Welcome Guest
You last visited December 5, 2016, 3:33 pm
All times shown are
Eastern Time (GMT-5:00)

# 649 formula

Topic closed. 109 replies. Last post 3 years ago by RJOh.

 Page 1 of 8
Horwood NL
Member #70613
February 6, 2009
296 Posts
Offline
 Posted: June 4, 2013, 9:04 am - IP Logged

Is there away to calculate the average number of games it would take for all the numbers from 1 to 49 to appear over the history of a 649 game or do you have to run down through the history 1 game at a time to check how long it takes. I'm thinking the information would be useful as a overall skip filter along with the individual number skip filters.

Kentucky
United States
Member #32652
February 14, 2006
7302 Posts
Offline
 Posted: June 4, 2013, 12:43 pm - IP Logged

Is there away to calculate the average number of games it would take for all the numbers from 1 to 49 to appear over the history of a 649 game or do you have to run down through the history 1 game at a time to check how long it takes. I'm thinking the information would be useful as a overall skip filter along with the individual number skip filters.

You'll never get an exact number because some number have gone 50, 60, or more drawings without being drawn and it's possible for all 49 numbers to be drawn in under 20 drawings. You should expect to see about 43 or 44 of the numbers drawn in 20 drawings. One or two of the other numbers might be in someone's pocket or you could guess to when those five numbers will be drawn.

Horwood NL
Member #70613
February 6, 2009
296 Posts
Offline
 Posted: June 4, 2013, 1:16 pm - IP Logged

You'll never get an exact number because some number have gone 50, 60, or more drawings without being drawn and it's possible for all 49 numbers to be drawn in under 20 drawings. You should expect to see about 43 or 44 of the numbers drawn in 20 drawings. One or two of the other numbers might be in someone's pocket or you could guess to when those five numbers will be drawn.

I know about 80% of the numbers hit in about 20 draws. What about getting an average number of times 1 to 49 are completed over the history of the draw , is that possible.

United States
Member #93947
July 10, 2010
2180 Posts
Offline
 Posted: June 4, 2013, 1:19 pm - IP Logged

I know about 80% of the numbers hit in about 20 draws. What about getting an average number of times 1 to 49 are completed over the history of the draw , is that possible.

Your question can easily be answered with a computer simulation, running it over millions of random draws.  It would give you a much more meaningful average and standard deviation because the available history in a 6/49 game is tiny compared to the number of possible outcomes.  This approach would also allow you to answer your question with as much precision as you desire by simply increasing the number of draws simulated.

Horwood NL
Member #70613
February 6, 2009
296 Posts
Offline
 Posted: June 4, 2013, 1:26 pm - IP Logged

Your question can easily be answered with a computer simulation, running it over millions of random draws.  It would give you a much more meaningful average and standard deviation because the available history in a 6/49 game is tiny compared to the number of possible outcomes.  This approach would also allow you to answer your question with as much precision as you desire by simply increasing the number of draws simulated.

But would the result be relevant to the draw you are playing since it is not useing your direct history?

United States
Member #93947
July 10, 2010
2180 Posts
Offline
 Posted: June 4, 2013, 1:39 pm - IP Logged

But would the result be relevant to the draw you are playing since it is not useing your direct history?

"But would the result be relevant to the draw you are playing since it is not useing your direct history?"

The result would be the average number of draws expected before all 49 numbers showed up.  This has no connection to any individual draw, the "one you are playing," or otherwise.  If the mechanism used to make the draws is fair, previous draws will have NO influence on current or future ones.  (And this includes Pre and Post tests!)

United States
Member #130795
July 25, 2012
80 Posts
Offline
 Posted: June 4, 2013, 2:07 pm - IP Logged

Is there away to calculate the average number of games it would take for all the numbers from 1 to 49 to appear over the history of a 649 game or do you have to run down through the history 1 game at a time to check how long it takes. I'm thinking the information would be useful as a overall skip filter along with the individual number skip filters.

Gwoof, this thread will degenerate into a lot of name-calling and other ad hominem comments very quickly.  I just want to forewarn you.

I am not familiar with any closed-end formulas for this.  Jimmy is mostly correct:  a Monte Carlo simulation might be the best way to go.  But I doubt that it requires "millions of random draws".  That depends on the degree of confidence that you want for the answer.  And that can be bootstrapped.  I'll have more details in a little while.

(It is a possible that an analysis of a history of several hundred drawings might be sufficient.  After all, that represents many hundred samplings.  But I agree with Jimmy:  I would prefer the structure of simulation, if only to validate conclusions from the historical drawings.)

But first, you need to buy into the concept that lottery drawings are completely random and completely independent.

That is indeed a fact.  But there are many vocal participants in these forums who do not believe that.  And there are others that accept the randomness of the process, but somehow they believe we can predict future draws from past draws -- or from silly arithmetic progression (which is really just a random number generator, usually a poor one).

