I live in Geelong, Victoria, Australia

Area		: 1,240 kmē
2006 		:205,000 (urban)
Density		: 165.3/kmē

What is the statistical chance of every running into them on any given day?

That is what is the likelihood of going into town, and walking down a
Street and seeing them in the same street, building etc(where ever I am)

Is there a math formula that can be applied to work out the
probability of this happening?

What are some of the variables that I can employ to minimise the
chances of this happening (obviously not going where they would
normally go), when you take in account that I have no idea where they
could be at any given time in relation to where I am.

The assumption the someone else or even yourself will be in a random spot
in a city is fairly silly.

While you can not guess which grocery store they will visit, chances
are they will visit one sometime during the week.

On the other hand a visit to the dump or the sewage works would seem
unlikely for you both.

(Density meant population or building density.?)

Sod's Law says that you will meet your Ex in the unlikeliest place and
in the most embarrassing circumstances imaginable.

It's probably best to grow a beard.

Or, if you've already got one, to shave it off.

(I am, of course, assuming that you are a man. Women can change their
appearances like nobody's business with a new hairdo or whatever.)

If you convince yourself that you will run into her
- and look forward to it

- then the chances are vastly reduced

There is also the 'skulk' factor, if she does not want to meet you,
she'll pretend not to see you, or hide behind a display in a
supermarket.

There is also the 'Stalk' Factor.

Maybe she does want to meet you again.

Hoping that just maybe ...

It's back to old times.

redfoxjumps-ga on 12 May 2006 00:44 PDT 
Density meant population or building density.?)

I assume "denisity" means people per square kilometer 165.3

It may sound silly, but i thought there might be a mathematical
approach to this quesion. (Maths is not my thing)

I think the chances are much greater than population density would
suggest, because you both will only go to certain areas of a city you
are familiar with, even if you both don't want to happen to run into
each other.  The skulk factor may help avoid this, but for my
thinking, if one of you spotted the other, that would count as an
incident of "the likelihood of going into town, and walking down a
Street and seeing them ..."

From: myoarin-ga on 12 May 2006 07:26 PDT 
"I think the chances are much greater than population density would
suggest, because you both will only go to certain areas of a city you
are familiar with, even if you both don't want to happen to run into
each other.  The skulk factor may help avoid this, but for my
thinking, if one of you spotted the other, that would count as an
incident of "the likelihood of going into town, and walking down a
Street and seeing them ..."

I guess thats just it, i am tied of "skulking" I was hoping for a
logical mathematical expression, like 1 in 1 million. (in fact 1 in
205,000)

but I do appreciate peoples comments

Let's say on a typical day in town, you see 500 people (walking down
the street, in a shop, whatever).  That is about 1/400 of the
population.  So 1 chance in 400 seems like an ok guess.

Sorry does not look like this is going to get answered. (Perhaps its a
silly question).

Is there anything I can do to fully qualify the question. Up the price?

I like the "sod's" law concept. I am trying to avoid it.
Perhaps probabilities, statistics, etc are not the answer.

G'day, again.
Increasing the price would maybe give you some interesting statistics,
but for your individual problem, I doubt that this would help you very
much.

You might try to anticipate what could occur if you meet and plan your
response, considering that it will probably be in public, maybe in the
presence of someone who knows you (no good trying to claim that you
are not the person she thinks).

Any Maths whizz'es care to have a stab!

You might consider running into them enough until it doesn't bother you anymore.

LOL It is funny to read no one is answering mathematically. Only
sociological perspectives.
Although it is mathematically challenging, even inpractical to
calculate. From the first sight, you gave the
area/population/residencial density (probably of night time). But all
those variables are irrelevant, of my opinion.

The reason behind my claim is that you need a variable where people
have available access, such as streets... or park. Let me assume the
only varialbe to consider is the streets. Applying the basic idea
Graph Theory, you can simplify your city. First you map the streets
with straight lines and crossings with intersections. (Since I checked
"Geelong, Victoria, Australia" is mesh streets, you may estimate by
drawing straight lines, like a mesh.) Suppose you start walking from a
intersection where you can choose 1 direction out of 4 direction. If
your selection is random, the probability of choosing 1 out of 4
direction is .25, leaving the probabilities of choosing other streets
in the city (other than those 4 to consider) be 0. At next
intersection, you find another probability of choosing one direction
to another, namely 1/n where n is the number of option and n is always
4 inside a mesh, and others be 0. After making decisions (X times) for
all the streets you walk. For a person you keep eye on meeting again,
he/she has 4^i options in total. By reasoning backward, you will see
that his/her last direction is definite. If he/she does not decide to
go back the street he/she came from, number of his/her choosing
streets must be (N+X-1,X), which is the combination of selections with
unlimited repetition. Thus the probability of him/her reaching the
street you chose at the end is (N+X-1,X)/4^i where N is the number of
all street blocks, and X is the number of times you chose.
The answer, of meeting the person again, is the product of two
probability. In this case, [(N+X-1,X)/4^i]*[PI_{k=1}^{X} :x:_{k}]
where :x: is the probability {Prob(choosing a street at x_{k})=:x:_{k} | k=1,...X}

Another variable which you may consider would be restricted area, like
office buildings, etc. The probability can be estimated by the product
of getting hired by the same employer and assigned to the same floor
as the person. Of course, it assumes the employment for both of you is
independent.

So it is really difficult question to answer mathematically but I
think you felt the sense.