Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.3 4.3bsd-beta 6/6/85; site harvard.UUCP Path: utzoo!linus!decvax!genrad!panda!talcott!harvard!greg From: greg@harvard.UUCP (Greg) Newsgroups: net.puzzle Subject: Re: Winning 1/3 of the Lottery (**SPOILER, NOT ROT13**) Message-ID: <540@harvard.UUCP> Date: Thu, 5-Dec-85 13:11:53 EST Article-I.D.: harvard.540 Posted: Thu Dec 5 13:11:53 1985 Date-Received: Sat, 7-Dec-85 15:58:31 EST References: <25@bbncc5.UUCP> <234@bbncc5.UUCP> Reply-To: greg@harvard.UUCP (Greg) Organization: Harvard Lines: 38 Summary: Boy, I am a real Dodo... (*spoiler, not rot13*) Mr. Denenberg has shown me the error of my ways...I thought that ten was the minimum number of tickets, but he says thta the answer is nine tickets. With the knowledge that the answer is nine, an example of a winning strategy is clear. The nine tickets are: #1: 1 2 3 4 5 6 #2: 7 8 9 10 11 12 #3: 13 14 15 16 17 18 #4: 19 20 21 22 23 24 #5: 19 20 21 25 26 27 #6: 22 23 24 25 26 27 #7: 28 29 30 31 32 33 #8: 28 29 30 34 35 36 #9: 31 32 33 34 35 36 Why does this work? Suppose the winning ticket does not have two numbers in common with any of the nine. It can have at most one number between 1 and 6, at most one between 7 and 12, and at most one between 13 and 18. That leaves at least three numbers between 19 and 36. There are two possibilities: At least two numbers are between 19 and 27, or at least two numbers on the winning ticket are between 28 and 36. If the former is true, then there are the following possibilities: The smaller # is between: The larger # is between: They are both in ticket: 19 and 24 19 and 24 #4 19 and 21 25 and 27 #5 22 and 24 25 and 27 #6 25 and 27 25 and 27 #6 If, on the other hand, there are two numbers between 28 and 36, then essentially the same thing happens, except with tickets #7-9 instead of #4-6. Not only that, eight tickets is definitely not enough, but the proof is long and tedious, and at the moment I don't have the stamina to type it up. -- gregregreg