(Note:  That is not to dismiss those who like to use the probabilities of patterns of drawings in order to reduce the numbers that they select.  Any such reduction is a personal choice.  And all such choices are valid, although some might be better than others statistically speaking.)

If you do buy into the concept of random and independent drawings, you might as well ignore anything that Jimmy or I have to say on the matter.

PS:  You also have to buy into the assertion that the random number generator used in the simulation is sufficiently well implemented.  Honestly, that is always debatable.  But IMHO, we have to take some things on faith until proven otherwise.

Horwood NL
Member #70613
February 6, 2009
296 Posts
Offline
 Posted: June 4, 2013, 3:07 pm - IP Logged

Gwoof, this thread will degenerate into a lot of name-calling and other ad hominem comments very quickly.  I just want to forewarn you.

I am not familiar with any closed-end formulas for this.  Jimmy is mostly correct:  a Monte Carlo simulation might be the best way to go.  But I doubt that it requires "millions of random draws".  That depends on the degree of confidence that you want for the answer.  And that can be bootstrapped.  I'll have more details in a little while.

(It is a possible that an analysis of a history of several hundred drawings might be sufficient.  After all, that represents many hundred samplings.  But I agree with Jimmy:  I would prefer the structure of simulation, if only to validate conclusions from the historical drawings.)

But first, you need to buy into the concept that lottery drawings are completely random and completely independent.

That is indeed a fact.  But there are many vocal participants in these forums who do not believe that.  And there are others that accept the randomness of the process, but somehow they believe we can predict future draws from past draws -- or from silly arithmetic progression (which is really just a random number generator, usually a poor one).

(Note:  That is not to dismiss those who like to use the probabilities of patterns of drawings in order to reduce the numbers that they select.  Any such reduction is a personal choice.  And all such choices are valid, although some might be better than others statistically speaking.)

If you do buy into the concept of random and independent drawings, you might as well ignore anything that Jimmy or I have to say on the matter.

PS:  You also have to buy into the assertion that the random number generator used in the simulation is sufficiently well implemented.  Honestly, that is always debatable.  But IMHO, we have to take some things on faith until proven otherwise.

I hope it doesn't end up that way. I'm not that good at math and I was just wondering if there was a simple formula which I could use to determine that average or maybe someone has already worked it out since it doesn't matter how many or what data base one uses. I just assumed it did made a differance the set of numbers you would use to determine the average number of games it would take to see all 49 numbers in the game one is playing.

United States
Member #130795
July 25, 2012
80 Posts
Offline
 Posted: June 4, 2013, 3:25 pm - IP Logged

I hope it doesn't end up that way. I'm not that good at math and I was just wondering if there was a simple formula which I could use to determine that average or maybe someone has already worked it out since it doesn't matter how many or what data base one uses. I just assumed it did made a differance the set of numbers you would use to determine the average number of games it would take to see all 49 numbers in the game one is playing.

I understand.  According to my simulation (I can post the code, if you like), on average, it takes about 220 draws to see all 49 numbers.  That is about 37 drawings of 6 numbers each.

With just 4000 simulated sets of draws, we can determine the average within about +/-1% (217 to 223) with 99% confidence.

PS:  It would be useful to compare that with the history of some 6/49 lottery, ideally with at least 667 drawings.  I have not done that (yet).

PPS:  Note that 220 is an average.  In the simulation, it took as few as 99 and as many as 552 draws to see all 49 numbers.  That is 17 to 92 drawings of 6 numbers each.  I would expect a range of 49 (-2.86sd) to 459 (+4sd) draws; that is 9 to 77 drawings of 6 numbers each.

United States
Member #130795
July 25, 2012
80 Posts
Offline
 Posted: June 4, 2013, 4:33 pm - IP Logged

I understand.  According to my simulation (I can post the code, if you like), on average, it takes about 220 draws to see all 49 numbers.  That is about 37 drawings of 6 numbers each.

With just 4000 simulated sets of draws, we can determine the average within about +/-1% (217 to 223) with 99% confidence.

PS:  It would be useful to compare that with the history of some 6/49 lottery, ideally with at least 667 drawings.  I have not done that (yet).

PPS:  Note that 220 is an average.  In the simulation, it took as few as 99 and as many as 552 draws to see all 49 numbers.  That is 17 to 92 drawings of 6 numbers each.  I would expect a range of 49 (-2.86sd) to 459 (+4sd) draws; that is 9 to 77 drawings of 6 numbers each.

Errata....  I made a mistake in the simulation.  Drawings (6 numbers at a time) are completely independent.  But of course, each 6 numbers drawn are dependent since they are drawn without replacement.

On average, it takes about 539 draws to see all 49 numbers; 90 drawings of 6 numbers each.

It takes about 7200 simulated sets of draws (1200 drawings of 6) to ensure that we can determine the average within about +/-1% (533 to 545) with 99% confidence.

