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# Are you good at Algebra?

Topic closed. 32 replies. Last post 1 year ago by Scientistman.

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Station C:\somewhere\in\space\time
Fiji
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February 27, 2012
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 Posted: August 19, 2012, 6:14 pm - IP Logged

x = (the (Z -2) root of 1246)) / 2y

What is z?

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

Hello, where is the guru understood A B E etc?, You are allowed to improve the formula
The goal, and classify each number of a lottery, and dpois put in ascending order in a line to see for the next drawing on sector line of all the lottery numbers
It is located most of the numbers, it is the end or the beginning of the line, you can improve the formula

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.

If it's not consistent, it's irrelevant :)

Pittsburgh, PA
United States
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 Posted: August 20, 2012, 1:15 am - IP Logged

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?

If it's not consistent, it's irrelevant :)

Station C:\somewhere\in\space\time
Fiji
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 Posted: August 20, 2012, 3:16 am - IP Logged

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
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June 9, 2010
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 Posted: August 20, 2012, 8:20 am - IP Logged

Hello, guru, the goal is to create a method of the frequencies of the numbers of a loteria.pois may be a relationship of numbers mathematics and physics, in the legend, are criteria for
Classify each number, and then put a line in ascending order by the value of the classification, and see if the next draw on that sector of the line is located most of the numbers drawn from a sample = 49/6 = 12,45,06, 25,35,39,42,12,09,10,36,28,17 ...... up to 49, then to classify each number by the formula, put a line in ascending order, but the value of frequency, bjetico and see where the line groups the numbers, what part do not you understand?
1 - raises the frequency of each of the numbers of the lottery
2 - Sorts the numbers in descending order of their frequencies.
From this point, what becomes interesting is the sort order of each numbers ie each number is associated with your order according to their frequency in the file.
3 - You take each result and it is the sum of the orders of their qualifying numbers
4 - this file is classified in ascending order of the sums. Thus, the
combinations that have the same numbers with higher frequencies are at the beginning of the file and the numbers with numbers lower frequencies are at the end of the file.

Of course, the assembly method makes the game more probable combinations move more towards the beginning or towards the end of the file.

One can thus conclude that using the same method and the same is efficiently significant, consistent, and sets tivem the same size, most likely combinations are in regions very close.
It is possible to test, you can improve the guru estrura the formula, crair other variables, ok

Pittsburgh, PA
United States
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July 20, 2012
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 Posted: August 20, 2012, 7:54 pm - IP Logged

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?

If it's not consistent, it's irrelevant :)

Pittsburgh, PA
United States
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July 20, 2012
37 Posts
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 Posted: August 20, 2012, 8:07 pm - IP Logged

Hello, guru, the goal is to create a method of the frequencies of the numbers of a loteria.pois may be a relationship of numbers mathematics and physics, in the legend, are criteria for
Classify each number, and then put a line in ascending order by the value of the classification, and see if the next draw on that sector of the line is located most of the numbers drawn from a sample = 49/6 = 12,45,06, 25,35,39,42,12,09,10,36,28,17 ...... up to 49, then to classify each number by the formula, put a line in ascending order, but the value of frequency, bjetico and see where the line groups the numbers, what part do not you understand?
1 - raises the frequency of each of the numbers of the lottery
2 - Sorts the numbers in descending order of their frequencies.
From this point, what becomes interesting is the sort order of each numbers ie each number is associated with your order according to their frequency in the file.
3 - You take each result and it is the sum of the orders of their qualifying numbers
4 - this file is classified in ascending order of the sums. Thus, the
combinations that have the same numbers with higher frequencies are at the beginning of the file and the numbers with numbers lower frequencies are at the end of the file.

Of course, the assembly method makes the game more probable combinations move more towards the beginning or towards the end of the file.

One can thus conclude that using the same method and the same is efficiently significant, consistent, and sets tivem the same size, most likely combinations are in regions very close.
It is possible to test, you can improve the guru estrura the formula, crair other 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.

If it's not consistent, it's irrelevant :)

bgonÃ§alves
Brasil
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June 9, 2010
1322 Posts
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 Posted: August 20, 2012, 11:05 pm - IP Logged

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 making the item 4 may leave out. The purpose of the formula of frequencies, is to see if a given time, join the frequencies in a sector of the line,
How you get the line, and applying the formula, the result of the registration of a lottery
An example of a lottery = 49/6, after seeing the value of each digit of the final formula, will give an example value = the number 26 after applying the formula given = 11.213 and 10.097 47 gave (is an example) in the list number 47 comes before is placed on a line in ascending order
The value numbers is then .... 47.26 ...
, the formula 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 done by multiplication, when you fully understand the formula, will be easy to do the rest, you will need to do some statistics of past results is logical. An example of the 49/6 but can be any lottery, the goal is to provide media formula of frequency ranges, each of numbers that goes to that cluster in a sector or at the beginning or end of the line, item 4 can delete,

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 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
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 Posted: August 21, 2012, 8:57 am - IP Logged

Sorry ScientestMan, not information for me to work off of.  But good luck in solving it yourself.

If it's not consistent, it's irrelevant :)

Pittsburgh, PA
United States
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 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.

If it's not consistent, it's irrelevant :)

bgonÃ§alves
Brasil
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June 9, 2010
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 Posted: August 21, 2012, 9:11 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.

Hello guru thanks for replying, the big problem is the creation of the code
Statistics. then the formula is easier then every number of a lottery
Analyzed by formula, have an average, which will in time join a sector of the line
Guru, you can also do to pick3 and pick 4, or a 49/6, do or spend the formula
Only the last digit where a 49/6 or similar, then placing the first part (random) until because ranges from 0 to 4 (first digit 49/6

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 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
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 Posted: August 23, 2012, 4:34 pm - 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.

Hello, thank you, mathh is easily the largest space of a number is when it came out (and was drawn back to be raffled, eg the number 17, was selected in the test returned 102 test
189, so he has space of 87 draws, since the smaller space he left the test came back at 189 and 194 he had a test space (distance of 5 draws), then each instance of a number 49/6. Each number see its largest space (distance output) with less space, so do have the statistic to see the biggest and best range to see media
Space, the final, example is when is when the number 17 gave 194, and 200 are on the test
So espaçp final (F) is 6 ok please says you understand this part?
Goal after classifying each formula put the number in ascending order in a line to see in the next section if dalinha is the beginning middle end of the line that will be
The next sweepstakes. Mathh tell me that you understand this part?

bgonÃ§alves
Brasil
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June 9, 2010
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 Posted: August 23, 2012, 4:54 pm - IP Logged

mathh hello, very good job!! in your previous post, then a number in the draw, he will cycle high (greater delay leaving) cycles and low (less delay exiting)
the final space (f), is the last test that came out for the last test drawn, because there seems to be the study of the behavior of a lottery number, that there is a relationship between mathematics and physics, which determines a logical reason for a number out, we need to make clear statistics of each number, to qualify. Mathh, you may be about to invent a great formula, frequencies and probabilities of the credit will be all yours, will be the best there is out there for is based on statistics of frequencies, this will to create, thanks mathh

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