tag:blogger.com,1999:blog-264226589944705290.post5829004933277481118..comments2023-11-05T03:45:25.001-08:00Comments on God Plays Dice: Probabilities on the circleMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-264226589944705290.post-50858187834547593362010-09-20T11:36:45.211-07:002010-09-20T11:36:45.211-07:00Was thinking about dropping n points on circle, an...Was thinking about dropping n points on circle, and it occurred to me that for any successful outcome, there exists exactly one of the points where a clockwise (WLOG) semi-circle contains all n points.<br /><br />We can then solve the trivial problem of finding the probability that the next n-1 points all fall in the clockwise semi-circle determined by the first point.<br /><br />1/2^(n-1) is this answer.<br /><br />Essentially we're realizing that the order of dropping doesn't matter, so we're just looking at the mass of the set of all successful outcomes, and realizing that we've counted 1/n of them. So, the total probability is just n/2^(n-1). <br /><br />This avoids conditional probability. We instead find a clever way to identify the set of outcomes in question, such that computing the probabilistic measure of each is trivial.Bradhttps://www.blogger.com/profile/10157211023456808058noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-71604963469783605502009-12-12T05:43:21.144-08:002009-12-12T05:43:21.144-08:00Interesting blog as for me. I'd like to read s...Interesting blog as for me. I'd like to read something more concerning this matter. Thanx for sharing this info.<br />Sexy Lady<br /><a href="http://www.baccaratgirls.com" rel="nofollow">English escort</a>Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-65333459807292791072008-07-30T02:01:00.000-07:002008-07-30T02:01:00.000-07:00This is very important site for math’s students. M...This is very important site for math’s students. Math is my favorite subject so I like this site very much.<BR/>___________________________<BR/>george<BR/><A HREF="http://www.widecircles.ca" REL="nofollow">widecircles</A>Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-40971869030508211522008-07-29T00:17:00.000-07:002008-07-29T00:17:00.000-07:00Sorry guys. I am not so good in maths.==========Sa...Sorry guys. I am not so good in maths.<BR/><BR/>==========<BR/>Sam<BR/><A HREF="http://www.globalinternetmarketing.net" REL="nofollow">Wide Circles</A>samhttps://www.blogger.com/profile/12290174624416774138noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-30015432588307805842007-11-23T09:42:00.000-08:002007-11-23T09:42:00.000-08:00Scrap that last argument. I've just realised what ...Scrap that last argument. I've just realised what i was doing wrong. thanks anywayAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-11708479655101666262007-11-23T09:29:00.000-08:002007-11-23T09:29:00.000-08:00Please tell me what's wrong with this argument:The...Please tell me what's wrong with this argument:<BR/><BR/>The first and second points are free to be anywhere on the circle.<BR/><BR/>Whether or not the three points lie in a semicircle depends on the placement of the third point. The proprtion of the circumference where this will happen will depend on the acute angle (theta) between between the first two points.<BR/><BR/>theta=180<BR/>For example if 1st point was at 0 degrees and the second at 180. Then the three points will all lie in a semicrcle with certainty.<BR/><BR/>theta=0<BR/>If both points lie at 0 degrees, again they all lie in semicirlce with certainty.<BR/><BR/>theta=90<BR/>If 1st point lies at 0 degrees and second at 90, then the three points will lie in the same semicircle with probability 3/4. <BR/><BR/>As the theta changes from 0 to 90, the probability decreases linearly from 1 to 3/4. As theta increases from 90 to 180 the probability increases from 3/4 back to 1.<BR/><BR/>The probability is therefore 7/8.<BR/><BR/>(Much easier to explain with diagrams)<BR/><BR/>ThanksAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-52514140953354845832007-10-10T17:47:00.000-07:002007-10-10T17:47:00.000-07:00Anonymous, close but wrong.The ans. = 1 - (exp(pro...Anonymous, close but wrong.<BR/><BR/>The ans. = 1 - (exp(prob.(third point is inside x)).<BR/><BR/>The prob that the angle = x is the same for every value of x: 0. The probability for the angle between the first two points is uniform. You gave the probability that the angle is <B>less than or equal to</B> x, which is x/pi. <BR/><BR/>The expectation of the probability of the third point being in that arc is simply the integral of (1/pi) * (x/2pi) from 0 to pi.<BR/><BR/>The integral is (1/pi) * (x^2/4pi). Evaluated at the endpoints this gives 1/4. One minus this expecation = 3/4.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-66697909213460131432007-10-10T16:22:00.000-07:002007-10-10T16:22:00.000-07:00i see your explanations - but somehow i am confuse...i see your explanations - but somehow i am confused.<BR/><BR/>please temme if this logic is sensible?<BR/><BR/>ans = prob(angle subtended by two points = x) * prob(third point is inside x)<BR/> = x/pi * x/2pi<BR/>integral of above from 0-pi<BR/><BR/><BR/> = pi/6Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-89868142063675283662007-10-10T10:49:00.000-07:002007-10-10T10:49:00.000-07:00I think the easiest way to see this is with some s...I think the easiest way to see this is with some simple diagrams and some conditional probability.<BR/><BR/>Since all three points are placed independently, let's imagine that two points are placed first. The distance between these points is uniformly drawn from [0, .5] circumference lengths. Ignore the endpoints for a moment and consider the (0, .5) cases: draw radii from the endpoints to the center. The first two points and radii through them form a wedge, varying in size from 0 to half the circle.<BR/><BR/>Now extend those radii to be diameters of the circle. It's easy to see that the points that are not within a semicircle of the original two points are the points that fall into the symmetrically opposite wedge. Anything within the larger portion of arc bounded by the diameters drawn from the first two points is within a semicircle of the first two points. The portion excluded is an arc that's symmetrically opposite, and of the same size as the distance between the first two points. This arc size will vary uniformly between 0 and 1/2 circumferences. <BR/><BR/>The conditional probability that the third point will fall in that arc therefore varies uniformly between 0 and 1/2. The expectation of this conditional probability is therefore 1/4. Since the placement of the three points is independent, this conditional probability that the third point does not fall into the same semicircle as the first two, based on the distance between the first two must be the same as the actual probability that the three points will not be in the same semicircle. We are interested in 1 - this probability, or the probability that all three points are within the same semicircle, which is, of course, 3/4, just as you said. <BR/><BR/>Finally, at the two endpoints of the distance between the first two points (which occur with probability zero), the portion of the circle exluded is either zero or 1/2, which averages to 1/4, just as the interior points do.Anonymousnoreply@blogger.com