Pittsburgh, PA United States Member #130598 July 20, 2012 37 Posts Offline

Posted: August 20, 2012, 1:12 am - IP Logged

Quote: Originally posted by dr san on August 19, 2012

Hello,whereistheguruunderstoodA B Eetc?,You are allowed toimprove theformula The goal,and classify eachnumber ofalottery, anddpoisput inascending orderin a lineto seefor thenext drawingonsectorline of allthe lotterynumbers It is locatedmost of the numbers,itisthe endor the beginningof the line,you can improve theformula

I'm sorry dr san but I really don't understand what you are trying to do nor how to set it up. I'll go back over what you've written when I have time and see if I can figure something out, but I think we have different approaches and I'm not sure I follow yours. I'll need some time.

Pittsburgh, PA United States Member #130598 July 20, 2012 37 Posts Offline

Posted: August 20, 2012, 1:15 am - IP Logged

Quote: Originally posted by SergeM on August 19, 2012

What is z?

Irrelevant. Z can be any arbitrary number (preferably an integer). The value of Z and the value of Y affect the value of X based on the formula you first posed. I don't really understand what these formulas are pointing to anyways. Do you need to know the answer of the formula or are you just throwing random formulas out there to test the acumen of the users? Can you solve the equations yourself?

Economy class Belgium Member #123700 February 27, 2012 4035 Posts Offline

Posted: August 20, 2012, 3:16 am - IP Logged

Quote: Originally posted by AlgorithmGuru on August 20, 2012

Irrelevant. Z can be any arbitrary number (preferably an integer). The value of Z and the value of Y affect the value of X based on the formula you first posed. I don't really understand what these formulas are pointing to anyways. Do you need to know the answer of the formula or are you just throwing random formulas out there to test the acumen of the users? Can you solve the equations yourself?

You did not isolate z. The question is what is z, so you have to isolate it and it is relevant.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2133 Posts Offline

Posted: August 20, 2012, 8:20 am - IP Logged

Hello,guru, the goalis to createa methodof the frequenciesof the numbersof aloteria.poismay be a relationshipof numbersmathematicsand physics,in the legend,are criteriafor Classify eachnumber,and then puta linein ascending orderby the valueof the classification,andsee ifthenext drawonthatsectorof the lineis locatedmost of thenumbers drawnfrom asample=49/6=12,45,06,25,35,39,42,12,09,10,36,28,17......up to 49,thento classifyeachnumberby the formula,puta linein ascending order,butthe value offrequency,bjeticoandsee wherethe linegroups thenumbers,what partdo notyou understand? 1 -raisesthe frequencyof each of thenumbers of thelottery 2 -Sortsthe numbersin descending orderoftheir frequencies. From this point,whatbecomesinteresting isthe sort orderof eachnumbersieeachnumberis associated withyour orderaccording to theirfrequencyin the file. 3 -You takeeach resultandit isthe sumof the ordersof theirqualifyingnumbers 4 -this fileis classifiedin ascending orderof the sums.Thus,the combinationsthat havethe same numberswith higher frequenciesareat the beginningof the file andthe numbers withnumberslowerfrequenciesare at the endof the file.

Of course, theassembly methodmakesthe gamemoreprobablecombinationsmovemore towards thebeginning ortowards the endof the file.

One canthus conclude thatusingthe samemethodand the sameisefficientlysignificant, consistent,and setstivemthe same size,most likelycombinationsarein regionsvery close. Itispossible to test,you can improve theguruestruratheformula,crairother variables,ok

Pittsburgh, PA United States Member #130598 July 20, 2012 37 Posts Offline

Posted: August 20, 2012, 7:54 pm - IP Logged

Quote: Originally posted by SergeM on August 20, 2012

You did not isolate z. The question is what is z, so you have to isolate it and it is relevant.

Actually the first post is in this forum said what is "X" if you know y and z. So I was solving for X. You didn't say you wanted a solution for Z. Besides I don't understand the benefit of doing your problems. Again I ask, what is your point, what are you getting at? And how are these particular equations relevant to anything other than an exercise in wit?

