I live in a city of roughly 8 million people. I see or run into about
a hundred people each day. The question: taking into account only
those two variables, what is the probability of accidentally running
into the same person twice within my lifetime (say 70 years)?

Alright man this one's pretty simple:

8 000 000 divided by 100 = 80 000 days of seeing 100 random people
each day in order to see each person once.

80 000 divided by 365 days (1 year) = 219.18 years

Unfortunately you'd have to live for 220 years based on these numbers.
Having said that, unless you get REALLY lucky, you'll see a lot of
people twice or thrice or more during your lifetime, however there
will be some that you will never see during your 70 years on planet
earth of the 8 million in your city.

This forum could be good for you if you have any other questions:

http://math436.phpnet.us/phpBB2/index.php

Habs, I don't think the math is quite that simple.

First you have to determine who this "single person" is.  It could be
any of these 3:
1) someone you have already met randomly and want to randomly see again
2) a truly random yet specific someone
3) or perhaps the more complicated choice... you want to know the
probability of running into any single person 2 times (ie if you see
100 people on day 1, then if you ever see any of those 100 people
again then you are happy)

Here are the odds for each:
1) Since you have already seen this person, you only need to see them
1 more time to be successful.  The odds of seeing this same person for
the second time on any given day is 100 in 8 million (or 0.0000125).
The odds of seeing this same person for the second time on the next
day is slightly less (because the next day could be the third random
meeting instead of the second) and it calculates to .00001249984375.
The formula to calculate the odds during your entire life (70 years) is:
1-(7,999,900/8,000,000)^(365*70) = 27.3398428%.

2) To meet a random specific person (we'll call him John), you must
randomly meet him once then randomly meet him again.  To do this, you
take the odds of meeting him once then multiply it by the odds of
meeting him once again.  This is:
27.3398428% * 27.3398428% = 7.47467%.

3) To meet anybody 2ce in your life, you take the odds of not meeting
a specific person (100% - 7.47467% = 92.52533%) and multiply that
times itself 8,000,000 times (for the 8 million people you could
possibly randomly meet 2ce).
92.52533% ^ 8,000,000 which is effectively 0.

I'm not 100% certain that I got all these formulas correct, but I'd
like to think I did a good job :)

Terribly sorry, I didn't finish #3.  The odds of not meeting anyone
twice is effectively 0% so the odds of meeting someone (anyone) twice
is effectively 100%.

Wasn't there a similar question like this posted earlier about the
odds of meeting someone again in a city? I'm sure it was posted within
the last few months.

Saem,
This one is similar in nature but a much different question:
http://answers.google.com/answers/threadview?id=221349

There might be a more similar one out there somewhere.

This is related to the famous Birthday problem: Among 23 random
people, the probability that two people have the same birthday is
greater than 50%. This is very surprising to many people.

http://en.wikipedia.org/wiki/Birthday_paradox

If this result is counterintuitive to you, consider that there are 23
choose 2 = 23*22/2 = 253 pairs of people, and each of those pairs may
be viewed as a chance for the birthdays to be the same. With 23
people, the number of chances 253 is on the same order of magnitude as
the number of days 365, and in fact, is just more than ln(2) * 365.

With 8 million people in your city, you will have a probability
greater than 50% of a repeat after only 3331 people, on day 34, even
if you are able to sample the people completely uniformly. If the
people you encounter are chosen independently, but not uniformly, this
increases the probability of a repeat.

After 70 years (including 18 leap years, 25568 days), with exactly 100
independent uniformly distributed people per day, the probability that
you have not encountered any repeats is

7999999/8000000 * 7999998/8000000 * 7999997/8000000 * ... 
* (8000000-2556799)/8000000
= 8000000!/((8000000-2556800)! 8000000^2556799)
~ 1.056 x 10^-200100

This is extremely close to 0, so the probability that you encounter at
least one repeat within 70 years is very close to 1, about .999...9
with 200100 9s.