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

Posted: June 7, 2013, 8:55 pm - IP Logged

Quote: Originally posted by mathhead on June 7, 2013

I cannot say with impunity what RJOh meant, but this is how I interpret his comment. My observations about about the 3,819,816 combinations of "regular" numbers for the MegaMillions 5/56. My data is a little out of date: 827 drawings from 6/24/2005 through 5/24/2013.

If I select a combination (for my ticket) that exactly matches a previous drawing, I would exclude it because there is only an 827-in-3,819,816 chance (0.02%) that the next drawing would repeat a previous drawing.

It's not that it cannot happen; all 5-tuple combinations are possible. Moreover, the chance of a duplicate will increase over time. But it's still very unlikely, to say the least.

What about 4-tuples (quads)?

Each ticket and drawing has C(5,4) = 5 quads. And there are C(56,4) = 363,175 unique quads total.

Empirically, among the past 827 drawings, there have been 4095 unique quads and 20 quads that appeared twice -- 4135 in all (= 827*5). So the current probability of a duplicate quad is 20-in-4135 (0.48%).

Again, that probability will increase over time. But since we have seen only about 1.12% of all 4-tuples (4115 / 367,290), I would guess that it will take some time for the probability of duplication quads to become significant.

So I would exclude a selection that includes a quad that has appeared in any previous drawing.

What about 3-tuples (triples)?

Each ticket and drawing has C(5,3) = 10 triples. And there are C(56,3) = 27,720 unique triples overall.

Empirically, we have see about 26% of all triples (7145 / 27,720) in the first 827 drawings. The breakdown is: 6133 triples appearing once, 905 triples appearing twice, 101 appearing 3 times, and 6 appearing 4 times. And again, the frequency of "duplicates" (including more than 2 times) will increase over time.

Also empirically, in the past year, 88% of the drawings have had 1 to 4 duplicate triples -- that is, triples that appeared in previous drawings. And 56% of the drawings have had 2 or 3 duplicates.

So I might (operative word) exclude a selection that has more than 4 duplicate triples (for now). I might even favor selections that have 2 or 3 duplicate triples (for now).

What about 2-tuples (pairs)?

I'll spare you the computational details. There are C(56,2) = 1540 pairs, and we have seen them all but one in the past 827 drawings. Each ticket and drawing has C(5,2) = 10 pairs. And in the past year, 94% of the drawings has 10 duplicate pairs -- that is, pairs that appeared in previous drawings; and the remainder have had 9 duplicate pairs.

So there is no point in trying to exclude drawings with duplicate pairs; it cannot be done. In fact, most drawings will have 10 duplicate pairs now. (One future drawing should have only 9 duplicate pairs. ;->)

Errata.... I wrote: ``Empirically, among the past 827 drawings, there have been 4095 unique quads and 20 quads that appeared twice -- 4135 in all (= 827*5). So the current probability of a duplicate quad is 20-in-4135 (0.48%).``

That conclusion is not right. And I see I failed to correct a typo in one place. "There's an excuse for that".

First, C(56,4) is 367,290. Second, in the first 827 drawings, there are no drawings with more than 1 duplicate quad -- that is, a 4-tuple that appears in a previous drawing. And in the past year, 94% of the drawings had no duplicate quads.

Again, the number of duplicates will increase over time. But for now, I would exclude a selection (ticket) that has more than 1 duplicate quad; and I might (operative word) exclude selections that have 1 duplicate quad.

It should be noted that any filtering of this sort is completely arbitrary. It does not increase the odds of a match. But we have to find a subset of the possible combinations somehow. And it just makes us feel good to choose from a subset that matches pattern(s) that arise a large percentage of time.

The only method that does increase the odds of a match is to purchase multiple tickets with no duplicate sub-combinations among them. At least, no duplicate triples and above; I ass-u-me you are not purchasing more than 27,720 tickets for one MegaMillions drawing. If you ensure there are no duplicate pairs (if you are purchasing less than 154 MM tickets), that also ensures there are no duplicate triples and above.