Pittsburgh, PA United States Member #130598 July 20, 2012 37 Posts Offline

Posted: August 20, 2012, 8:07 pm - IP Logged

Quote: Originally posted by dr san on August 20, 2012

Hello,guru, the goalis to createa methodof the frequenciesof the numbersof aloteria.poismay be a relationshipof numbersmathematicsand physics,in the legend,are criteriafor Classify eachnumber,and then puta linein ascending orderby the valueof the classification,andsee ifthenext drawonthatsectorof the lineis locatedmost of thenumbers drawnfrom asample=49/6=12,45,06,25,35,39,42,12,09,10,36,28,17......up to 49,thento classifyeachnumberby the formula,puta linein ascending order,butthe value offrequency,bjeticoandsee wherethe linegroups thenumbers,what partdo notyou understand? 1 -raisesthe frequencyof each of thenumbers of thelottery 2 -Sortsthe numbersin descending orderoftheir frequencies. From this point,whatbecomesinteresting isthe sort orderof eachnumbersieeachnumberis associated withyour orderaccording to theirfrequencyin the file. 3 -You takeeach resultandit isthe sumof the ordersof theirqualifyingnumbers 4 -this fileis classifiedin ascending orderof the sums.Thus,the combinationsthat havethe same numberswith higher frequenciesareat the beginningof the file andthe numbers withnumberslowerfrequenciesare at the endof the file.

Of course, theassembly methodmakesthe gamemoreprobablecombinationsmovemore towards thebeginning ortowards the endof the file.

One canthus conclude thatusingthe samemethodand the sameisefficientlysignificant, consistent,and setstivemthe same size,most likelycombinationsarein regionsvery close. Itispossible to test,you can improve theguruestruratheformula,crairother variables,ok

I understand a little better thank you for the clarification. I guess I still don't see how the intial formula you described of "F = ( A - B + F ) / 3" fits into the schema. Let me clarify again though your process. You are saying:

1. Find the frequency of each possible number in the set (i.e. 1-49 in a 49/6 game)

2. Create a frequency value that is incremented the more frequent the number appears.

3. List the numbers with highest frequency first in a list (or file/table whatever)

4. take each result and it is the sum of the orders of their qualifying numbers.... this I don't understand. What is the qualifying numbers? If the number drawn was "32" would the qualifying numbers be "3" and "2" and the sum "5"?

5. List the sum of the qualifying numbers in ascending order in a seperate table.

(Or am I supposed to determine which numbers have the highest frequency and the lowest sum? Similar to alphabetizing names, start with the last name, then the first. So a number with a high frequency and low sum is at the top of the list? and are you talking about one list or two?)

Then once you have the list, how do you calculate which numbers to play? Is it simply the topmost numbers? This to me is a little illogical and part of the gamblers fallacy. The gamblers fallacy has two sides. One side says what happens least is most likely to happen next and the other side says what happens most is most likely to happen next. I personally believe the frequency of numbers is virtually irrelevant. Unless there is demonstrable proof that certain numbers are favored in which case those numbers MAY be biased. However a data sample of only a few hundred numbers (even a few thousand) may not be enough of a sample to indicate bias. Considering if there is a bias it is most likely uninentional and very small. If the bias wasn't small someone would have found it by now and be taking full advantage of it. Please further explain your initial formula, part 4 of your system, the final usage of the computations and what data pool you are using and I'll see about working on it. Thanx.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2133 Posts Offline

Posted: August 20, 2012, 11:05 pm - IP Logged

Quote: Originally posted by AlgorithmGuru on August 20, 2012

I understand a little better thank you for the clarification. I guess I still don't see how the intial formula you described of "F = ( A - B + F ) / 3" fits into the schema. Let me clarify again though your process. You are saying:

1. Find the frequency of each possible number in the set (i.e. 1-49 in a 49/6 game)

2. Create a frequency value that is incremented the more frequent the number appears.

3. List the numbers with highest frequency first in a list (or file/table whatever)

4. take each result and it is the sum of the orders of their qualifying numbers.... this I don't understand. What is the qualifying numbers? If the number drawn was "32" would the qualifying numbers be "3" and "2" and the sum "5"?

5. List the sum of the qualifying numbers in ascending order in a seperate table.

(Or am I supposed to determine which numbers have the highest frequency and the lowest sum? Similar to alphabetizing names, start with the last name, then the first. So a number with a high frequency and low sum is at the top of the list? and are you talking about one list or two?)

Then once you have the list, how do you calculate which numbers to play? Is it simply the topmost numbers? This to me is a little illogical and part of the gamblers fallacy. The gamblers fallacy has two sides. One side says what happens least is most likely to happen next and the other side says what happens most is most likely to happen next. I personally believe the frequency of numbers is virtually irrelevant. Unless there is demonstrable proof that certain numbers are favored in which case those numbers MAY be biased. However a data sample of only a few hundred numbers (even a few thousand) may not be enough of a sample to indicate bias. Considering if there is a bias it is most likely uninentional and very small. If the bias wasn't small someone would have found it by now and be taking full advantage of it. Please further explain your initial formula, part 4 of your system, the final usage of the computations and what data pool you are using and I'll see about working on it. Thanx.

