[quote name=\'Fedya\' post=\'177801\' date=\'Feb 13 2008, 02:32 PM\']
Probability question for the calculus students: New York used to have a lotto that was pick 6 out of 54 numbers, meaning there were 54C6, or 25,827,165 possibile combinations. They also had a "quick pick" that would randomly pick the combination for your ticket. Assuming that the "quick picks" are fully random (and not pseudorandom) in that on any given quick pick, each and every combination is exactly as likely to come up as any other combination, if you play 25,827,165 quick picks, what's the probability that you won't have the winning combination?
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This is a binomial problem, as there are only two outcomes... win or lose.
P(win) = p = 1/x,
P(lose) = q = 1-(1/x)
x=54C6 ("54 choose 6", or 54!/(6!*48!).
You will purchase n tickets, where n=54C6 also.
X is the random variable representing the number of winning tickets bought.
P(X=0) = nCx * p^n * q^(n-x)
Since n=x, nCx is simply 1. Similarly, (n-x) = (n-n) = 0, and q^0 = 1. So this simplifies to just:
P(X=0) = p^n, or (1/54C6)^(54C6). That comes to 0.3679.
Now let's talk about the P of being the only holder of a winning ticket...