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View Full Version : Could you buy all the numbers in the lottery?

dalovindj
12-28-2002, 03:34 AM
How many tickets would you have to buy to get all possible combinations in a lottery with 6 balls drawn (numbers 1-32) where the powerball is the 6th number?

Bryan Ekers
12-28-2002, 03:41 AM
There are 906,192 combinations. If you could buy one ticket a second, it would take you over 10 days to get them all.

Alereon
12-28-2002, 03:42 AM
I believe the proper formula is: 32*(32-1)*(32-2)*(32-3)*(32-4)*(32-5)
This gives me: 652,458,240 tickets.
This assumes that numbers can't be duplicated.

12-28-2002, 03:47 AM
hmmm... which is it...

Bryan Ekers
12-28-2002, 03:52 AM
Originally posted by FDISK
I believe the proper formula is: 32*(32-1)*(32-2)*(32-3)*(32-4)*(32-5)
This gives me: 652,458,240 tickets.
This assumes that numbers can't be duplicated.

Actually, your formula (32!/(32-6)!) is incomplete. You're generating the number of permutations, which means sets of 6 numbers in different orders are considered different sets. Thus; (1,2,3,4,5,6) is considered different than (6,5,4,3,2,1). Since for the purposes of lottery tickets, the two are the same, what you need is the combinations formula:

32! / [(32-6)! * 6!]

Desmostylus
12-28-2002, 04:05 AM
Wait a minute, what's the significance of "powerball is the 6th number"?

Cabbage
12-28-2002, 04:08 AM
There would be 6,444,032 tickets [32 * (32!)]/[(5!) * (27!)].

A group of people from Australia tried to do this (buy up all the tickets) in the Virginia lottery several years ago (44 pick 6, for a total of 7059052 tickets). Each person in the group was responsible for buying large blocks of tickets (I don't know the details on that). Actually, they only managed to buy up about 75% of the tickets, but they did win, and, luckily for them, they didn't have to split it with anyone. I don't remember the jackpot exactly, it was between 10 and 14 million, IIRC, but I could be wrong there. It was a huge risk for them, and I doubt they really came out ahead in the end (and I would figure they were probably in the hole) when you figure in all the taxes, and the fact that the jackpot is paid out as a 20 year annuity.

sweepkick
12-28-2002, 05:16 AM
Good question...

Powerball lists the odds of winning as being 1 in 120,526,770.

That is 5 unique numbers out of a single barrel of 53 numbers, *plus* a single ball out of a barrel of 42 balls. That kind of math makes my brain hurt, so let's give Powerball the benefit of the doubt. Let's also say that it's 1 in 120 Mil even.

With the last jackpot being at ... what .. 313 million (over a quarter of a BILLION dollars ... for chrissakes!, effin rediculous)... How can I make this work for me?

Can I spend 120 million bucks, buy every combination (120 million tickets), claim 119,999,999 as gambling losses to the IRS, and only pay taxes on the money that I actually came out ahead on?

Granted, the cash option would be at worse case HALF of the total prize, but hell... turning 120 mil into 150 mil wouldn't be too bad. How long would it take you to turn 120 mil into 150 mil by conventional means?

The only problem... (and here's the kicker) ... having to share the jackpot with some schmo who happened to hit the same winning numbers. You are *not* guaranteed that this would not happen. Thus, if you bought every ticket available thinking you'd be the only one winning, you could be in for quite the shocker, and one b*tch of a loss.

Urban Ranger
12-28-2002, 05:33 AM
Yeah, wait till the IRS laughs in your face.

Desmostylus
12-28-2002, 05:46 AM
The answer to the OP could be:

32!/(26!x6!) = 906,192

32x32!/(27!x5!) = 6,444,032

27x32!/(27!x5!) = 5,437,152

I repeat: What's the significance of "powerball is the 6th number"?

