

Subject:
Chances of meeting a friend in a foreign city
Category: Science > Math Asked by: macaonghusga List Price: $5.00 
Posted:
24 Jun 2003 17:56 PDT
Expires: 24 Jul 2003 17:56 PDT Question ID: 221349 
I know a number of people, say 100. Every year I visit 3 cities (Vienna, Paris and Cairo) once. In each of those cities there are say 250 places I could be in, and on any one visit I will probably visit 125 of them (eg the train station, 5 cafes, 10 shops etc, parts of a street). Let us define a place as a circle of radius 10 metres. Likewise each of my 100 known people. They visit those three cities (their visits have no relation to mine, or to each others). We may or may not be in those cities at the same time, and if we are we may or may not be in the same place within that city at any time. Our trips are totally independent and we do not know each others travel plans. What are the chances of me bumping into any one of my known people  ie that I and any one of those 100 will, within the same 60 seconds, be in one of those 125 places of one of those 3 cities. (I would like to see the calculation, not just the number). Thank you.  
 


There is no answer at this time. 

Subject:
Re: Chances of meeting a friend in a foreign city
From: tutuzdadga on 24 Jun 2003 20:23 PDT 
I can't help you with the calculation, but I CAN tell you that it happened to ME, not once, but TWICE  in the same day  within one minute's time! I attended rural high school in the US, graduating with a class of only 24 students. Three years later I found myself living and working in Germany. One day while shopping in a grocery store in a town I was touring about 60 miles from town I lived in, I rounded the corner and literally ran into a man I went to high school with. He was also living and working in Germany (unbeknownst to me) and just happened to be touring the same town, on that same day, more than 100 miles from where he lived in Germany. We shook hands and marveled at the coincidence, then I quickly turned to fetch my wife and literally bumped into a girl with whom I had attended the same high school. I was dumbfounded  then my friend told me that he had married this girl before coming to Germany, and of course, she came with him. What are the chances of that? Weird huh? Good luck; tutuzdadga 
Subject:
Re: Chances of meeting a friend in a foreign city
From: tlspiegelga on 24 Jun 2003 21:41 PDT 
No help with calculation here either, but it also happened to me twice. Not on the same day but on two separate trips to Israel. Strange but true! tlspiegel 
Subject:
Re: Chances of meeting a friend in a foreign city
From: probonopublicoga on 25 Jun 2003 00:01 PDT 
It's never happened to me. Therefore, the odds are 2 to 1. 
Subject:
Re: Chances of meeting a friend in a foreign city
From: politicalguruga on 25 Jun 2003 02:31 PDT 
LOL Probono Actually, calculating the chances for this only proves that statistics is not that accurate science. If we calculated the chances that 1/100X would be in one of 3Y and would be in one of 750 places (or 250 per Y); and the possibilities of such an event, we sould get a certain numbe. However, it seems that in reallife (TM) there's more probablity you'll meet people of your own milleu, because they are more likely to travel to the same places as you. 
Subject:
Re: Chances of meeting a friend in a foreign city
From: nelsonga on 25 Jun 2003 12:56 PDT 
I'll be a wise acre. 50%  It either happens or it doesn't. :) 
Subject:
Re: Chances of meeting a friend in a foreign city
From: manukaga on 23 Jul 2003 22:53 PDT 
Well, I think it's time for someone to actually have a bash at this... just don't attach too much meaning to the answer! First of all, the length of trips is very important. This directly impacts the chances of you and your friends being in the same city at the same time, which is an important precondition for meeting. ;) Thus, if you are in city X on a particular day, and one friend visits that city for an nday stay in the given year, they have an n/365 chance of being in city X on that day. Since you've said 17 days, I'll assume that stays of 1, 2, 3, ..., 7 days are equally likely, both for you and your friends. Thus the median trip would be 4 days. Now, you've said that on a typical trip you might visit 125 (of 250 possible) places. It clearly doesn't make sense to assume that this is independent of the length of the trip, so I'll assume a constant rate of visitations of 30/day. Again, this applies both to you and your friends. Now, the next part is one of the most crucial. We need to describe the distribution of time spent in each place. You've said that this can be from 1 minute to 3 hours, but obviously they can't be uniformly distributed; at 30 visits per day, with probably not much more than 15 hours a day allocated to them, we need to have a mean duration preferably no more than 30 minutes. Perhaps not surprisingly, I haven't been able to come up with a terribly convincing piecewise smooth function to represent this (I've only played around with functions in two pieces; I suspect three are needed). So here's a sample distribution I made up out of thin air: t(min) Ave # per day   1 5 3 5 5 5 10 4 15 3 30 2 45 1.5 60 1.25 90 1 120 1 150 0.75 180 0.5 This gives an average duration of just under 25 minutes, or around 12.5 hours' worth of visits per day. One difficulty with this is that the duration of your stay in each place is likely to be well correlated with the duration of your friends' visits to the same place. If you spend five minutes at the train station and three hours in a restaurant, it's much more likely that your friends will do the same than the reverse. (I suppose their train could break down, and they might decide not to eat at the restaurant (especially after seeing the prices?)  