Mega Millions
| Winners | Probability |
|---|---|
| 0 | 0.64% |
| 1 | 3.23% |
| 2 | 8.16% |
| 3 | 13.74% |
| 4 | 17.36% |
| 5 | 17.54% |
| 6 | 14.77% |
| 7 | 10.66% |
| 8 | 6.73% |
| 9 | 3.78% |
| 10 | 1.91% |
Quote: DocI think one complicating factor might be the rules under which the lottery seeds the initial jackpot.
The states certainly lose out on the initial draw which has a guaranteed jackpot of $12 million.
| Drawing | Date | Jackpot (26 payments) | Difference | Millions tickets (31.8%) | |
|---|---|---|---|---|---|
| 707 | 30-Mar-12 | $640,000,000 | $277,000,000 | 871 | |
| 706 | 27-Mar-12 | $363,000,000 | $73,000,000 | 230 | |
| 705 | 23-Mar-12 | $290,000,000 | $49,000,000 | 154 | |
| 704 | 20-Mar-12 | $241,000,000 | $41,000,000 | 129 | |
| 703 | 16-Mar-12 | $200,000,000 | $29,000,000 | 91 | |
| 702 | 13-Mar-12 | $171,000,000 | $23,000,000 | 72 | |
| 701 | 9-Mar-12 | $148,000,000 | $21,000,000 | 66 | |
| 700 | 6-Mar-12 | $127,000,000 | $19,000,000 | 60 | |
| 699 | 2-Mar-12 | $108,000,000 | $14,000,000 | 44 | |
| 698 | 28-Feb-12 | $94,000,000 | $11,000,000 | 35 | |
| 697 | 24-Feb-12 | $83,000,000 | $11,000,000 | 35 | |
| 696 | 21-Feb-12 | $72,000,000 | $11,000,000 | 35 | |
| 695 | 17-Feb-12 | $61,000,000 | $10,000,000 | 31 | |
| 694 | 14-Feb-12 | $51,000,000 | $10,000,000 | 31 | |
| 693 | 10-Feb-12 | $41,000,000 | $9,000,000 | 28 | |
| 692 | 7-Feb-12 | $32,000,000 | $9,000,000 | 28 | |
| 691 | 3-Feb-12 | $23,000,000 | $8,000,000 | 25 | |
| 690 | 31-Jan-12 | $15,000,000 | $3,000,000 | ? | |
| 689 | 27-Jan-12 | $12,000,000 | ? |
| Drawing | Date | Jackpot (26 payments) | Difference | Millions tickets (31.8%) | |
|---|---|---|---|---|---|
| 688 | 24-Jan-12 | $71,000,000 | $11,000,000 | 35 | |
| 687 | 20-Jan-12 | $60,000,000 | $10,000,000 | 31 | |
| 686 | 17-Jan-12 | $50,000,000 | $8,000,000 | 25 | |
| 685 | 13-Jan-12 | $42,000,000 | $9,000,000 | 28 | |
| 684 | 10-Jan-12 | $33,000,000 | $8,000,000 | 25 | |
| 683 | 6-Jan-12 | $25,000,000 | $10,000,000 | 31 | |
| 682 | 3-Jan-12 | $15,000,000 | $3,000,000 | ? | |
| 681 | 30-Dec-11 | $12,000,000 | ? |
Quote: WizardThat doesn't add up. Assuming we believe the 640M, that would mean a jackpot increase of 277M since the last draw. There are about 2.6154 tickets sold per dollar increase, which would mean 724M tickets sold.
I wonder if they are looking to raise the estimated prize again this afternoon as the ticket sales roll in.
You'd think that they can see the speed at which the machines are churning out tickets to come up with a reliable estimate.Quote: slytherI wonder if they are looking to raise the estimated prize again this afternoon as the ticket sales roll in.
Except there's no history for this size drawing, to help them estimate how much of a surge there will be after people get out of work. And you know there's gonna be a surge.....
Quote: 98Clubs888 million sold thats about 5 winners on the average *considering full coverage of all ways*.
