The numbers in our lottery go up to 49. You are allowed to choose 6 numbers.
So how many tickets would have to be bought to have every combination of numbers?
This was a debate on another board I went to not long ago and nobody seemed to agree on the answer.
The only thing I know is that the number of tickets you’ll have to get will be, well, VERY VERY high.
But does anybody know the exact number? And how to reach it?
Maybe I’m missing the debate here but:
49 choices for 6 spots should be 49^6, right?
So:
13 841 287 201
13-14 billion sound right?
The number of unique, non-ordered ways to choose 6 numbers without replacement from 49 numbers is 49 factorial divided by (43 factorial x 6 factorial), or 13,983,816.
If I remember my statistics correctly, we’re looking at a combination problem.
You have 49 numbers drawn in groups of 6. So…
49c6 = 13,983,816
So almost 14 million tickets. If you’ve got $13.9 million already, why bother playing the lottery?
You would need (49x48x47x46x45x44)/720 tickets. In other words, 13,983,816 tickets.
Oh, and don’t think buying every possible ticket will work. Even with a jackpot above $14 million, even if it’s a Canadian lottery (no taxes and you get the entire jackpot up front) the odds are that with a large jackpot, someone else - possibly multiple someone elses - will win as well, so you’ll have to share the jackpot and get screwed.
I’m going to take a stab and say that there are 13,983,816 possible combos.
For some strange reason the number 13,983,816 just popped into my head.
If you’re not an individual, but a syndicate and the draw is a double or triple rollover…
Bah! It doesn’t matter what you do, and how hard you work at something, there’s always some idle slob doing it the easy way and getting away with it! :mad: Why can’t these cheap freeloaders buy their own 13,983,816 tickets?
Let’s not forget, for practical purposes, the extreme and possibly insurmountable logistic difficulties involved. In the lotteries I’m familiar with, you don’t get to choose the drawing date of the tickets you purchase - you buy them for the next drawing. Most have drawings once or twice a week. For the sake of simplicity, we’ll say once a week.
The machines that print the tickets have physical limitations. All the ones I’ve seen employ dot matric printers, which are comparitively slow. I’ve never actually timed one, but they seem to be able to print a ticket in about 3 seconds. Let’s use that figure. Hell, let’s be generous and use 2 seconds. So, for a single machine to print out 13,983,816 tickets, it would take 2 times that or 27,967,632 seconds, which is 323.7 days - nearly a year. Far too slow. In order to get our tickets printed in 7 days in time for the next drawing, we’d need to employ over 46 machines. And of course, the stores would each have to be open 24 hours a day, every day.
And that’s assuming that each store’s management will let us hang out in his store 24 hours a day for a week, and that the stores have sufficient print stock and ink to handle the load, and that allow us to monopolize their lottery machine(s) to the exclusion of all other customers. All of which seems extraordinarily unlikely to me. You could spread your efforts out over more lottery machines, to lighten the load of each one, but then you’d need increasing amounts of manpower, which presents its own set of problems - for example, there’s always the possibility that the person who actually bought the winning ticket might decide the prize is rightfully his and give you legal trouble over it. Realistically, I just don’t think it’s possible.
I don’t know. If I owned a small shop and someone told me they planned to spend a dollar every two seconds on my premises, I’d be open to persuasion. I’d maybe even take on more staff and lock the doors.
Either way, if I had that kind of money, syndicate or not, I’m sure I could find a lot safer and smarter ways of investing it.
Encore!
Always someone trying to be practical. (Spoil sport or someone of the genus)
You beat me to it.
I just thought there would be a way to pay off some official. I mean write a contract with the state so that you give them $13,983,816, and they just take it as a given that you are the winner of the lottery, and give you your winnings, minus taxes and whatever you had to split with other freeloaders.
the correct number is what you get from 494847…
an excerpt from my old math book:
“In February of 1992, an Australian investment group came very close to cornering the market on all possible combinations of six numbers from one to ffourty-four in the Virginia state lottery. The group bought 5 million of the possible 7.05 million tickets in a lottery with a $27 million jackpot. Only time prevented them from getting the other 2 milion tickets. The group was the only winner; their profit: approx. $22 million. Since then, the Virginia State Lottery COmmision (and other states) have written regulation to rpevent individuals and groups from cornering the market.”
13,983,816 . Didn’t this happen once. In the 1960’s. For twenty minutes or so?
In addition to nailing the six-out-of-six grand prize, I figure you’d get:
258 wins on the “five out of six”
25830 wins on the “four out of six”
Lots and lots of wins on the “three out of six” (I’m too lazy to do the math)
These secondary prizes are smaller, but they do add up.
[Guinness Dude]
Brilliant!
[/GD]
No, the correct formula is n!/k!(n-k)!, with n=49 and k=6. That gives you the answer of 13,383,816 (look familiar?). You can also go into google and type “49 choose 6” (the standard way of reading a combination problem like this aloud) and get the same answer.
You’re using the permutation formula, n!/(n-k)!, which yields the answer 10,068,347,520. This is the answer if the order of the numbers is relevant, which it is not in this case.
I’ve never timed one, but those machines do indeed use a dot matrix printer, and they are fairly slow. Where I worked, we had a regular that would come in a few times a week and generally buy $50-100 of Pick 3 tickets. Those tickets have to print one play per ticket, unlike Powerball, which can do up to 10 quick picks, but only up to 5 picked numbers (due to the limitation of the paper ticket), unless you can get back there and enter every number directly in the machine (but I’ve never tried entering 10 sets of picked numbers on one ticket.) Anyway, it would generally take 15-30 minutes, depending on other traffic, to run all his slips and print all his tickets. So, assuming 5 plays per ticket, you’re looking at 2,796,764 tickets to be printed up and the same amount of slips to fill out to begin with. Assuming you could continously run the machine without having to stop for maintainance and optimal speed (and let’s assume you can do 50 tickets in 15 minutes, so $1000/hr), I still think it would take nearly 3,000 hours. Assuming a once-a-week draw (which probably isn’t realistic), you’re looking at monopolizing 16,648 machines. Of course, it’d be even higher if you assume that you have to work around a twice-weekly drawing and a draw break. Assume that you could do this for the Wednesday Powerball (and that the original number would be correct.) Let’s assume you could start selling at midnight on Sunday (time estimated because I can’t remember how long the draw break lasts.) You’d have to stop selling at 10 PM on Wednesday. That gives you 94 hours. You’re looking at monopolizing nearly 30,000 machines (29,753 to be exact.)
In any case, I agree it can’t be done, even if you can use Q.E.D.'s time assumption.