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Path: utzoo!linus!philabs!cmcl2!harvard!bbnccv!bbncc5!akhanna
From: akhanna@bbncc5.UUCP (Atul Khanna)
Newsgroups: net.puzzle
Subject: Re: Winning 1/3 of the Lottery (*spoiler*)
Message-ID: <43@bbncc5.UUCP>
Date: Thu, 21-Nov-85 11:04:41 EST
Article-I.D.: bbncc5.43
Posted: Thu Nov 21 11:04:41 1985
Date-Received: Sat, 23-Nov-85 10:21:02 EST
References: <25@bbncc5.UUCP> <516@harvard.UUCP>
Reply-To: akhanna@bbncc5.UUCP (Atul Khanna)
Organization: Bolt Beranek and Newman, Cambridge, MA
Lines: 35

In article <516@harvard.UUCP> greg@harvard.UUCP (Greg) writes:
>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.
>-- 

Not quite - the problem states that the 6 numbers are distinct, i.e.
order does not matter.  That makes the problem more interesting.


-- 
Atul C Khanna
BBN Communications Corporation, Cambridge MA