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        riddles >> medium >> Meeting probability
        
(Message started by: crocodile on Oct 21^st, 2003, 3:31am)

Title: Meeting probability
Post by crocodile on Oct 21^st, 2003, 3:31am

Two persons agreed to meet in a definite place between noon and one o'clock. If either person arrives while the other is not present, he or she will wait for up to 15 minutes. Calculate the probability that the meeting will occur, assuming that the arrival times are independent and uniformly distributed between noon and one o'clock.

Title: Re: Meeting probability
Post by william wu on Oct 21^st, 2003, 3:40am

Classic problem; it was introduced to me as "Romeo and Juliet" :)

Title: Re: Meeting probability
Post by towr on Oct 21^st, 2003, 4:02am

I didn't know this one yet..

::[hide]I get 7/16th, taking one person as the reference, they will meet if the other is within 15 minutes earlier or later, at the start of the hour there's only 15 minutes left of that interval, at the end as well, and in the middle half hour there's a half hour window of opportunity, so we have
((15+30)/2*1/4 + 30* 1/2 +(30+15)/2*1/4)/60=7/16[/hide]::

Title: Re: Meeting probability
Post by william wu on Oct 29^th, 2003, 3:46am

It's the right answer, although I find the reasoning a little vague -- while it's true that at those particular arrival times for the first person, you have those corresponding windows of opportunity, it's not clear to me how that translates to considering all possible arrival times. I did it graphically:

Title: Re: Meeting probability
Post by towr on Oct 29^th, 2003, 3:51am

on 10/29/03 at 03:46:23, william wu wrote:

It's the right answer, although I find the reasoning a little vague

Well, it's pretty much the same thing you did, but I eliminated the pink area under the gray area.. and thus just looked at the height of the gray area in the y direction. (which first grows to 30, then stays constant for 30 minutes, then decreases to 15 again)

Title: Re: Meeting probability
Post by Sir Col on Oct 29^th, 2003, 5:09am

My approach is a little cumbersome, but using rectangle and triangle probability distributions...

Let X and Y be the times that each of the people arrive: X~R(0,1) and Y~R(0,1).

Let Z be the difference in their times: Z=X–Y, so Z~T(-1,1).

By considering the graph (see below), the area under the triangle must be 1, and as the base is 2 units, the height (on the y-axis) will be 1.

As P(-1/4 [le] Z [le] 1/4) = 2P(0 [le] Z [le] 1/4), P(0 [le] Z [le] 1/4) = (3/4+1)(1/4)/2 = 7/32.

Hence P(people meet) = P(-1/4 [le] Z [le] 1/4) = 7/16.

Title: Re: Meeting probability
Post by birbal on Jul 25^th, 2010, 11:24pm

Nice problem...;)
Will the answer change if we take is to discretization level of seconds ?

Title: Re: Meeting probability
Post by towr on Jul 26^th, 2010, 1:36am

Not much, if any.
If you take Williams figure as starting point, instead of straight lines between the gray and pink areas you'd have a line made up of steps.