Hello,guru,thanks for makingtheitem 4may leaveout.The purpose of theformulaof frequencies,is to see ifa giventime,jointhe frequenciesin asectorof the line, Howyou getthe line,and applying theformula,theresultof theregistrationof a lottery An exampleof a lottery=49/6,after seeingthe valueof each digitof thefinalformula,willgive anexample value=thenumber26 afterapplying theformulagiven=11.213and10.09747gave(isan example)inthelistnumber47comes beforeis placedona linein ascending order Thevaluenumbersisthen....47.26... , theformula for the division seems to be the best = She'll provide the average between the highest and lowest range of occurrence of number, wGuruhich is not necessarily equal to the average of all intervals,when donebymultiplication,when youfully understand theformula,will beeasy to dothe rest,you will needto do somestatisticsof past resultsis logical.An example of the49/6but can be anylottery,thegoalis to providemediaformulaoffrequencyranges,eachofnumbersthatgoes tothat clusterin asectororat the beginning orend of the line,item 4can delete,

United States Member #123632 February 25, 2012 156 Posts Offline

Posted: August 21, 2012, 3:52 am - IP Logged

Ok hi this is scientistman I am the one who started the discussion. Thank you all for your input.

I will attempt to break it down somewhat.

I am only attempting to see if algebra might work for my formula.

I am not wanting to give my formual away so I hope if someone wins with it they might remember me.

I should have included more than x y and z so I will now.

We will use v w x y and z if this is ok to do in algebra

I know (v) and I know (w) and I know (x) I just dont know (y) and if I know (y) then I know (z)

(y) is somewhat unpredictible yet might be possibly attainable through algebra. Maybe?

If I can attain (y) I can attain the winning numbers in any lottery every single time.

So for the example

lets use 2 7 8 and 10 11 18

so it would look like this

02 07 08 is (v)

10 11 18 is (w)

_____________

= 8 4 10 (x)

so

10 - 2 = 8

11 - 7 = 4

18 - 8 = 10

I still dont know ( y )

so lets say (y) is 12 15 19, now 12 15 19 is not the answer its (y) and (y) is what you need to arrive at the answer, with (y) you can attain (z) or any winning lottery numbers anywhere in the world before it occurs.

but to arrive at (z) the answer! I need (y)

So if you know (y) you take (x) and (y) and subtract (v)

and this gives you (z) The EXACT Lottery numbers Everytime! Without fail.

It's Right on the Money every time in any drawing anywhere in the world.

Without (y) it doesnt work, so you have to know (y) to arrive at (z)

so How would I arrive at (Y) ?

Or am I asking the impossibly question? I'm not sure cause some people are brainiacs.

I just recall a story where a math teacher handed out the tests and had two mathimatical equations on the chalk board.

One student handed in his test and did one of the problems on thechaulk board, the teach freaked out what he saw because the 2 questions on the chaulk bard where not part of the test they were the 2 UNSOLVABLE math equations ofall time and the stuent figured out one of them thinking it was just part of the test.

I hope I am not confusing anyone. I took algebra and got straight A's but I think that was just because the math teacher felt sorry for me because of a family death in my family and the whole school including the teachers felt sad for me so I think the teacher passed me, so I forgot most of what I knew.

Im just wondering if (y) can be attained if you know that (v) has to be subtracted from it to arrive at (z) or do you need the actual answer (z) first to figure out the example ?

If the answer is IMPOSSIBLE without me giving you (z) then I understand I just wanted to throw it out there. Thanks

Pittsburgh, PA United States Member #130598 July 20, 2012 37 Posts Offline

Posted: August 21, 2012, 8:58 am - IP Logged

Dr san, I understand, unfortunately I'm not there yet. I plan on doing statistical analysis at some later point in time but its not what I'm focused on right now. When I get there, if I come up with something I'll definitely let you know. Thanx for all the input. Good luck.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2133 Posts Offline

Posted: August 21, 2012, 9:11 am - IP Logged

Quote: Originally posted by AlgorithmGuru on August 21, 2012

Dr san, I understand, unfortunately I'm not there yet. I plan on doing statistical analysis at some later point in time but its not what I'm focused on right now. When I get there, if I come up with something I'll definitely let you know. Thanx for all the input. Good luck.