It does occur to me that if you did this game online, you could keep a barometer running of the number of un-chosen combinations. People would work themselves into a Pavlovian frenzy trying to guess the remaining combinations.
I think people would be stupid enough to pay a premium fee of $1 to have the computer randomly choose one of the final combinations that are not picked yet (possibly when the unchosen combinations are less than a million).
Quote: pacomartin
I think people would be stupid enough to pay a premium fee of $1 to have the computer randomly choose one of the final combinations that are not picked yet (possibly when the unchosen combinations are less than a million).
I think only the smarter players would do this. I would much rather have a unique number today than to face splitting the pot with 4 or 5 other people. Even with 888M tickets sold, there will still be more than 1M tickets unchosen, so I think the threshold would need to be higher, perhaps 20 million.
Quote: boymimboLikely, folks, you don't know the formula of what goes into prizes. When you get into uncharted territory like we have here, they might break formula and put more into the prize pool.
Well, there are two ways to handle the situation. You could pay a premium to get a unique number at the present time, or you could block the number so that no one else can choose it at a later time. I was thinking of the former situation.
The lottery organizers may find that it turns people off if their first half dozen guesses at their favorite numbers are all blocked.
Quote: CrystalMathI think only the smarter players would do this. I would much rather have a unique number today than to face splitting the pot with 4 or 5 other people. Even with 888M tickets sold, there will still be more than 1M tickets unchosen, so I think the threshold would need to be higher, perhaps 20 million.
I had heard this morning on the radio (they could have been wrong) that every combination had been taken. I don't know who the announcer would have heard that from, but it certainly would change things if you KNEW any tickets bought today would 100% be split with at least 1 more person, but anything yesterday could be unique.
Quote: cclub79I had heard this morning on the radio (they could have been wrong) that every combination had been taken. I don't know who the announcer would have heard that from, but it certainly would change things if you KNEW any tickets bought today would 100% be split with at least 1 more person, but anything yesterday could be unique.
That would be interesting, if it is/was in fact true. Of course, I would suspect that it could be determined that all combinations were played. Now, why would the lottery leak that information. It might have an impact that would reduce the amount of ticket sales (can't get it all for myself.) But, why then would the lottery not leak information that the winning numbers had not been played (say for last Tuesday's drawing) to increase sales? I think that rules would probably stipulate not giving out that information, but I am not a lottery worker, so I can't say with any certainty there is such a rule. Interesting question to ask the lottery officials.
my life. People bothering you all the time, the security
risk for the family. Huge money like this falling into
your lap ruins just as many lives as it saves.
Quote: cclub79I had heard this morning on the radio (they could have been wrong) that every combination had been taken.
If there are X million of possible combinations and Y million tickets have been sold, where Y > X, the simple mind concludes all possible combinations have been sold (wow, I can speak in math!)
Quote: NareedIf there are X million of possible combinations and Y million tickets have been sold, where Y > X, the simple mind concludes all possible combinations have been sold (wow, I can speak in math!)
It was a news-ish guy talking, so I didn't expect a simple mind, but who knows, you could be right! Plus, do we know for a fact that the QuickPick (or whatever) is truly random? And do all of the numbers get issued from one central location in the USA, or each STATE's mainframe?
Quote: NareedIf there are X million of possible combinations and Y million tickets have been sold, where Y > X, the simple mind concludes all possible combinations have been sold (wow, I can speak in math!)
And that is true, the simple mind can conclude that, but to outright say it is not fact. Multiple tickets with the same combinations, as alluded to in many previous posts. Just sayin'. I for one wish they had said that the winning ticket was sold on Wednesday morning, in the city where I live. It could be mine, it might reduce the amount of potential duplicates sold (I don't want to share, but will, if I must, he says with a smile.)
Quote: bushmanAnd that is true, the simple mind can conclude that, but to outright say it is not fact. Multiple tickets with the same combinations, as alluded to in many previous posts. Just sayin'. I for one wish they had said that the winning ticket was sold on Wednesday morning, in the city where I live. It could be mine, it might reduce the amount of potential duplicates sold (I don't want to share, but will, if I must, he says with a smile.)
