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Eyad
 2 years ago
Best ResponseYou've already chosen the best response.1An ant stands in the middle of a circle (3 metres in diameter) and walks in a straight line at a random angle from 0 to 360 degrees. Problem is, it can only walk one metre before it needs a break. The ant has the memory of a fish and forgets what direction it has just walked in.. Anyway, after the break, it gets all dizzy and thus chooses another random direction from 0 to 360 in an attempt to escape the circle again. As you can well imagine, it could escape the circle after just 2 walks (just one break needed). Or... it could take 20,000 walks (19,999 breaks needed)!! There might even be the very slim possibility it might take 20,000^20,000 walks.... What is the average amount of walks required for the ant to escape the circle?

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.1@satellite73 ,@Calcmathlete

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0there is no snap solution to this problem i think try googling it and you will come up with many many posts, since apparently this is a common problem no good solutions though

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0am I on the completely wrong track to say that since he has a remaining 0.5 to go to escape the circle that \(\cos\theta\ge\frac12\implies\frac\pi6\le\theta\le\frac\pi6\)

amrit110
 2 years ago
Best ResponseYou've already chosen the best response.0not completely wrong....ur on the right track...:)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0I mean \[\frac\pi3\le\theta\le\frac\pi3\]

amrit110
 2 years ago
Best ResponseYou've already chosen the best response.0yes...now u just compute the average

amrit110
 2 years ago
Best ResponseYou've already chosen the best response.0u do realize the ant neednt exactly go by ur condition!!

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0so that is 1/3 of a circle... so there is a 1/3 chance he gets out on move 2 where do I include the average, I still don't quite get it :/

amrit110
 2 years ago
Best ResponseYou've already chosen the best response.0true...good enough. avearge is when u consider a particular no. of test cases. so if u consider 100, there is 33% chance of getting out on the 2nd move. now all u have to do is calculate the remaining chances for the other angles he might take....now although this is quite random, the probability for each angle range decreases n goes like a geometric series

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0yeah that's where the problem for me is... a geometric series is a good idea if we can set one up, but since the exact angle that he returns on is variable that won't be easy

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0like he could godw:1341369002782:dw(move 1)

amrit110
 2 years ago
Best ResponseYou've already chosen the best response.0yup...its a tedious problem...but i think a general solution is difficult but not impossible to obtain.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1341369028085:dwthat leaves him with a 0% chance of leaving the circle in the next move, ordw:1341369061211:dwin which case he has some other odds of leaving the circle in the next move

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0@Zarkon probability problem, interested?

amrit110
 2 years ago
Best ResponseYou've already chosen the best response.0for all u know, there is even the possiblity like the question itself that the ant might not come out at all if he keeps taking angles that put him back inside.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0that was suggested by @Outkast3r09 on an earlier post, but @Eyad says there is a definite answer, so...

amrit110
 2 years ago
Best ResponseYou've already chosen the best response.0http://math.stackexchange.com/questions/133851/antinacircle. this page has a good solution.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0well that still is not as elegant as I was hoping where's the actual answer now...?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.011.4 is the "expected value" so I guess that's the answer

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.1Ok Then Ty everyone :) Ty @TuringTest ,@amrit110 :ty for the site its useful :)
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