Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10 beta 3/9/83; site fluke.UUCP Path: utzoo!linus!decvax!tektronix!uw-beaver!ssc-vax!fluke!prodeng From: prodeng@fluke.UUCP (Jim Hirning) Newsgroups: net.math Subject: Re: circle probability Message-ID: <704@vax2.fluke.UUCP> Date: Fri, 16-Sep-83 14:43:32 EDT Article-I.D.: vax2.704 Posted: Fri Sep 16 14:43:32 1983 Date-Received: Sat, 17-Sep-83 18:19:05 EDT References: <1141@emory.UUCP> Organization: John Fluke Mfg. Co., Everett, Wash Lines: 87 No answers yet to what I thought was a fairly interesting problem: What is the probability that 3 points chosen at random on a circle lie within a semicircle (i.e. all within 180 degrees)? That's not so hard, so then answer the same question for n points. If that's too easy, try it on the surface of an m-sphere. (I don't know the answer for the m-sphere - if anyone can give me a reference I'd be grateful (or a solution)). ----------------------------------------------------------- John Pedersen {sb1,akgua}!emory!jfp I am really bothered by this problem. Correct me if I'm wrong, but I had the impression that probability dealt with descrete objects, not continuous ones. A circle is most definitely continuous. Even more to the point, how can we construct ratios of (ways we can get 3 points to lie within a semicircle) to (ways we can pick 3 points at random) when there are uncountably many ways to do both? (Remember the points on a circle are uncountable.) HOWEVER, ignoring both of these problems, it seems to me that this may be the way John Pedersen looks at the problem: If we choose 1 point at random, it certainly is in the same semicircle with itself. (probability is 1) If we choose 2 points at random, the smaller arc between them must be <= a semicircle. So, again probability is 1. Once 2 points have been chosen, the probability that a third point will lie within a semicircle containing the other two points is 3/4. Here is how I come up with the number 3/4: Let A and B be the first two points. It is equally likely (?????) that the arc between them will encompass any angle between 0 and 180 degrees. To find out what part of the circle would place a third point within a semicircle containing A and B, consider two semicircles containing A and B: one with endpoint A, the other with endpoint B. If we take the union of these two semicircles, we will encompass the largest portion of the circle in which the third point may fall and still share a s.c. (semicircle) with A and B. If you draw a diagram (my terminal isn't too hot at circles :-) ) you will see that the union of these two s.c.'s covers an arc of angle 360 degrees - (size of small arc AB in degrees). Thus if A and B are very close, the third point C may fall almost anywhere in the circle. If arc AB is 180 degrees, this fails, but since this is only one of an infinitely many (in fact uncountably many) cases, it shouldn't amount to a hill of beans. (I hope!!!) So, if we consider all AB arcs between 0 and 180 degrees equally likely (a big IF), then allowable arcs for C between 360 and 180 are equally likely. So, we average (over an uncountable set! I told you this problem REALLY bothers me) to get an available arc for C of 270 degrees, or 3/4 of the circle. So, on random choice of C, there is a 3/4 chance of it being in the same s.c. with A and B. WHEW! When we consider more points than 3, I believe the same strategy applies, but instead of considering the arc between A and B, consider the arc encompassing all the points considered so far. Assume that all are within one s.c., because if they weren't, no method of choosing a next point will make them all in one s.c. Again, the probability at each stage is 3/4. So, multiplying all the probabilities together, for n points, we get ****(3/4)^(n-2)****. (3 points, (3/4)^(3-2)=(3/4)^1=3/4) For n < 3 the probability is 1. I won't be on the net at this address beyond September 23, so please don't send me mail if you don't think it will get here before then. Please do comment to the net, I'd appreciate anyone shedding some light on this subject. Thanks. Debbie Smit fluke!prodeng (until Sept 23)