I think that this was a reporter who was saying that the number of tickets that had been sold is equal to or greater than the number of all possible combinations.
Remember, the odds of splitting a jackpot with a Megaball of 1-31 is higher than the much higher numbers, as many people will play birthdays, lucky numbers, etc. Also, how many tickets were sold with the "Lost" numbers? Factor those into play (which become more important when non-lottery players are playing) and my guess is there is still a handful of combinations that will not be selected.
My 10 Megaballs are all different, so I'm happy...
Quote: bushmanAnd that is true, the simple mind can conclude that, but to outright say it is not fact.
Of course.
If more tickets have been sold than there are possible combinations, then it is possible all combinations have been sold. But without a tally you can't know it.
Quote:I for one wish they had said that the winning ticket was sold on Wednesday morning, in the city where I live. It could be mine, it might reduce the amount of potential duplicates sold (I don't want to share, but will, if I must, he says with a smile.)
But no one knows what the winning tiket will be.
When I bought lotto, the people in my pool sometimes would tell me "pick the winner this time!" I replies "If I knew what the winner was going to be, I would buy it myself and not split the prize" >:)
Quote: cclub79Plus, do we know for a fact that the QuickPick (or whatever) is truly random?
When I did buy lotto my MO was to buy about 15-20 or so of the same combinations every time, and the rest in quick picks. I dind't do a statistical analysis, but I noticed some fo the quick picks repeated two or three numbers from some combinations.
One combo I recall involved the numbers 8, 12 and 15. Say three quick picks out of ten would have 8 and 15 or 12 and 15. One lotto agent claimed she was told the terminal picks the random numbers themselves. I suppose an RNG per terminal would be a small matter.
Quote: WizardA problem seems to be that the ratio of meter increase to tickets sold is not consistent from drawing to drawing.
Quote: pacomartinThe states certainly lose out on the initial draw which has a guaranteed jackpot of $12 million.
Given that:
- 50% of sales is definitely paid back to players
- The contribution rate is inversely proportional to the ticket sales
- The target contribution rate is 31.8% if the non-jackpot return is 18.2%
- The reseed is always $12,000,000 (is it?)
I think the reason for the contribution change is so they can have a much higher contribution rate when the jackpot is low and a lower rate when it's high. The per-dollar value of increasing the jackpot diminishes as the jackpot grows. An additional $10 million makes a huge difference when the jackpot is at $12 million but even another $100 million doesn't matter much once it's at $600 million.
We have a similar setup on many of the progressives at our casino. When the jackpot is low the contribution rate to the main jackpot is high. Once it passes a certain threshold the main jackpot rate lowers while the reseed rate increases. For example:
| Current Jackpot | Meter Rate | Reseed Rate |
|---|---|---|
| < $10,000 | 25% | 5% |
| < $25,000 | 20% | 10% |
| < $50,000 | 10% | 20% |
| >= $50,000 | 5% | 25% |
Quote: NareedOf course.
If more tickets have been sold than there are possible combinations, then it is possible all combinations have been sold. But without a tally you can't know it.
But no one knows what the winning tiket will be.
When I bought lotto, the people in my pool sometimes would tell me "pick the winner this time!" I replies "If I knew what the winner was going to be, I would buy it myself and not split the prize" >:)
I got you. I am just saying it was not responsible reporting if the reporter stated that the winning ticket had been purchased (I do not know what was stated in the report.) Yes, logic deems that it has been purchased. I am not expecting to spend any money on Tuesday's drawing, as I believe someone (probably more than one) has a winner. But, then again, I am willing to pony up the few sheckels if it does not hit tonight.
I like how you answered your pool mates. ;>)
Quote: pacomartinThe states certainly lose out on the initial draw which has a guaranteed jackpot of $12 million.