but it's surely not as likely.) So I've drawn up a possible joint distribution of times and put it on the web at http://home.iprimus.com.au/scarletmanuka/friends.htm for your edification (or whatever)  it was a bit too awkward to insert here. Note that the table entries should be divided by 30 to give a probability distribution. The significance of this table is that it represents how long each of you are visiting a place that you both visit. Of course, the numbers here are also just made up. Now we're beginning to be in a position to start answering the question! The time of day that you and your friends visit certain places will also be somewhat correlated in real life, but I am going to ignore this, except to confine visiting activities to a 15 hour period instead of around the clock. So, assuming that you and a given friend are both in the same city on the same day and decide to visit the same place, what are the chances that you'll be there at the same time? Well, if you visit for x minutes and your friend for y minutes, there are xy possible combinations of starting times (more or less  this is where a continuous distribution would simplify things) that will lead to a meeting, out of (15*60)^2 = 810,000 possible combinations, assuming 1minute granularity to time. ;) Now if we divide the table at the link I gave earlier by 30 to get a probability distribution, and multiply each element by the respective value of xy/810000, the sum of all the table entries gives us the probability that you meet your friend, given that you both visit the same place on the same day. Excel tells me that for these figures this probability is 0.00258 (of course, given our assumptions, there's a considerable fudge factor here). So, if you and your friend are in the same city on the same day, what's the expected number of places that you'll both visit that day? You're both visiting 30 places out of 250  they won't actually be a uniformly random selection, but I'll pretend they are. The probability of you both visiting a fixed place is then (30/250)^2 = 9/625, so the expected number of places you both visit is (9/625)*250 = 3.6; so the expected number of meetings is 3.6*0.00258 = 0.0093. OK, so what is the expected number of days that you will both be in a given city? Let's assume your visit to the city is N days long and your friend visits for M days. Then the expected number of overlaps is MN/365; since M varies uniformly from 1 to 7 you can expect to meet your friend (4*N/365)*0.0093 = 0.00010*N times during your stay. Now we've got the average behaviour for any given friend, so we multiply by 100 to say that your chance of meeting any friend during your visit is 0.010*N. From there it's simple: the average visit is 4 days and you make three visits per year, so you can expect around 0.12 meetings per year, or roughly one every eight years. So there you go! Feel better now?  The Scarlet Manuka 
Subject:
Re: Chances of meeting a friend in a foreign city
From: probonopublicoga on 23 Jul 2003 23:03 PDT 
Actually, there are more confounding factors than Manuka has supposed. How good are their eyesights? How good their memories? What are their sexes? (Women change their appearance 10 times a day, so that even their nextdoor neigbours fail to recognise them.) How gorgeous are they? In fact, the more I think about it ... The more I realise that it's just impossible. 
Subject:
Re: Chances of meeting a friend in a foreign city
From: manukaga on 24 Jul 2003 18:25 PDT 
Hey, he only asked for the chances of meeting them, not for the chances of actually recognising them when he does! ;) Of course this does add some difficulty to the task of verifying my ohsorigorous derivation... <grin> 
Subject:
Re: Chances of meeting a friend in a foreign city
From: macaonghusga on 25 Jul 2003 09:16 PDT 
Thank you all, especially obviously manuka.... 
Subject:
Re: Chances of meeting a friend in a foreign city
From: probonopublicoga on 25 Jul 2003 23:25 PDT 
I just remembered that, a few years ago, I attended a party in The Hague ... During the course of the evening, I spoke separately to two Dutch women each of whom told me that she had used to live in Surinam. Later, I introduced them to each other and (would you believe?) they had even known each other when they had lived in Surinam ... but they had failed to recognise each other that evening! That must prove something .... Perhaps that all statisticians are mad? 
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