Blown & Injected
12-28-2002, 06:20 AM
my probability class was about 12 years ago - but if the chance of winning, according to AOL, is about 125 million to one. Then couldent one buy all the combinations with about 125 million tickets and the payoff is - was 315 million = about 170 million if only one winner gets to split the take?

AWB
12-28-2002, 07:16 AM
Originally posted by Desmostylus
The answer to the OP could be:

32!/(26!x6!) = 906,192

32x32!/(27!x5!) = 6,444,032

27x32!/(27!x5!) = 5,437,152

I repeat: What's the significance of "powerball is the 6th number"?

Cabbage was correct: the answer is 6,444,032 tickets

The formula is:

nCr * X or

n!/(n-r)!r! * X

where n is the size of the range of numbers on the balls, r is the number of balls drawn, and X is the range of the Powerball. In this OP, n is 32, r is 5, and X is 32.

The significance of the Powerball is that the 6th ball is independent of the other 5. It's drawn from a completely separate bin of balls. That why the answer is 32C5 * 32, not simply 32C6.

Desmostylus
12-28-2002, 07:19 AM
Thankyou.

aahala
12-28-2002, 07:27 AM
If the 120 million figure is approximately correct, then if you purchased one number per second, 24/7, then it would take about 1390 days to purchase every number.

If you used cash, then I would suggest the largest bill in US circulation, \$100. Those packs of bills we see on TV with the paper ban is 100, so we would need 12,000 packs.

I'm guessing we could at maximum get only about 250 packs in a typical briefcase and the thing would be very heavy to tote. About 48 of those would be needed.

The volume of paper you would get in return would be significantly greater than the \$100 bills, so you better plan on having additional boxes.

AWB
12-28-2002, 07:28 AM
Also, the problem with buying all the tickets in any lottery is that you have to fill out all the cards to be scanned by the lottery machine. You can't just ask for 6,444,032 tickets. After a while, it'd be not unlikely for it to randomly pick a line of numbers that it had previously picked.

So, besides filling out 1,288,807 cards with all the unique combinations, you'd have to find quite a few clerks willing to scan thousands of your cards. And you'd have to incur the wrath of the customers behind you that you'd displace from buying their own tickets.

(The infamous Australians who did this for the Virginia Lottery bent the rules and asked some lottery retailer to scan their tickets "behind the scenes". Even doing that, they were only able to get about half of them scanned, reducing their chance to 1 in 2. Yep, they could've lost after all that. Since then, VL official rules state that tickets must be bought at retail cashiers, not behind the scenes.

The Austrailans only won about \$28,000,000, which they took as an annuity. They have some complicated payback schedule for their investors which didn't get their initial money back to them for 10 years. A good 7% investment would've done this for them without getting cramped fingers from filling out all the cards. :D )

flex727
12-28-2002, 10:12 AM
Originally posted by Cabbage
There would be 6,444,032 tickets [32 * (32!)]/[(5!) * (27!)].

A group of people from Australia tried to do this (buy up all the tickets) in the Virginia lottery several years ago (44 pick 6, for a total of 7059052 tickets). Each person in the group was responsible for buying large blocks of tickets (I don't know the details on that). Actually, they only managed to buy up about 75% of the tickets, but they did win, and, luckily for them, they didn't have to split it with anyone. I don't remember the jackpot exactly, it was between 10 and 14 million, IIRC, but I could be wrong there. It was a huge risk for them, and I doubt they really came out ahead in the end (and I would figure they were probably in the hole) when you figure in all the taxes, and the fact that the jackpot is paid out as a 20 year annuity.

Why can't you just go to the authority running the lottery and hand them a check for \$7,059,052.00 and ask for a share of the winnings (100% share if there are no other winners). I don't see how this would be dishonest or illegal in any way and the lottery would certainly take in far more money than it would have otherwise.

jacobsta
12-28-2002, 12:36 PM
Just a note that the REAL powerball formula is going to be 53!/48!5!*42, giving us the expected odds per the lottery = 120526770
or just slightly worse than 1 in 120million, just as the lottery says

BrandonR
12-28-2002, 01:51 PM
Is there a straight answer yet? Too many numbers... So many formulas... Such different answers.

mangeorge
12-28-2002, 02:12 PM
Sure there's a straight answer. Go see a good movie. Ferget the lotto.
Peace,
mangeorge

Church Key Kid
12-28-2002, 02:48 PM
Yes, it is possible to buy all the combinations for a lottery.

