To win a lottery, a player must correctly match six different numbers from 1 to 42. If a computer randomly assigns six number for the lottery, how many tickets would a person have to buy to have a 1% chance of winning?
Any ideas?
2007-01-17 02:32:49 · 10 answers · asked by sahsjing 7 in Science & Mathematics ➔ Mathematics
The answer in the book: 52, 722 tickets.
2007-01-17 02:44:53 · update #1
My approach:
Probability to wind: P = (6/42)(5/41)(4/40)(3/39)(2/38)(1/37) = 1.906E-7
# of tickets to buy = 1/(100P) = 5.246E4
2007-01-17 02:58:05 · update #2
It seems that the "correct" answer you are given does not assume that the tickets are all different.
Yes, the probability of winning with one ticket is 1 / 5245786.
If all n tickets are different. the probability of winning with n tickets is indeed n / 5245786, so you need 5245786 / 100 tickets.
But if you have no control over the combinations given to you, they are allowed to be the same and you're in the "without replacement" case. The probability of winning at least once is
1 - p(not winning at all with n tickets) =
1 - (5245785 / 5245786)^n.
Making this equal to 1/100 and solving for n, you get
n = log 0.99 / log(5245785 / 5245786)
= 52722 after rounding up.
2007-01-17 03:24:11 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Many of the others are close if not right. I will change one assumption. My assumption is that all of your tickets are different from each other. The number of combinations is similarly calculated to be 42!/((42-6)!6!)=5245786=a. Basically 42*41*40*39*38*37 possibilities, then divide by 6! because the drawing order doesn't matter. The probability of the first ticket winning is 1/a, but assuming the second ticket is different from first the probably is 1/(a-1), so the total probability of winning with two tickets is 1/a+1/(a-1). The probability of winning with n tickets is summation over i of 1/(a-i) for i=0 to n-1. It turns out the number of tickets needed to win is 52197, which is close to the other number 52458. It is a little less which makes sense because you are using the inclusion of extra information into the problem. Either answer could be right depending on how your tickets are chosen. If they are chosen at random with possibility of repeats 52458 is correct, if repeats are not possible in the tickets you have then 52197 is possible.
2007-01-17 03:49:35 · answer #2 · answered by Josh G 2 · 1⤊ 0⤋
the possibility of 6 numbers taken from 42 is co mbinations of 42 elements taken per 6. = 42*41*40*39*38*37/720>= 5,245,786 . number of tickets/5,245,786 = 1/100
number of tick= 52,458 ( must be whole number)
2007-01-17 04:05:55 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
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2014-10-09 09:07:31 · answer #4 · answered by Anonymous · 0⤊ 0⤋
If the numbers were ordered then
The first number would have 42 possibilities,
the 2nd number would have 41
the 3rd would have 40
..
the 6th would have 37
The total number of ordered combinations is obtained by multiplying
42*41*40*39*38*37 = 42!/(42-6)! = 3,776,965,920 ordered combinations.
But we have to divide this by the number of unordered combination of 6 numbers which 6!
So the total number of unordered combinations is
3,776,965,920/6! = 3,776,965,920/720= 5,245,786
To have 1% chance you would need 1% of this number
so 52,457.86 = 52,458 tickets.
2007-01-17 02:42:03 · answer #5 · answered by catarthur 6 · 1⤊ 1⤋
Morninginfo has the right answer. It is 52,458 tickets to give you a 1% chance of winning.
The person who has a LARGE 3 billion+ combination answer is forgetting one thing. The position of the number does not matter and hence the formula is more of a combination and not permutation. Hence, the number of possibilities is 42! / ((42-6)! * (6)!)
2007-01-17 03:03:12 · answer #6 · answered by Chandra 2 · 0⤊ 1⤋
There are 5,245,786 possible results from the draw. So you need to buy 52,458 tickets to have a 1% chance.
Unless the payout (lump sum, after taxes) is a lot more than 5 million dollars, this is a bad idea.
2007-01-17 02:41:00 · answer #7 · answered by morningfoxnorth 6 · 0⤊ 1⤋
Tell your teacher that is a terrible question, because no matter how many lottery tickets you buy, each one has the same chance of winning, and the more you spend on them the less you're going to win (if you do win).
2007-01-17 02:37:17 · answer #8 · answered by surfinthedesert 5 · 1⤊ 2⤋
Every ticket has a 50% chance of winning or losing. You don't necessarily have a 1% chance of winning, if you buy 1% of the total number of tickets and/or combinations of numbers. Whoever has the winning combination has 100% chance of winning. ALL of the other tickets, have 0% chance of winning. So, if you bought every ticket possible, except the winning ticket, ALL of your tickets would have 0% chance of winning. The only real way you have a 1% chance of winning is if you are one of 100 people that buy the winning ticket, and you have agreed, in advance, to share the winnings equally. If each of the other 99 persons decides to let you have his/her share, you COULD get 100% of the jackpot, but you will surely get 1%.
2007-01-17 02:42:54 · answer #9 · answered by Anpadh 6 · 0⤊ 5⤋
one ticket if your lucky, but if not 252 tickets give you a better chance
2007-01-17 02:43:50 · answer #10 · answered by passion 1 · 0⤊ 0⤋