Subject: Re: How to win the lottery (sort of)
Date: Wed, 15 Oct 1997 08:51:12 -0400
From: Alex Bogomolny

Dear Lubna:

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.)

Best regards,
Alexander Bogomolny

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