However, you need to take into account the following things:

Let's use the Ohio Lottery for example. The odds of getting 6 out of 6 numbers are 1 in 13,983,816.

To buy each combination, it would cost \$13,983,816 (not counting your time (time to fill out all the cards, and time to scan them all in), labor (you would have to hire people to go to other stores to help you can them all in), and supplies (assuming 10 picks to a card, you don't think that the Lottery Commission is just going to give you 1,398,382 cards to fill out, do you?).

Ignoring the time, labor, and supplies, and assuming that you did not want the annuity option, but instead wanted the lumpsum option, the Ohio Lottery would need to be at least \$41 Million in order for you to break even, after taxes (the lump sum payment would be \$14,247,500). The Ohio Lottery rarely goes that high, but even if it were, you would need to take into account several other factors.

To choose that many numbers, you would need to fill out 1,398,382 cards. Lets pretend you could build a machine that would calculate the next 10 combinations in the series and fill out the card for you. Assuming it could fill out one card per second, it would take 16.2 days to fill out the cards (of course, you could always fill them out ahead of time and keep them in storage until you need them). I don't even want to think of how long it would take to hire people to fill them out accurately by hand.

Let's assume that it takes 5 seconds to feed the card through the lotto machine and for it to print out your ticket. Ignoring the time it takes to change the ink and paper, and assuming that you were the only one in line, it about 1942.2 man hours to scan them all. In the Ohio Lottery, you realistically have 4 days from the Saturday drawing to the Wednesday drawing. However, you can only scan during a certain timeframe (which I believe is 6am-11pm). Let's pretend you start scanning at 8pm on Sat night, which gives you 3 hours on Saturday, 17 hours each on Sun, Mon, and Tues, and 13.5 hours on Weds (Lottery scanning stops at 7:30PM). That means you have only 67.5 manhours per person to scan. Therefore, you would need to have 29 people, including yourself, to help you scan for those 5 days.

You also have to take into account that if the lottery were \$41 million, chance are that 2 or more people playing will win. That means that you would win, and at least one other person would win. Therefore, your winnings would only total \$7.1 million, after taxes. You just spent \$13,983,816 to win \$7.1 million!

So, yes, you could do this. However, I doubt you have \$13,983,816 just laying around that you could spend on the lottery. If you did, I doubt you would need to play the lottery.

Even if you had that much, I don't think it would be worth it to risk that much money on the small chance that you would be the only one winning.

(Phew! Does that seem as cumbersome to read as I think? :) )

Robot Arm
12-28-2002, 04:54 PM
All the state lotteries I know of also pay smaller prizes for tickets which match some, but not all of the numbers. One state used to give a free ticket for matching three-out-of-six, and cash for matching four- or five-out-of-six. I don't know the formulae used to determine the amounts.

To calculate whether it would be a worthwhile wager to buy every possible ticket, you'd need to know how many other tickets were sold (to know the chances of splitting the prize with other winners), and add up the expected payoffs of all your winning tickets. If you really want to be precise when you figure the odds, the math would get very hairy indeed.

handy
12-28-2002, 07:03 PM
Powerball Lottery: Odds of Winning a Prize:
http://mathforum.org/library/drmath/view/56122.html

Or:

Q: How do you calculate the odds of winning the Powerball jackpot?

To play Powerball, the player must pick five numbers from a field of 53 numbers and one number from a field of 42 numbers. The following basic odds formula can be used to calculate the odds of winning the Powerball jackpot (1:120,526,770):
http://www.idaholottery.com/faqs.html

dalo, where do you get a lottery with only 32 numbers?