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Path: utzoo!linus!philabs!cmcl2!harvard!greg
From: greg@harvard.UUCP (Greg)
Newsgroups: net.puzzle
Subject: Re: Winning 1/3 of the Lottery (*spoiler*)
Message-ID: <516@harvard.UUCP>
Date: Wed, 20-Nov-85 13:43:51 EST
Article-I.D.: harvard.516
Posted: Wed Nov 20 13:43:51 1985
Date-Received: Sat, 23-Nov-85 08:32:29 EST
References: <25@bbncc5.UUCP>
Reply-To: greg@harvard.UUCP (Greg)
Organization: Harvard
Lines: 27

In article <25@bbncc5.UUCP> larry@bbn.UUCP (Larry Denenberg) writes:
>The Massachusetts State Megabucks Lottery works like this:  A ticket costs
>$1 and consists of six distinct numbers of your choosing between 1 and 36
>inclusive.  You may buy as many tickets as you like.  On Lottery Day
>the Authorities choose six numbers.  If those six are the same as your
>six, you win!
...
>What is the minimum number of tickets that you must buy to ensure that
>at least one of your tickets matches two or more of the winning numbers?

Well, if you're incredibly unlucky, you may end up buying every single
ticket which has only one number in common with the winning ticket, and also
every single ticket which has no numbers in common with the winning ticket.
The question remains, how many such tickets are there?

Let A1, A2, ..., A6 be the winning ticket.  On a given unlucky ticket, there
are 35 possibilities (anything other than A1) for the first number, 35
possibilities for the second number (anything other than A2), and so on.
This gives us 35^6 unlucky combinations so far.  But in addition, there are
the combinations which have one number in common with the winning ticket.
In that case, there are 6*35^5 possibilities, because there are six positions
for the matching numbers and 35^5 combinations for the nonmatching numbers.

Thus, if you want to be sure of matching two numbers with the winning ticket,
you must buy 41 * 35^5 + 1 tickets.
-- 
gregregreg