United States Member #93947 July 10, 2010 2180 Posts Offline

Posted: June 7, 2013, 11:50 pm - IP Logged

mathhead,

You said, "It should be noted that any filtering of this sort is completely arbitrary. It does not increase the odds of a match. But we have to find a subset of the possible combinations somehow. And it just makes us feel good to choose from a subset that matches pattern(s) that arise a large percentage of time."

It makes us feel good. That tells it all!

It should also be noted that, as with other popular filtering criteria, lotteries don't pay any more when you win with a combination that won before, for example, than they do for a "fresh" one. If they did, there might be justification for tediously poring over draw histories. But they don't, and yesterday's winner is just as likely to hit today, as any other.

United States Member #5599 July 13, 2004 1185 Posts Offline

Posted: June 8, 2013, 1:27 am - IP Logged

Quote: Originally posted by jimmy4164 on June 7, 2013

mathhead,

You said, "It should be noted that any filtering of this sort is completely arbitrary. It does not increase the odds of a match. But we have to find a subset of the possible combinations somehow. And it just makes us feel good to choose from a subset that matches pattern(s) that arise a large percentage of time."

It makes us feel good. That tells it all!

It should also be noted that, as with other popular filtering criteria, lotteries don't pay any more when you win with a combination that won before, for example, than they do for a "fresh" one. If they did, there might be justification for tediously poring over draw histories. But they don't, and yesterday's winner is just as likely to hit today, as any other.

--Jimmy4164

Hi Jimmy,

I hope you are not dead before you read this. Afterall anything is possible.

But, given your long history of survival at the LP, you probably will read this. *S*

Even though the draws are not physically connected, the results in some criteria from one draw to the next are connected. Once in a blue moon those off the wall possibilities occur, however, historically the historical data shows that they don't happen very often. Either you can believe that historically you will be alive tommorrow or that you will any at any minute now, your choice.*L*

Live long and prosper. *L*

You are a slave to the choices you have made. jk

Even a blind squirrel will occasioanlly find an acorn.

I hope you are not dead before you read this. Afterall anything is possible.

But, given your long history of survival at the LP, you probably will read this. *S*

Even though the draws are not physically connected, the results in some criteria from one draw to the next are connected. Once in a blue moon those off the wall possibilities occur, however, historically the historical data shows that they don't happen very often. Either you can believe that historically you will be alive tommorrow or that you will any at any minute now, your choice.*L*

Live long and prosper. *L*

JKING wrote: ``Even though the draws are not physically connected, the results in some criteria from one draw to the next are connected. Once in a blue moon those off the wall possibilities occur, however, historically the historical data shows that they don't happen very often.``

I hope I did not contribute to this misunderstanding. Jimmy is right; and I am right. We are answering different questions.

Consider the following analogy.... A family has 3 female children. What is the probability that the next child is female, assuming equal probabilities for male and female?

The probability of having 4 female children is 1 in 16 (6.25%) if we consider order. (This is a common debate, which we can ignore for my puproses.) But the probability that the next child is female is 1 in 2 (50%), which is unchanged by history.

The probability that a particular quad (4-tuple) will be drawn is 1 in 367,290 for a 5/56 game like MegaMillions. That probability is unchanged by history.

But the probability that a drawing will include any of the previously-drawn quads is currently about 1 in 17.85 (5.6%) for the MegaMillions game from 6/24/2005 through 5/24/2013. That is, 5*4115/376,290 because there are 5 quads in each drawing. That probability will change over time; in particular, it will become more likely.

But I was addressing the question: how many previously-drawn quads might be included in a drawing? That, too, will change over time. Honestly, I am not taking the time to compute it. Instead, I look at "recent" history to observe that it is zero about 94% of the time (for now); it is 1 for the remaining 6%. Over time, it will become more likely that each drawing will include 1, then 2 eventually up to 5 previously-drawn quads.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2124 Posts Offline

Posted: June 8, 2013, 8:01 am - IP Logged

Hellomathheadasthe formulacould workconsideringthatwe Poderiamsseparatelotteryin twodigit=initial andfinal digit, andthenhowtojointwo parts,one cansee howa party maygiveevidenceof the other andviceversa,eg,aftera studypredicted thatifthelastdigits(0-9)favorswere2,3,4,5,8,9(a lotteryfor49/6)thenthrough thesecan giveclues totheinitialdigits From0-4or viceversa,digits=000,112initialexamplemay giveevidenceof thelastdigit Whendivided into 4sectors(4 sets) to handlethecentral stripof the greatestprobability of75% to 80%, OK

