hyperdimension on December 26, 2004Hi,The right person to answer your question is Mr. Ion Saliu,I find an interesting article about binomial distribution, with the next example:Poisson DistributionIn extreme cases, very small p so that the standard deviation is not much less than the mean, the Gaussian Distribution is not appropriate, but a different approximation is: the Poisson Distribution. Going back to the Binomial Distribution (which is still exact), we only nee... [ More ]

Reply #10

Hyperdimension — Sun, 26 Dec 2004 21:41:55 GMT

Hi,The right person to answer your question is Mr. Ion Saliu,I find an interesting article about binomial distribution, with the next example:Poisson DistributionIn extreme cases, very small p so that the standard deviation is not much less than the mean, the Gaussian Distribution is not appropriate, but a different approximation is: the Poisson Distribution. Going back to the Binomial Distribution (which is still exact), we only need to worry about values of n much smaller than N.In the Indiana... [ More ]

Reply #9

Bertil — Sun, 26 Dec 2004 13:24:51 GMT

Quote: Originally posted by Hyperdimension on December 26, 2004Hi,Ion Saliu has a program call FORMULA.exe and Superformula, both programs calculate the Standard deviation for an dvent of probability p in N of binomial dvents,I'll use Superformula for the next example,The program calculates p as a fraction of 2 values, 6 in 49 in this case,1st element of the fraction p = 62nd element of the fraction p = 49Enter the number of trials, N =2000Results:The standard deviation for an dvent of probabili... [ More ]

Reply #8

Hyperdimension — Sun, 26 Dec 2004 06:14:06 GMT

Hi, Ion Saliu has a program call FORMULA.exe and Superformula, both programs calculate the Standard deviation for an dvent of probability p in N of binomial dvents,I'll use Superformula for the next example,The program calculates p as a fraction of 2 values, 6 in 49 in this case,1st element of the fraction p = 62nd element of the fraction p = 49Enter the number of trials, N =2000Results: The standard deviation for an dvent of probability p = .12244898 in 2000 binomial experiments is

Reply #7

Bertil — Sat, 25 Dec 2004 23:00:40 GMT

Quote: Originally posted by Hyperdimension on December 25, 2004Hi,13.85641 is very close to 14.1, the difference is common probability vs actual statisticsRegards Hi Hype, The value 13.856 must come from SQRT 192, which comes from (49-1)^2/12. But this formula apllies to a continuous uniform distribution. But we are here dealing wwith a discrete unif.distr. for which the formula is (N^2-1)/2, which wold yield 14.14. But that value refers to a single integer sam... [ More ]

Reply #6

Hyperdimension — Sat, 25 Dec 2004 18:55:46 GMT

Hi, 13.85641 is very close to 14.1, the difference is common probability vs actual statisticsRegards

Reply #5

Bertil — Sat, 25 Dec 2004 14:14:53 GMT

Quote: Originally posted by Nick Koutras on December 24, 2004Quote: Originally posted by Bertil on December 22, 2004 For each lotto game one can add the numbers in a set and get a sum, which will form a nearly perfect normal curve if all combinations are used. But each set has a mean and a std.dev. , which will not form a normal curve because there will be more of the lowest value than of the highest. E.g. in the 5/53 lotto part of Powerball the lowest std.dev. will be 1... [ More ]

Reply #4

Bertil — Sat, 25 Dec 2004 14:07:18 GMT

Quote: Originally posted by Fenix on December 24, 2004How did you get 1.58 for the 1-2-3-4-5 set? I got the numbers from my hand-held calculator. If you calculate the s.d. by hand you will get sqrt 10/4 as a sample but sqrt 10/5 as a population. Here we are dealing with samples. Bertil

Reply #3

Nick Koutras — Sat, 25 Dec 2004 02:15:15 GMT

Quote: Originally posted by Bertil on December 22, 2004 For each lotto game one can add the numbers in a set and get a sum, which will form a nearly perfect normal curve if all combinations are used. But each set has a mean and a std.dev. , which will not form a normal curve because there will be more of the lowest value than of the highest. E.g. in the 5/53 lotto part of Powerball the lowest std.dev. will be 1.58 for the set 1-2-3-4-5 and all othe r sets of five c... [ More ]

Reply #2

straightchaser — Fri, 24 Dec 2004 22:32:16 GMT

Bertil,There is no formula that I know of to predict the mean of the standard deviation. One can try to bootstrap the mean of the standard deviation. You seem to be quite adept at simulating random sets. Bootstrapping allows you to simulate the distribution of the standard deviation from which you can estimate the mean standard deviation. Happy bootstrapping.'chaser

Reply #1

Fenix — Fri, 24 Dec 2004 22:18:55 GMT

How did you get 1.58 for the 1-2-3-4-5 set

Standard deviation of lotto sets

Bertil — Wed, 22 Dec 2004 12:33:17 GMT

For each lotto game one can add the numbers in a set and get a sum, which will form a nearly perfect normal curve if all combinations are used. But each set has a mean and a std.dev. , which will not form a normal curve because there will be more of the lowest value than of the highest. E.g. in the 5/53 lotto part of Powerball the lowest std.dev. will be 1.58 for the set 1-2-3-4-5 and all othe r sets of five consecutive numbers, of which there must be 48, while the