Helloguruthanksforreplying,the big problemisthe creationof the code Statistics.then theformulaiseasier theneverynumber ofa lottery Analyzedbyformula,havean average, which willin timejoina sectorof the line Guru,you can alsodo topick3andpick4,or a49/6,door spend theformula Onlythelastdigitwhere a49/6or similar,thenplacingthefirstpart (random)untilbecauseranges from 0to 4 (firstdigit49/6

United States Member #130795 July 25, 2012 80 Posts Offline

Posted: August 23, 2012, 8:13 am - IP Logged

dr san wrote: ``04 = (A)> (greater than) space between the draws. 05 = (B) <less than) space between the draws``

I am having trouble understanding what you want. But if it has something to do with determining the spacing between drawing the same number (or set of numbers), perhaps the following will be helpful.

Consider the "mega" number of the Mega Millions lottery. There are 46 "mega" numbers. The probability of the same "mega" number appearing k drawings later (and not before) is given by the binomial distribution. Specifically: (1/46)*(1 - 1/46)^(k-1), where the operator "^" means "to the power of". (For example, 4^3 = 4*4*4 = 64.)

For example, suppose the most-recently drawn "mega" number is 30 (21 Aug 2012). The probability that 30 will be the "mega" number in the next drawing is 1/46. The probability that 30 will be the "mega" number in 2 drawings later (and not before) is (1/46)*(1-1/46). In 3 drawings later (and not before): (1/46)*(1 - 1/46)^2.

If we compute that probability for k=1,2,3,..., we find that 50% of the time, 30 (really any particular "mega" number) will be appear again 13 to 63 drawings later (that is, the middle two quartiles: 25% to 75% cumulative probability).

I have analyzed the 742 Mega Million drawings from 24 June 2005 through 31 July 2012 (I'm a little behind). And in fact, the same "mega" number appears again 11 to 58 drawings later. Pretty close to expectations based on statistical theory.

The statistical theory is similar for the "distance" between drawing a particular number among the 5 "regular" numbers (non-megas). There are 56 "regular" numbers. So the probability of including a particular "regular" number in the next drawing is COMBIN(55,4) / COMBIN(56,5), or about 8.93%.

So I would expect a particular number from the last drawing to appear in the next drawing with a probability of 8.93%. In 2 drawings later (and not before) with probability 8.93%*(1 - 8.93%). In 3 drawings later (and not before) with probability 8.93%*(1 - 8.93%)^2.

So 50% of the time, I would expect a particular number from the last drawing to appear again 4 to 8 drawings later (again, the middle two quartiles: 25% to 75% probability). I have not vetted that against the actual historical drawings.

But that is just one particular number. I find it more useful to ask: given a selection of 5 "regular" numbers, how recently should we expect to have seen __any__ 1, 2 or 3 of our selected numbers in past drawings.

Based on statistical theory, I would expect any 1 of 5 selected "regular" numbers to appear 2 to 4 drawings earlier 50% of the time. Any 2 numbers to appear 6 to 13 drawings earlier. Any 3 numbers to appear 88 to 212 drawings earlier.

Analyzing the 742 historical Mega Millions drawings, I find that any 1 of 5 selected "regular" numbers appears 1 to 4 drawings earlier 50% of the time. Any 2 numbers appear 6 to 25 drawings earlier. Any 3 numbers appear 46 to 225 drawings earlier. It is very rare for any 4 or 5 numbers to appear in a previous drawing. Again, pretty close to expectations based on statistical theory.

So in choosing 5 numbers to play in the next drawing, we might reject any combination that matches 4 or 5 numbers of any previous drawings, or that matches 3 numbers within the last 45 drawings or no recently than 226 drawings earlier. Et cetera.

Of course, such a choice does not alter the probabililty of matching some or all numbers in the next drawing.

It is just a rationalization for choosing ("filtering") among the 3.8 million 5-of-56 combinations that we might play. It just makes us feel better about our choices.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2133 Posts Offline

Posted: August 23, 2012, 4:34 pm - IP Logged

Quote: Originally posted by mathhead on August 23, 2012

dr san wrote: ``04 = (A)> (greater than) space between the draws. 05 = (B) <less than) space between the draws``

I am having trouble understanding what you want. But if it has something to do with determining the spacing between drawing the same number (or set of numbers), perhaps the following will be helpful.

Consider the "mega" number of the Mega Millions lottery. There are 46 "mega" numbers. The probability of the same "mega" number appearing k drawings later (and not before) is given by the binomial distribution. Specifically: (1/46)*(1 - 1/46)^(k-1), where the operator "^" means "to the power of". (For example, 4^3 = 4*4*4 = 64.)