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

Posted: June 8, 2013, 11:18 am - IP Logged

Quote: Originally posted by mathhead on June 8, 2013

JKING wrote: ``Even though the draws are not physically connected, the results in some criteria from one draw to the next are connected. Once in a blue moon those off the wall possibilities occur, however, historically the historical data shows that they don't happen very often.``

I hope I did not contribute to this misunderstanding. Jimmy is right; and I am right. We are answering different questions.

Consider the following analogy.... A family has 3 female children. What is the probability that the next child is female, assuming equal probabilities for male and female?

The probability of having 4 female children is 1 in 16 (6.25%) if we consider order. (This is a common debate, which we can ignore for my puproses.) But the probability that the next child is female is 1 in 2 (50%), which is unchanged by history.

The probability that a particular quad (4-tuple) will be drawn is 1 in 367,290 for a 5/56 game like MegaMillions. That probability is unchanged by history.

But the probability that a drawing will include any of the previously-drawn quads is currently about 1 in 17.85 (5.6%) for the MegaMillions game from 6/24/2005 through 5/24/2013. That is, 5*4115/376,290 because there are 5 quads in each drawing. That probability will change over time; in particular, it will become more likely.

But I was addressing the question: how many previously-drawn quads might be included in a drawing? That, too, will change over time. Honestly, I am not taking the time to compute it. Instead, I look at "recent" history to observe that it is zero about 94% of the time (for now); it is 1 for the remaining 6%. Over time, it will become more likely that each drawing will include 1, then 2 eventually up to 5 previously-drawn quads.

Errata.... I wrote: ``The probability that a particular quad (4-tuple) will be drawn is 1 in 367,290 for a 5/56 game like MegaMillions. That probability is unchanged by history.``

Arrgghh! In an attempt to KISS, I misstated the statistic.

We can indeed form 367,290 quads with 56 numbers -- C(56,4). So there is a 1-in-367,290 chance of randomly generating a particular quad.

But the probability that a particular quad is drawn is 52 in 3,819,816, which is the same as 1 in 73,458, because there are 52 possible drawings that contain that quad.

(And for anyone who might wonder, that is different from the odds of matching any 4 in a drawing -- i.e. winning the 3rd or 4th tier prize of the MegaMillions game -- because there are 5 quads in each drawing, and those odds exclude the chance of matching all 5.)

The point is: those odds are constant and independent of history. In contrast, the probabilities I was quoting originally were related to matching a particular number of previously-drawn quads. Of course, that does depend on history; and it changes over time.

mid-Ohio United States Member #9 March 24, 2001 19830 Posts Offline

Posted: June 9, 2013, 12:07 pm - IP Logged

This thread started out asking about 6/49 games, how did it get to answering questions about MegaMillions that weren't asked?

Posters are getting like politicians with their answers, when someone ask a question it gets answered by what ever the poster want to talk about.

There are plenty of 6/49 games around the country with decent jackpots that if LP members had any special knowledge or skills to win a jackpot, one of them should have won one by now. I came close back in 2002 and earlier this year in February with a 5of6, but close doesn't win a jackpot. I'm still trying occasionally but I keep getting lured away by the attraction of the multi-state games jackpots.

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

This thread started out asking about 6/49 games, how did it get to answering questions about MegaMillions that weren't asked?

Posters are getting like politicians with their answers, when someone ask a question it gets answered by what ever the poster want to talk about.

There are plenty of 6/49 games around the country with decent jackpots that if LP members had any special knowledge or skills to win a jackpot, one of them should have won one by now. I came close back in 2002 and earlier this year in February with a 5of6, but close doesn't win a jackpot. I'm still trying occasionally but I keep getting lured away by the attraction of the multi-state games jackpots.

Don't get your panties in a bunch! Gwoof's 6/49 question had already been answered. I was responding to a question from Jimmy4164, which I interpreted as conceptual in nature even though it was specific to a comment about the OCL that you made. I used the MM as an example of the concepts, since that is what I had numbers for off-hand. I thought Jimmy might appreciate the numerical details.

