View Question
 Question
 Subject: Simple math question Category: Science > Math Asked by: rurikbird-ga List Price: \$5.00 Posted: 18 Jul 2006 23:34 PDT Expires: 17 Aug 2006 23:34 PDT Question ID: 747629
 ```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)?```
 There is no answer at this time.

 Subject: Re: Simple math question From: habsfan1433-ga on 18 Jul 2006 23:54 PDT
 ```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.```
 Subject: Re: Simple math question From: habsfan1433-ga on 18 Jul 2006 23:59 PDT
 ```This forum could be good for you if you have any other questions: http://math436.phpnet.us/phpBB2/index.php```
 Subject: Re: Simple math question From: jack_of_few_trades-ga on 19 Jul 2006 06:40 PDT
 ```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 :)```
 Subject: Re: Simple math question From: jack_of_few_trades-ga on 19 Jul 2006 06:41 PDT
 ```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%.```
 Subject: Re: Simple math question From: saem_aero-ga on 19 Jul 2006 07:39 PDT
 ```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.```
 Subject: Re: Simple math question From: jack_of_few_trades-ga on 19 Jul 2006 10:23 PDT
 ```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.```
 Subject: Re: Simple math question From: pzhon-ga on 21 Jul 2006 10:37 PDT
 ```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.```