Based on 6 numbers from 1-40 with $1.00 per ticket.
2006-08-21 08:32:44 · 34 answers · asked by butnozzle 2 in Science & Mathematics ➔ Mathematics
It's completely impossible. Because more than one person can have the winning numbers, so you'd only get half/a third/a quarter of the jackpot.
2006-08-21 08:38:42 · answer #1 · answered by Anonymous · 1⤊ 1⤋
Yes, it is approzimately $3.8million, but it would still be impossible to buy all the tickets. If you had a dedicated machine that could print out 1 ticket/second, every second of every day, it would take nearly 44 days just to print out the tickets. Now, of course, that assumes you had all the tickets marked beforehand and did not make any mistakes in marking them.
When the NY Lotto first came out, over 10,000 people bought the numbers 1, 2, 3, 4, 5, & 6. each week. With the $1 million dollar prize at that time, each winner would have gotten $100 before taxes. Another fact, in all the lotteries in all the world, the numbers 1, 2, 3, 4, 5, and 6 have never come up.....Hmm.
Maybe I should buy them, they are due...
2006-08-25 05:56:16 · answer #2 · answered by jimbo_wizard 5 · 0⤊ 0⤋
Well to find the answer you need to know whether the order of the numbers chosen are important. For example, if the order is important then 5 26 3 3 11 is different then 26 3 5 11 3 even though it has the same numbers.
If the order of the numbers are important then you would use a permutation to get 40!/34! = 2.76 trillion dollars. Thats a lot of money!
If the order of the numbers were not important you would use the binomial coefficient which is 40!/(6!34!) = 3.8 million dollars.
I Hope this was helpful.
2006-08-22 05:22:31 · answer #3 · answered by Elim 5 · 0⤊ 0⤋
I'm not sure how many numbers you countries lottery has, but based on the UK lottery, it's 6 numbers correct from 6 choices of 49 numbers:
The odds are 1 in 13983816 - So statistically speaking you would have to buy 13983816 tickets, each with a different number sequence. Assuming you can buy one ticket every 10 secs - that should take you about 4.5 years
Good Luck!
2006-08-21 08:49:55 · answer #4 · answered by the_big_v 5 · 1⤊ 0⤋
Making a bunch of assumpitons, you will need to spend $3,838,380 to guarantee a win on the jackpot (if the order of the numbers did not matter!!). Depending on the jackpot size, number of winners and number of 'secondary' prizes this would determine how much the jackpot woulf need to be. WIth the smaller odds (in Canada, the lottery is 1 in 13 million) the chances of a split of the jackpot is better. Also, another question is would you be able to create and process all of those slips in time for the lottery, because even if you could do one every second, it would take almost 45 days to do.
2006-08-21 08:50:43 · answer #5 · answered by Nice Guy 3 · 0⤊ 1⤋
Ther are 3,838,380 possible combinations, it would cost you $3,838,360 dollars to play every combination. HOWEVER, bear in mind that some people will win 3, 4, and 5 number payouts, someone ELSE may also win the jackpot and you would have to split it, to to take a lump sum cash payout usually means a reduced amount due to present values and the prevailing interest rate, AND last but not least, lottery winnings are taxable.
To the person who answered that there are billions of combinations, sorry, but you are wrong. The formula is NOT a permutation, it's a combination formula: 40! / (6!)(34!)
2006-08-21 08:43:22 · answer #6 · answered by Anonymous · 2⤊ 0⤋
If you are the sole winner, you will need 3.83838 million dollars to have inevitable win. method 40 factorial / 6 factorial. if there is no repeating number allowed, this is the correct answer, trust me, or find out yourself from your maths teacher, the real cleaver one of course.
if you really want to get rich, i think buying a lottery is a way going no where, how about robbing a bank, the chance of you successfully robbing a bank and get away with it should be higer than wining jackpot in lottery but don't do it because it's illegal, ok? ^^
2006-08-21 08:42:36 · answer #7 · answered by lippy19850528 3 · 1⤊ 0⤋
The odds vary slightly because of the parameters of individual games (eg. Powerball), but even if you use the conservative example of 4 million to one odds, and even if you could front that much money, were the sole winner and devise a system where no number could possibly be missed with a game terminal you could monopolize you still would have the obstacle of the processing time for all of those tickets exceeding the time interval until the next game.
2006-08-21 08:52:03 · answer #8 · answered by Traveller 3 · 1⤊ 0⤋
It's rather the other way round, because in an ideal lottery the jackpot is all the money paid by those playing, so if you play every number, and one other person plays, playing one non-winning number, you make a profit of $1.00. If 100,000,000 other people all play one non-winning number, you make $100,000,000! But, if even one other person gets the winning number, ...
2006-08-21 08:51:18 · answer #9 · answered by Sangmo 5 · 0⤊ 0⤋
I have read about people doing this when the jackpot gets big enough. They buy a liquor store and run it day and night printing tickets with all numbers
If I did the math right with 40 possible number picked 6 times you would only need 3.8million.
2006-08-21 08:44:41 · answer #10 · answered by rscanner 6 · 1⤊ 0⤋
it's pretty simple to calculate, actually. you just need to run these numbers to get the total possible number combinations:
40*39*38*37*36*35 = 2,763,633,600
since you make it easy at $1 per ticket, the pot would have to be more than the number listed above (in dollars) for you to win. HOWEVER, there are ALSO prizes for matching partial numbers and this assumes that you wouldn't have to split the pot with another grand prize winner. Not to mention that lottery pots don't generally (or ever) go into the three billion dollar range
2006-08-21 08:43:21 · answer #11 · answered by promethius9594 6 · 1⤊ 1⤋