I am sorry that my incessant embellishments went so far off-topic, although I'm sure this is not the first time that a thread digressed in "different" directions. I will see if Todd is willing to delete my responses.

mid-Ohio United States Member #9 March 24, 2001 19830 Posts Offline

Posted: June 9, 2013, 2:38 pm - IP Logged

Quote: Originally posted by mathhead on June 9, 2013

Don't get your panties in a bunch! Gwoof's 6/49 question had already been answered. I was responding to a question from Jimmy4164, which I interpreted as conceptual in nature even though it was specific to a comment about the OCL that you made. I used the MM as an example of the concepts, since that is what I had numbers for off-hand. I thought Jimmy might appreciate the numerical details.

I am sorry that my incessant embellishments went so far off-topic, although I'm sure this is not the first time that a thread digressed in "different" directions. I will see if Todd is willing to delete my responses.

Sorry, my fruit of the loom are not in a bunch, please continue in the direction you were headed. No need to bother Todd, I'm he got better things to do.

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

Kentucky United States Member #32652 February 14, 2006 7314 Posts Offline

Posted: June 9, 2013, 5:08 pm - IP Logged

Quote: Originally posted by mathhead on June 9, 2013

Don't get your panties in a bunch! Gwoof's 6/49 question had already been answered. I was responding to a question from Jimmy4164, which I interpreted as conceptual in nature even though it was specific to a comment about the OCL that you made. I used the MM as an example of the concepts, since that is what I had numbers for off-hand. I thought Jimmy might appreciate the numerical details.

I am sorry that my incessant embellishments went so far off-topic, although I'm sure this is not the first time that a thread digressed in "different" directions. I will see if Todd is willing to delete my responses.

"Don't get your panties in a bunch! Gwoof's 6/49 question had already been answered."

You're correct because even your wrong answer is still an answer!

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

I believe the majority of us acknowledge we just made a mistake. Was your mistake a "panties in a bunch" type of mistake?

This thread started out asking about 6/49 games, how did it get to answering questions about MegaMillions that weren't asked?

Posters are getting like politicians with their answers, when someone ask a question it gets answered by what ever the poster want to talk about.

There are plenty of 6/49 games around the country with decent jackpots that if LP members had any special knowledge or skills to win a jackpot, one of them should have won one by now. I came close back in 2002 and earlier this year in February with a 5of6, but close doesn't win a jackpot. I'm still trying occasionally but I keep getting lured away by the attraction of the multi-state games jackpots.

"Posters are getting like politicians with their answers, when someone ask a question it gets answered by what ever the poster want to talk about."

Asking on average how many drawing cycles should it take before each of 49 numbers drawn at 6 in each drawing isn't what I would call complex math. Just like with politicians, we're seeing complex answers that never really answer the question.

What is so difficult about saying in the majority of drawing cycles all 49 numbers should be drawn in 35 drawings?

United States Member #93947 July 10, 2010 2180 Posts Offline

Posted: June 9, 2013, 6:07 pm - IP Logged

Quote: Originally posted by Stack47 on June 9, 2013

"Posters are getting like politicians with their answers, when someone ask a question it gets answered by what ever the poster want to talk about."

Asking on average how many drawing cycles should it take before each of 49 numbers drawn at 6 in each drawing isn't what I would call complex math. Just like with politicians, we're seeing complex answers that never really answer the question.

What is so difficult about saying in the majority of drawing cycles all 49 numbers should be drawn in 35 drawings?

It's difficult because it's an insufficient and misleading statement. Mathhead, RJOh, and myself went to the trouble to do the calculations; all you did was come to kibitz.

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

Posted: June 9, 2013, 6:42 pm - IP Logged

Quote: Originally posted by Stack47 on June 9, 2013

"Don't get your panties in a bunch! Gwoof's 6/49 question had already been answered."

You're correct because even your wrong answer is still an answer!

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

I believe the majority of us acknowledge we just made a mistake. Was your mistake a "panties in a bunch" type of mistake?

