http://tech.groups.yahoo.com/group/mathforfun/ MATH for FUN - A place to post and work math problems. Tue, 17 Nov 2009 17:06:30 GMT video_ranger http://tech.groups.yahoo.com/group/mathforfun/message/13942 http://tech.groups.yahoo.com/group/mathforfun/message/13942 ... Thanks - yes picking an odd number near the center or even at the exact center (if it's odd) would be the thing to do to get the game over with quickly Mon, 16 Nov 2009 20:10:20 GMT clooneman2000 http://tech.groups.yahoo.com/group/mathforfun/message/13941 http://tech.groups.yahoo.com/group/mathforfun/message/13941 Brilliant, sir. Of course, the disclaimer is that the optimal strategy doesn't guarantee victory, but it does increase your chances. And, if you have to choose Mon, 16 Nov 2009 19:55:57 GMT clooneman2000 http://tech.groups.yahoo.com/group/mathforfun/message/13940 http://tech.groups.yahoo.com/group/mathforfun/message/13940 The game described is one between two players - you and your opponent, for example. So yes, it is a scenario where, by definition, one player will win and one Mon, 16 Nov 2009 16:16:22 GMT video_ranger http://tech.groups.yahoo.com/group/mathforfun/message/13939 http://tech.groups.yahoo.com/group/mathforfun/message/13939 Peter's explanation for choosing a number on one of the extremes looks reasonable - normally after an incorrect guess there's some helpful information saying Mon, 16 Nov 2009 13:24:52 GMT bogaduck http://tech.groups.yahoo.com/group/mathforfun/message/13938 http://tech.groups.yahoo.com/group/mathforfun/message/13938 I am thinking that it is either you or your opponent who wins. Since any number is equally likely to occur, it doesn't matter which number you choose - your Sun, 15 Nov 2009 22:14:57 GMT clooneman2000 http://tech.groups.yahoo.com/group/mathforfun/message/13937 http://tech.groups.yahoo.com/group/mathforfun/message/13937 Hmmm... I like it. Even when I was wondering to myself that perhaps choosing from the extremities of the selection was the best way to go. I won't be attmpting Sat, 14 Nov 2009 19:18:57 GMT video_ranger http://tech.groups.yahoo.com/group/mathforfun/message/13936 http://tech.groups.yahoo.com/group/mathforfun/message/13936 Claim (I have a proof but maybe someone else wants to try to prove or disprove it): Say there are N numbers left, this is a generalization from those cases Fri, 13 Nov 2009 12:42:25 GMT clooneman2000 http://tech.groups.yahoo.com/group/mathforfun/message/13935 http://tech.groups.yahoo.com/group/mathforfun/message/13935 Nope; what I have in mind is two players in competition with one another, not one person taking repeated guesses. Fri, 13 Nov 2009 12:35:00 GMT clooneman2000 http://tech.groups.yahoo.com/group/mathforfun/message/13934 http://tech.groups.yahoo.com/group/mathforfun/message/13934 That's exactly the approach I was thinking of. When there are two numbers to choose from, it doesn't matter which number you choose; you have a 50-50 chance of Thu, 12 Nov 2009 22:33:50 GMT video_ranger http://tech.groups.yahoo.com/group/mathforfun/message/13933 http://tech.groups.yahoo.com/group/mathforfun/message/13933 Yes this looks like a more difficult and interesting problem - thanks Clooneman. I haven't solved it yet but consider that after the first player chooses, the Thu, 12 Nov 2009 22:02:09 GMT Rick http://tech.groups.yahoo.com/group/mathforfun/message/13932 http://tech.groups.yahoo.com/group/mathforfun/message/13932 The course numbering system is different at every college, so "Math 20" is meaningless outside of your college. Look up the course outline on the college web Thu, 12 Nov 2009 21:22:12 GMT Amal Abdallah http://tech.groups.yahoo.com/group/mathforfun/message/13931 http://tech.groups.yahoo.com/group/mathforfun/message/13931 hi, I need to take a math 20 placement test at my home community collage, I took math back home in Lebanon and was instructed in French and therefore I am not Thu, 12 Nov 2009 20:56:12 GMT Edward Moore http://tech.groups.yahoo.com/group/mathforfun/message/13930 http://tech.groups.yahoo.com/group/mathforfun/message/13930 I only quickly scanned through that site, but this looks slightly different than the well-known puzzle - this looks like there are three people, one who picks Thu, 12 Nov 2009 19:53:15 GMT Rick http://tech.groups.yahoo.com/group/mathforfun/message/13929 http://tech.groups.yahoo.com/group/mathforfun/message/13929 Yes, there is. It is well-known in computer science. See http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game -- Rick ... From: Thu, 12 Nov 2009 19:46:04 GMT clooneman2000 http://tech.groups.yahoo.com/group/mathforfun/message/13928 http://tech.groups.yahoo.com/group/mathforfun/message/13928 Here's one for you all to put your brains together about. A number is chosen between 1 and... say, 10, *inclusive*. Two players then take it in turns to guess