Subject: Re: How to win the lottery (sort of)
Date: Wed, 15 Oct 1997 08:51:12 -0400
From: Alex Bogomolny
If you are bent on playing lotery you should probably count more on your luck than math. However, the question is quite legitimate.
I'll use (m n) for the binomial coefficient: (m n) = m!/(n!(m-n)!).
There are (6 3) ways to choose three numbers out of six. Each of this may be matched with (47 3) triples (selecting 3 numbers out of remaining 50-3=47.) So the total is (6 3)(47 3). However, some 6-tuples have been counted more than once. Apply the idea of the inclusion-exclusion principle: remove from that number the number of ways to select 4 elements out of 6 times the number of ways to select 2 out of the remaining 46. But now you overdid it. Add to this the number of ways to select 5 out of 6 times the number of ways to select 1 out of the remaining 45. Add one.
(6 3)(47 3) - (6 4)(46 2) + (6 5)(45 1) - (6 6)(44 0)
It's 309044 but you may want to verify this. This is how many 6-guesses match 3 numbers. To get the probability, divide this by the total (50 6) which is 15890700 (You may want to check this one too.)