I usually ignore your postings (certainly in this thread) as a waste of hot air. But I suppose that nonsensical remark deserves a response, even though it reflects more on your character than on mine.

I posted the following at http://www.lotterypost.com/thread/262010/3100463 on June 5: ``Actually, that demonstrates that neither of my "passes" is correct. I'll have to look for a programming error later. [....] Sorry for the misdirection.``

Moreover, I did not say that __I__ answered Gwoof's question. I simply said it "had already been answered".

In fact, in my June 5 response, I acknowledged that RJOh's conclusions based on OCL data are essentially correct. I wrote: ``that is consistent with my analysis of the 31-year(!) data for the Canada 6/49 lotto: 19 to 64 drawings for all 49 numbers to appear in the first 6 numbers (not considering the bonus number), with a mean of 36 +/-2 rounded (with 95% confidence).``

I also acknowledged that Jimmy1464 demonstrated that a correct simulation, unlike mine, reaches the same conclusions. I wrote: ``the mean of your simulation is consistent with my analysis of the Canada 6/49 past numbers.``

Finally, I did answer Gwoof's follow-up question based on real analysis, in contrast to your hand-waving gibberish (click here). I wrote: ``when the bonus number is included (i.e. a draw of 7 numbers from the same pool of 49), there is only a small difference, to wit: 18 to 63 drawings with a mean of 32 +/-2 rounded (95% confidence).``

Kentucky United States Member #32652 February 14, 2006 7314 Posts Offline

Posted: June 10, 2013, 1:51 am - IP Logged

Quote: Originally posted by mathhead on June 9, 2013

I usually ignore your postings (certainly in this thread) as a waste of hot air. But I suppose that nonsensical remark deserves a response, even though it reflects more on your character than on mine.

I posted the following at http://www.lotterypost.com/thread/262010/3100463 on June 5: ``Actually, that demonstrates that neither of my "passes" is correct. I'll have to look for a programming error later. [....] Sorry for the misdirection.``

Moreover, I did not say that __I__ answered Gwoof's question. I simply said it "had already been answered".

In fact, in my June 5 response, I acknowledged that RJOh's conclusions based on OCL data are essentially correct. I wrote: ``that is consistent with my analysis of the 31-year(!) data for the Canada 6/49 lotto: 19 to 64 drawings for all 49 numbers to appear in the first 6 numbers (not considering the bonus number), with a mean of 36 +/-2 rounded (with 95% confidence).``

I also acknowledged that Jimmy1464 demonstrated that a correct simulation, unlike mine, reaches the same conclusions. I wrote: ``the mean of your simulation is consistent with my analysis of the Canada 6/49 past numbers.``

Finally, I did answer Gwoof's follow-up question based on real analysis, in contrast to your hand-waving gibberish (click here). I wrote: ``when the bonus number is included (i.e. a draw of 7 numbers from the same pool of 49), there is only a small difference, to wit: 18 to 63 drawings with a mean of 32 +/-2 rounded (95% confidence).``

"Moreover, I did not say that __I__ answered Gwoof's question. I simply said it "had already been answered"."

Saying you didn't answers Gwoof's question doesn't change the fact you gave a wrong answer.

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

Even Jimmy noticed you said it.

"If you check back with Mathhead's revised simulation results you will discover that using 35-39 as your expected average is quite likely going to put you into bankruptcy at some point. 90 is closer to the expected average."

Jimmy even quoted your wrong answer based on your second simulation.

"I also acknowledged that Jimmy1464 demonstrated that a correct simulation, unlike mine, reaches the same conclusions."

You must be talking about your third or was it a fourth attempt?

"Actually, that demonstrates that neither of my "passes" is correct. I'll have to look for a programming error later."

IMO, the real problem is you don't understand 6/49 lotto games are only drawn twice or three times a week, less than 8000 in 50 years of play. What is the point of running a simulation when the actual play only is 0.0572% of all the possible results in 50 years?

"Finally, I did answer Gwoof's follow-up question based on real analysis, in contrast to your hand-waving gibberish"

Maybe you missed where Gwoof asked Jimmy1464 "But would the result be relevant to the draw you are playing since it is not useing your direct history?".

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

The good news, for what it's worth, you impressed the heck out of Jimmy1464!