Drawing Date Jackpot (26 payments) Difference Millions tickets (31.8%) 707 30-Mar-12 $640,000,000 $277,000,000 871 706 27-Mar-12 $363,000,000 $73,000,000 230 705 23-Mar-12 $290,000,000 $49,000,000 154 704 20-Mar-12 $241,000,000 $41,000,000 129 703 16-Mar-12 $200,000,000 $29,000,000 91 702 13-Mar-12 $171,000,000 $23,000,000 72 701 9-Mar-12 $148,000,000 $21,000,000 66 700 6-Mar-12 $127,000,000 $19,000,000 60 699 2-Mar-12 $108,000,000 $14,000,000 44 698 28-Feb-12 $94,000,000 $11,000,000 35 697 24-Feb-12 $83,000,000 $11,000,000 35 696 21-Feb-12 $72,000,000 $11,000,000 35 695 17-Feb-12 $61,000,000 $10,000,000 31 694 14-Feb-12 $51,000,000 $10,000,000 31 693 10-Feb-12 $41,000,000 $9,000,000 28 692 7-Feb-12 $32,000,000 $9,000,000 28 691 3-Feb-12 $23,000,000 $8,000,000 25 690 31-Jan-12 $15,000,000 $3,000,000 ? 689 27-Jan-12 $12,000,000 ?
Drawing Date Jackpot (26 payments) Difference Millions tickets (31.8%) 688 24-Jan-12 $71,000,000 $11,000,000 35 687 20-Jan-12 $60,000,000 $10,000,000 31 686 17-Jan-12 $50,000,000 $8,000,000 25 685 13-Jan-12 $42,000,000 $9,000,000 28 684 10-Jan-12 $33,000,000 $8,000,000 25 683 6-Jan-12 $25,000,000 $10,000,000 31 682 3-Jan-12 $15,000,000 $3,000,000 ? 681 30-Dec-11 $12,000,000 ?
I think the math is wrong. You don't use the $640MM, you use the increase in the CASH number.
For the 26-annual payments, the cash pool is invested in US T-Bills and paid out yearly, so that would be an incorrect number to use.
Quote: bushmanI got you. I am just saying it was not responsible reporting if the reporter stated that the winning ticket had been purchased (I do not know what was stated in the report.)
Absolutely. I was just explaining how math fails people, or people fail math, when confronted with large but simple numbers. The Wizard already explained how some numbers are repeated.
Quote:But, then again, I am willing to pony up the few sheckels if it does not hit tonight.
Nothing wrong with that, as long as you know what your chances are (and they're absolute zero if you don't play). And nothing wrong with fantsizing what you'd do with the money, either. Thats' the fun of playing lotto.
Quote:I like how you answered your pool mates. ;>)
Thank you. I really dont' mind sueprstitions, when people keep them to themselves. If you want to chant all the way to the lotto agency, have fun doing it. But don't try to get me to join you :)
Ask 11 people to pick a number between 1 and 10. What are the odds that all 10 numbers are chosen?
Quote: boymimboNareed, think of this:
Ask 11 people to pick a number between 1 and 10. What are the odds that all 10 numbers are chosen?
No idea, and I don't know how to work it out. I know there is a chance all numbers were picked (and if you'd said nine people, there's no chance all numbers were picked), because there are more people than numbers. But there's also a chance one or more numbers get more than one pick. In fact, even if you had 1,000 people pick one number from 1 to 10, chances are still not 100% that all numbers get picked. Close, I guess, but not necessarily 100%.
That's what comes from understanding probability at a qualitative level only :)
Quote: NareedIf there are X million of possible combinations and Y million tickets have been sold, where Y > X, the simple mind concludes all possible combinations have been sold (wow, I can speak in math!)
I was responding to this, Nareed. It doesn't matter whether Y=11 and X=10, or Y = 880,000,000 and X = 175,000,000. The probabilities are calculable.
Them again, you said "the simple mind", and you certainly don't fall into this class (a compliment).
Quote: boymimboThe probabilities are calculable.
And apparently it's a real pain-in-the-butt calculation. I don't think anyone in this forum has devised a solution yet.
Quote: boymimboI was responding to this, Nareed. It doesn't matter whether Y=11 and X=10, or Y = 880,000,000 and X = 175,000,000. The probabilities are calculable.
