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DevonFangs
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Joined: November 9th, 2009, 1:49 pm

Two guys meeting

March 21st, 2010, 8:30 pm

They asked me this one- it's pretty easy but nice.Two guys have an appointment, but they don't remember what time. The both of them just remember the hour (say 4pm), but not the minute at which the appointment is (i.e., they don't remember if it's 4.15pm or 4.32pm or whatever else), and so the both of them decide to get there at a random time during that hour (so at a random moment between 4pm and 5pm) and just wait there for 10 minutes.What is the probability that they meet?
 
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dunrewpp
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Two guys meeting

March 22nd, 2010, 3:05 am

Let X and Y be iid u[0,1]. Ans. P{|X-Y|<(1/6)}.QuoteOriginally posted by: DevonFangsThey asked me this one- it's pretty easy but nice.Two guys have an appointment, but they don't remember what time. The both of them just remember the hour (say 4pm), but not the minute at which the appointment is (i.e., they don't remember if it's 4.15pm or 4.32pm or whatever else), and so the both of them decide to get there at a random time during that hour (so at a random moment between 4pm and 5pm) and just wait there for 10 minutes.What is the probability that they meet?
 
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dunrewpp
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Two guys meeting

March 22nd, 2010, 3:10 am

A modified problem: Same situation, except that each guy waits a random time not exceeding a certain amount of time w. Now Find the probability that they meet. For what value of w are the two probabilities from the original version and the modified version the same?
 
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FritzJacob
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Two guys meeting

March 22nd, 2010, 5:44 am

I am assuming both w(x) & w(y) iid u[0,w]I also assume w is in hours.Then p itself is random. p = Pr{-w(y) < Y-X < w(x)}p = 1 - 0.5(1 - w(x))^2 - 0.5(1 - w(y))^2 & p <= 1E[p] = 1 - 0.5E[(1 - w(x))^2] - 0.5E[(1 - w(y))^2] & p <= 1
Last edited by FritzJacob on March 21st, 2010, 11:00 pm, edited 1 time in total.
 
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FritzJacob
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Two guys meeting

March 22nd, 2010, 6:05 am

E[p] = w - (w^2)/3 & p <=1 {if w > 1, this formula changes, with the integration done with new boundaries..anyone interested can try}For initial problem p = 11/36solving the quadratic equation:We get w which gives 11/36 as p.
Last edited by FritzJacob on March 21st, 2010, 11:00 pm, edited 1 time in total.
 
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DevonFangs
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Two guys meeting

March 22nd, 2010, 7:56 am

Fancy modification!The original question was easy because they wanted me to find a number quickly. The question was kinda "can you imagine a graphic way of evaluating this, which is particularly simple when the waiting time (i.e. 10 mins) is a divisor of the total time (i.e. 1 hour)? "
Last edited by DevonFangs on March 21st, 2010, 11:00 pm, edited 1 time in total.
 
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FritzJacob
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Two guys meeting

March 22nd, 2010, 7:59 am

DevonFangs, I edited my answer..can u check.Btw I am not 100% sure if my answer and method is right, waiting for others to comment.
Last edited by FritzJacob on March 21st, 2010, 11:00 pm, edited 1 time in total.
 
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DevonFangs
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Two guys meeting

March 22nd, 2010, 8:21 am

I'm sorry, page not refreshed and I didn't see your answer - which, however, is correct.
 
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wileysw
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Two guys meeting

March 22nd, 2010, 9:52 pm

not sure about the exact "graphic way" DevonFangs has in mind, but one geometrical interpretation of the problem is to calculate the area of a strip along the diagonal in a unit square (defined by |y-x|<=w, where w is the waiting time as a fraction of the total time. w=1/6 here).one way to get the area is to subtract two triangles from the square which gives 1-2*(1-w)^2/2=2*w-w^2. w=1/6 results 11/36 as FritzJacob gave.[one could generalize to n people, all waiting for w, and consider the volume of a region along the diagonal of the hypercube defined by |max-min|<=w. the result is n*w^(n-1)-(n-1)*w^n]if the waiting time for each person could be different, the strip is not symmetric w.r.t. the diagonal any more. still simple for two person, but very messy for the general case.for dunrewpp's question, FritzJacob's answer is correct, and one could easily see the geometrical meaning of each term in p. the final result will be slightly larger than 1/6*2*60=20mins (exact result is w=[3/2-2/sqrt(3)]*60~20.7min), as w^2 is convex, and has a slightly larger average than the average squared.----- ----- ----- ----- -----there is a very nice discussion of the general case
Last edited by wileysw on March 23rd, 2010, 11:00 pm, edited 1 time in total.
 
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FritzJacob
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Two guys meeting

March 23rd, 2010, 2:25 pm

QuoteOriginally posted by: FritzJacob{if w > 1, this formula changes, with the integration done with new boundaries..anyone interested can try}E[p] = 1 - (1/3w)
 
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DevonFangs
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Two guys meeting

March 23rd, 2010, 5:09 pm

wileysw,exactly what I had in mind.(The fancy thing is that they asked me a graphical rep during a phone interview!)