For example, suppose the most-recently drawn "mega" number is 30 (21 Aug 2012). The probability that 30 will be the "mega" number in the next drawing is 1/46. The probability that 30 will be the "mega" number in 2 drawings later (and not before) is (1/46)*(1-1/46). In 3 drawings later (and not before): (1/46)*(1 - 1/46)^2.

If we compute that probability for k=1,2,3,..., we find that 50% of the time, 30 (really any particular "mega" number) will be appear again 13 to 63 drawings later (that is, the middle two quartiles: 25% to 75% cumulative probability).

I have analyzed the 742 Mega Million drawings from 24 June 2005 through 31 July 2012 (I'm a little behind). And in fact, the same "mega" number appears again 11 to 58 drawings later. Pretty close to expectations based on statistical theory.

The statistical theory is similar for the "distance" between drawing a particular number among the 5 "regular" numbers (non-megas). There are 56 "regular" numbers. So the probability of including a particular "regular" number in the next drawing is COMBIN(55,4) / COMBIN(56,5), or about 8.93%.

So I would expect a particular number from the last drawing to appear in the next drawing with a probability of 8.93%. In 2 drawings later (and not before) with probability 8.93%*(1 - 8.93%). In 3 drawings later (and not before) with probability 8.93%*(1 - 8.93%)^2.

So 50% of the time, I would expect a particular number from the last drawing to appear again 4 to 8 drawings later (again, the middle two quartiles: 25% to 75% probability). I have not vetted that against the actual historical drawings.

But that is just one particular number. I find it more useful to ask: given a selection of 5 "regular" numbers, how recently should we expect to have seen __any__ 1, 2 or 3 of our selected numbers in past drawings.

Based on statistical theory, I would expect any 1 of 5 selected "regular" numbers to appear 2 to 4 drawings earlier 50% of the time. Any 2 numbers to appear 6 to 13 drawings earlier. Any 3 numbers to appear 88 to 212 drawings earlier.

Analyzing the 742 historical Mega Millions drawings, I find that any 1 of 5 selected "regular" numbers appears 1 to 4 drawings earlier 50% of the time. Any 2 numbers appear 6 to 25 drawings earlier. Any 3 numbers appear 46 to 225 drawings earlier. It is very rare for any 4 or 5 numbers to appear in a previous drawing. Again, pretty close to expectations based on statistical theory.

So in choosing 5 numbers to play in the next drawing, we might reject any combination that matches 4 or 5 numbers of any previous drawings, or that matches 3 numbers within the last 45 drawings or no recently than 226 drawings earlier. Et cetera.

Of course, such a choice does not alter the probabililty of matching some or all numbers in the next drawing.

It is just a rationalization for choosing ("filtering") among the 3.8 million 5-of-56 combinations that we might play. It just makes us feel better about our choices.

Hello, thank you,mathhiseasilythe largestspaceof anumberiswhen it came out(andwas drawnbackto be raffled,egthe number17,was selectedin the testreturned102test 189,so he hasspaceof 87draws, sincethesmaller spacehe leftthe testcame back at189 and194hehad atestspace(distanceof 5draws),then eachinstanceof anumber49/6.Eachnumberseeits largestspace(distanceoutput)withlessspace, sodohavethestatisticto seethe biggest and bestrangeto seemedia Space,the final,exampleis whenis whenthe number17 gave194,and200areon the test Soespaçpfinal(F)is 6okpleasesaysyou understandthis part? Goalafter classifyingeachformulaputthenumberin ascending orderin a lineto seeinthe nextsectionifdalinhaisthe beginningmiddleend oftheline thatwill be The nextsweepstakes.Mathhtell methat you understandthis part?

bgonÃ§alves Brasil Member #92564 June 9, 2010 2133 Posts Offline

Posted: August 23, 2012, 4:54 pm - IP Logged

mathhhello,verygood job!!in yourprevious post,then anumber inthe draw,hewillcyclehigh (greaterdelayleaving)cyclesandlow(lessdelayexiting) the final space(f),isthelast testthat came outfor thelast testdrawn, becausethere seems to bethe studyof the behavior ofalotterynumber,thatthere is a relationshipbetween mathematicsandphysics, whichdetermines alogicalreasonfora numberout,we need to makeclearstatisticsof eachnumber,toqualify.Mathh, youmay be about toinventa greatformula,frequenciesand probabilitiesofthe creditwill be all yours,will bethebestthere isout thereforis based onstatisticsof frequencies,thiswillto create, thanksmathh