If X<Y then we know not all combinatiosn have been purchased ;)
Seriously, yes I know the odds can be calculated. I just don't know how.
Quote:Them again, you said "the simple mind", and you certainly don't fall into this class (a compliment).
Likely tell ;)
I should have said "the unsophisticated mind," or even more charitably "the unschooled mind."
Quote: winmonkeyspit3I think the math is wrong. You don't use the $640MM, you use the increase in the CASH number.
I did use the increase in the CASH number.
$277 million = 31.8% * 871 million tickets @ $1 apiece.
The point of showing the previous run was to emphasize that their is no massive jumps until the Jackpot reaches levels of hysteria.
I think it is possible that the on the occasions when someone hits the $12 million right away that the States do not make any money on the game. The entire amount of money collected goes to pay the jackpot, the secondary prizes, and the overhead.
But if they can reach hysteria level once a year, that more than makes up for the times when someone hits early. But the question is how much of this money is new, and how much is redirected from the normal instant games that make up the bulk of the revenue for lottery organizations.
I suspect Florida does not participate, because they lose money on the early winners, and they feel that it would take away from the more profitable games when it gets big.
==========================
I doubt that reporter knew that all the numbers had been picked. They just thought all the numbers had probably been picked, given the amount of tickets sold.
In the unlikely event that no one wins, can you imagine the jackpot going over a billion dollars by Tuesday.
Quote: cclub79I had heard this morning on the radio (they could have been wrong) that every combination had been taken.
Assuming every ticket is randomly chosen, then the expected number of tickets sold to cover every number would be 3,362,905,623. There have been only about 26% that many sold. It would require a simulation to determine the probability of covering all of them in 888M tickets (my estimate of tickets sold), but I think it would be under 1%.
I think JB was asking about this type of problem earlier. I'm afraid there is no easy formula, that I know of, for the probability of covering x combinations in y draws.
Quote: WizardI think JB was asking about this type of problem earlier. I'm afraid there is no easy formula, that I know of, for the probability of covering x combinations in y draws.
The precise way to do it is with an absorbing Markov transition matrix. It is easy enough to do it with a six sided dice. The probably of covering all six combinations in n rolls of the dice is:
n, probability all 6 sides thrown
6, 1.54% = 5/(18*18)
7, 5.40%
8, 11.40%
9, 18.90%
10, 27.18%
11, 35.62%
12, 43.78%
...
17, 74.46%
...
28, 96.38%
...
55, 99.974%
The Markov matrix for the dice is follows. You must raise it to the (n-1) power and look at state 6 (upper right hand corner). It's an absorbing matrix, because once you hit every side of the dice, there are no other states to advance toward.
| state | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 1/6 | 5/6 | 0 | 0 | 0 | 0 |
| 2 | 0 | 1/3 | 2/3 | 0 | 0 | 0 |
| 3 | 0 | 0 | 1/2 | 1/2 | 0 | 0 |
| 4 | 0 | 0 | 0 | 2/3 | 1/3 | 0 |
| 5 | 0 | 0 | 0 | 0 | 5/6 | 1/6 |
| 6 | 0 | 0 | 0 | 0 | 0 | 1 |
But you can't do a matrix with 176 million rows and columns. There must be an approximation, but I don't know it.
But judging by the dice, if you sample 3-4 times as many possibilities, then there is still a small percentage chance that you haven't covered every possibility.
With the dice example you do not even have to raise it to any power to get answers like the average number of times each state is visited.Quote: pacomartinThe precise way to do it is with an absorbing Markov transition matrix. It is easy enough to do it with a six sided dice.
You must raise it to the (n-1) power and look at state 6 (upper right hand corner). It's an absorbing matrix, because once you hit every side of the dice, there are no other states to advance toward.
But you can't do a matrix with 176 million rows and columns. There must be an approximation, but I don't know it.
But judging by the dice, if you sample 3-4 times as many possibilities, then there is still a small percentage chance that you haven't covered every possibility.
By arranging the absorbing matrix into its canonical form, one can apply well know simple formulas to answer many questions.
