tag:blogger.com,1999:blog-4115025577315673827.post5813664195511634148..comments2019-10-23T18:01:26.365+05:30Comments on CSE Blog - quant, math, computer science puzzles: Choose the maximum of two numberUnknownnoreply@blogger.comBlogger10125tag:blogger.com,1999:blog-4115025577315673827.post-41476318603352584782014-10-02T15:22:49.417+05:302014-10-02T15:22:49.417+05:30If the chosen number is p say bigger with the prob...If the chosen number is p say bigger with the probability (p-a)/b-a. If the other number is q, probability of your winning is 1/2*((p-a) + (b-q))/(b-a) [In case q is smaller than p] and 1/2*((b-p) + (q-a))/(b-a) [In case p is smaller than q]. In both cases probability of your winning is 1/2 + 1/2*|p-q|/(b-a)arpit jainhttps://www.blogger.com/profile/01777952825810370264noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-74717952185926591352012-11-02T17:47:27.912+05:302012-11-02T17:47:27.912+05:30Isn't this the same as choosing my threshold t...Isn't this the same as choosing my threshold to be 0.75? after all my expectation for the limit is 0.75 if I am choosing uniformly.mohithttps://www.blogger.com/profile/10678573228651642137noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-80625492470586054622010-02-09T16:21:13.642+05:302010-02-09T16:21:13.642+05:30@Ramdas..
Choosing mid point does not work. Note t...@Ramdas..<br />Choosing mid point does not work. Note that I have two numbers chosen by me in the two hands. I can choose numbers such that you always lose. Only the hands are chosen by random. :P and not the numbers. So, when you choose k randomly you are doing good. :)Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-71223505955478267162010-02-08T21:25:40.550+05:302010-02-08T21:25:40.550+05:30if the first number is k, say that the other numbe...if the first number is k, say that the other number is smaller with probability (k-a)/(b-a). i think this is the best probability of winning that u can get and is definitely greater than 50%. eg. a=0, b=10, k=3.<br />Say that the other number is smaller with probability 30%. you have a 70 % probability of winning.kphttps://www.blogger.com/profile/06763630657499914791noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-32341768564477501262010-02-08T15:11:58.986+05:302010-02-08T15:11:58.986+05:30That's as good as saying, choose the midpoint ...That's as good as saying, choose the midpoint of the interval. If one number (the chosen one) lies on one side of the midpoint, then probably the other number lies on the other side, but I'm going to be EVEN safer by saying it is just greater than the picked one. Since it is a uniform distribution of course. Trivial?<br /><br />There are 4 cases of where the numbers lie wrt the midpoint - LL,RL,LR,RR - each with equal probability. I'm obviously correct in the LR and RL case. Which is 0.5 prob. I'm also correct in the LL case when I chose the smaller of the two. And I'm also correct in the RR case which I chose the larger of the two.<br /><br />Now, the problem doesn't seem as impressive or as much fun :PYa, I Am A Dastardhttps://www.blogger.com/profile/04386601441575337262noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-55041898740259685042010-02-05T22:46:09.291+05:302010-02-05T22:46:09.291+05:30@Ramdas.. had a discussion with a junior yesterday...@Ramdas.. had a discussion with a junior yesterday... came to the conclusion that the point raised by you is indeed correct. Sorry for the trouble. Don't know how Winkler did this mistake. :(<br /><br />Hence, changing the problem statement to all numbers in the range [a,b]<br />ThanxPratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-13359619241528157132009-12-10T12:55:07.879+05:302009-12-10T12:55:07.879+05:30@ramdas..
Relabeling the post under unsolved categ...@ramdas..<br />Relabeling the post under unsolved category until someone can answer ur point in a more convincing way :)<br /><br />Any takers??Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-61467347614813343752009-12-10T12:20:27.818+05:302009-12-10T12:20:27.818+05:30just a point to add, the probability that a random...just a point to add, the probability that a randomly chosen point in the reals lies in a finite interval (a,b) is _zero_. that is if you can define probability over an interval on the reals (which u cant, I think). so it doesnt even make sense to say "randomly choose a point on the real line" - how? uniformly? u cant! :P anyway, ditch :PYa, I Am A Dastardhttps://www.blogger.com/profile/04386601441575337262noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-48600574694199054272009-12-10T12:17:35.058+05:302009-12-10T12:17:35.058+05:30@ramdas..
As written in the post, source of the pr...@ramdas..<br />As written in the post, source of the problem and solution is the book by P. Winkler (The only godfather of puzzle alive probably :P).. So, to your question, you cannot say that the solution is non-sense. :P <br /><br />I also had the same reaction but then it said that the question appeared in some math olympiad. Those guys know measure theory well :P :PPratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-27883730477199823032009-12-10T11:42:12.618+05:302009-12-10T11:42:12.618+05:30if I may, this solution is nonsense?!?! (as i foun...if I may, this solution is nonsense?!?! (as i found on discussing it :P) As good as saying if the number is positive, keep it (assume the random number u chose in ur head is 0). Is it really true that the probability of a number lying between -N and +infi is different from -N and -infi? :P Kya fart hai :P I guess we all need to learn some basic measure theory first :PYa, I Am A Dastardhttps://www.blogger.com/profile/04386601441575337262noreply@blogger.com