Again, that is an average.  In the simulation, it took as few as 175 and as many as 1969 draws to see 49 numbers.  That is 30 to 329 drawings of 6 numbers each.  The expected range is 49 (-2.75sd) to 1251 (+4sd) draws; that is 9 to 209 drawings of 6 numbers each.

(Actually, there were 10 simulated sets -- about 0.15% -- that exceeded 2004 draws, the arbitrary max number for each simulated set.)

mid-Ohio
United States
Member #9
March 24, 2001
19825 Posts
Online
 Posted: June 4, 2013, 4:55 pm - IP Logged

Is there away to calculate the average number of games it would take for all the numbers from 1 to 49 to appear over the history of a 649 game or do you have to run down through the history 1 game at a time to check how long it takes. I'm thinking the information would be useful as a overall skip filter along with the individual number skip filters.

I checked the last 260 cycle of Ohio's Classic Lotto (6/49) to see how long it took for all 49 numbers to appear and it varied for 19-56 drawings.

0: 0  10: 0  20: 1  30: 15  40: 10 50: 4
1: 0  11: 0  21: 1  31: 14  41: 8  51: 3
2: 0  12: 0  22: 3  32: 13  42: 8  52: 3
3: 0  13: 0  23: 3  33: 13  43: 6  53: 3
4: 0  14: 0  24: 6  34: 12  44: 5  54: 2
5: 0  15: 0  25: 7  35: 11  45: 4  55: 2
6: 0  16: 0  26: 9  36: 11  46: 4  56: 1
7: 0  17: 0  27: 11 37: 10  47: 4  57: 0
8: 0  18: 0  28: 11 38: 10  48: 3  58: 0
9: 0  19: 1  29: 14 39: 11  49: 3  59: 0

The most recent one took 30 drawings

* you don't need to buy more tickets, just buy a winning ticket *

United States
Member #130795
July 25, 2012
80 Posts
Offline
 Posted: June 4, 2013, 5:12 pm - IP Logged

I checked the last 260 cycle of Ohio's Classic Lotto (6/49) to see how long it took for all 49 numbers to appear and it varied for 19-56 drawings.

0: 0  10: 0  20: 1  30: 15  40: 10 50: 4
1: 0  11: 0  21: 1  31: 14  41: 8  51: 3
2: 0  12: 0  22: 3  32: 13  42: 8  52: 3
3: 0  13: 0  23: 3  33: 13  43: 6  53: 3
4: 0  14: 0  24: 6  34: 12  44: 5  54: 2
5: 0  15: 0  25: 7  35: 11  45: 4  55: 2
6: 0  16: 0  26: 9  36: 11  46: 4  56: 1
7: 0  17: 0  27: 11 37: 10  47: 4  57: 0
8: 0  18: 0  28: 11 38: 10  48: 3  58: 0
9: 0  19: 1  29: 14 39: 11  49: 3  59: 0

The most recent one took 30 drawings

Good input!  That's within about -2.39sd to -1.14sd of what I would expect.  I would expect to have to look at 1130 to 1200 drawings of 6 numbers each in order to know the historical average with any reasonable degree of confidence.  That's about 11.5 years (!).

mid-Ohio
United States
Member #9
March 24, 2001
19825 Posts
Online
 Posted: June 4, 2013, 5:31 pm - IP Logged

Good input!  That's within about -2.39sd to -1.14sd of what I would expect.  I would expect to have to look at 1130 to 1200 drawings of 6 numbers each in order to know the historical average with any reasonable degree of confidence.  That's about 11.5 years (!).

If you have a data file with all the data in order by date of drawing you can just do a physical count back or write a routine to do it, that what I did.   Ohio has three drawings a week and 977 drawings represent drawings going back to 01/22/2007 when the game was started.

* you don't need to buy more tickets, just buy a winning ticket *

bgonÃ§alves
Brasil
Member #92564
June 9, 2010
2122 Posts
Offline
 Posted: June 4, 2013, 6:42 pm - IP Logged

hello, mathead if I split the lottery 49/6 or any other, in two digits starting
Go 0-4 and final digits 0-9 in the case of 49/6, as one might use the medium to come in favor of the gambler, and knew the probabidade is average, this engulfing the punter
So instead of predicting numbers we provide small segments (patterns) or the number or not this this. Because you need the numbers is an event very narrow with many variants
Being unique event, now small segments (small tens of rows and columns) in different matrices and so without thinking of the numbers, of course you have more bets, but increases confinça when playing in groups as one example of groups of mathematicians who one is teamed won jacpot

Kentucky
United States
Member #32652
February 14, 2006
7302 Posts
Offline
 Posted: June 4, 2013, 10:21 pm - IP Logged

"But would the result be relevant to the draw you are playing since it is not useing your direct history?"

The result would be the average number of draws expected before all 49 numbers showed up.  This has no connection to any individual draw, the "one you are playing," or otherwise.  If the mechanism used to make the draws is fair, previous draws will have NO influence on current or future ones.  (And this includes Pre and Post tests!)

"This has no connection to any individual draw"

IMAO!

 Page 1 of 8