what is the probability of winning when only buying 1 ticket? What is the probability when buying 100 tickets?

Picking 6 numbers out of 30, where you make exactly 6 picks to get those 6 winning numbers, where all 30 possible numbers are different, picked numbers can't be repeated, and the order in which the numbers are picked doesn't matter:

P(first picked number is a winner) = 6/30

P(2nd picked number is a winner) = 5/29

P(3rd picked number is a winner) = 4/28

P(4th picked number is a winner) = 3/27

P(5th picked number is a winner) = 2/26

P(6th picked number is a winner) = 1/25

P(all 6 of the above events happen) = 6/30 * 5/29 * 4/28 * 3/27 * 2/26 * 1/25

P(all 6 of the above events happen) = (6 * 5 * 4 * 3 * 2 * 1) / (30 * 29 * 28 * 27 * 26 * 25)

P(all 6 of the above events happen) = 24! * 6! / 30!

P(all 6 of the above events happen) = "30 choose 6"

P(all 6 of the above events happen) = 30 nCr 6   (on a calculator)

P(all 6 of the above events happen) = 0.00000168413  (about 1 in 600,000)

Buying 100 tickets and you want to win at least once:

P(at least 1 win) = 1 - (1 - 0.00000168413)^100

P(at least 1 win) = 0.00016839896   (about 1 in 5,938.3)

Buying 100 tickets and you want to win exactly once (not twice or more):

P(win exactly once) = P(win on a single draw) * ( P(lose on a single draw) )^99 * (# ways to arrange 1 win and 100 losses)

P(win exactly once) = 0.00000168413 * (1 - 0.00000168413)^99 * 100

P(win exactly once) = 0.00016838492  (about 1 in 5,938.8)

by Top Rated User (103k points)