Any basic Markov Chain tutorial will show how to do this. (If I had more time, I could show this)
For the question at hand, I bet $5 there is a calculus solution but my BF and I are off to the Final 4!
Go Wildcats!
2 4 23 38 46 - 23
Hurley's numbers were close (closer than any of mine!)
EDIT: Funny though..I picked 5 tickets on the same ticket (full card) and amazingly, I got all six numbers out of the five tickets...what I did was pick numbers based on the 25+5 numbers that have hit the most..out of those 30 numbers, I got all of them right (just on different tickets, LOL; but I had the right idea I guess)
Quote: PopCan
We have a similar setup on many of the progressives at our casino. When the jackpot is low the contribution rate to the main jackpot is high. Once it passes a certain threshold the main jackpot rate lowers while the reseed rate increases.
Quote: FoxnewsOn Friday, the lottery estimated that total ticket sales for this jackpot, which has been building up since Jan. 28, will be about $1.46 billion, said Kelly Cripe, a spokeswoman for the Texas Lottery Commission.
It is difficult to guess precisely what the lotteries do to get the seed started. But this statement that the total sales since the last jackpot was won have been 1.46 billion tickets. That would lead me to believe that ticket sales for this final go round are in the low 600 millions, instead of the high 800's.
It would be incredible if nobody picked the combination.
Quote: TIMSPEEDMy best out of 10 tix, was 2 numbers plus the mega...ohh for the other three numbers...
EDIT: Funny though..I picked 5 tickets on the same ticket (full card) and amazingly, I got all six numbers out of the five tickets...what I did was pick numbers based on the 25+5 numbers that have hit the most..out of those 30 numbers, I got all of them right (just on different tickets, LOL; but I had the right idea I guess)
You should have bought all 53,130 combinations of those 25 numbers.
Quote: CrystalMathYou should have bought all 53,130 combinations of those 25 numbers.
True...I would have only spent $53,130 to win $640M (without anyone splitting it hopefully)..even WITH a split, still in the millions..
Quote: WizardAssuming every ticket is randomly chosen, then the expected number of tickets sold to over every number would be 3,362,905,623
What an interesting number. With the following prime factors 3,362,905,623 = 3*7*7*11*2079719
Can you explain where that number came from?
Quote: pacomartinCan you explain where that number came from?
I wrote a computer program to sum 175711536/i for i=1 to 175711536. The answer is a floating point number, so may not be exactly right. The prime factorization I think is a coincidence. I didn't know you had such an interest in the math-heavy topics.
Quote: pacomartin
Can you explain where that number came from?
The Wizard's post got me thinking about it. Using the poission distribution, you can calculate the average number of unpicked tickets given the total quantity sold. Use this and calculate the number of draws required to get down to 1 remaining ticket. This turns out to be combos*ln(combos). From that point, you should need combos/2 draws on average to pick the last number. My number doesn't match the Wizard exactly, but it's pretty close.
Edit: the last number will take the number of ticket combos, not combos/2.
Quote: CNNSomeone in Maryland is about to have a wildly euphoric weekend. A winning ticket for the record-breaking $640 million Mega Millions jackpot was sold in Baltimore County, the Maryland Lottery said Saturday.
Lottery officials are waiting for information on potential jackpot winners from other Mega Millions states.
...and the Maryland Lottery website implies that there were multiple winners:
Quote: Maryland LotteryWinning Ticket Sold in Baltimore County one of several nationwide
i know i was hoping to win it all myself,
so in lieu of that i wish the same for the
winner from "bawlmer"
Quote: WizardI wrote a computer program to sum 175711536/i for i=1 to 175711536. The answer is a floating point number, so may not be exactly right. The prime factorization I think is a coincidence. I didn't know you had such an interest in the math-heavy topics.
I have a Master's of Science in Applied Mathematics, but my undergraduate degree was pure mathematics.
By the math, with the last prize being 363 million, that's a delta of 277 million dollars.
277/652.306 = 42.5% going into the grand prize pool.
Source = http://www.megamillions.com/numbers/drawDetails.asp?drawDate=3/30/2012
