Quote from: Steffi on December 22, 2011, 12:10:55 AM
Sorry if this seems pedantic but.... What??? That maths seems highly suspect to me.
If you pass 99.99% then you pass to 9,999 out of 10,000 people, therefore one person in 10,000 clocks you which is a percentage of 0.01 ( which is the same 0.01% needed to make the 99.99% up to 100%)
When the number of people goes up 10 times to 100,000, the number of clocks goes up 10 times as well so then 10 out of 100,000 clock you - the percentage however remains the same which is the main feature of percentages and ratios
I think you confused the notions of "being clocked by 1 person in 10,000 on average" (99.99% = the probability that you pass in front of ONE single person) and "the probability of not being clocked by even 1 person in 10,000" ( = probability of passing in front of 10,000 random people)

See it this way... since on average you get clocked by 1 person out of 10,000, you would expect it to be fairly likely that at least one person out of a random group of 10,000 people has clocked you, right? (36.8%)
And since on average you get clocked by 10 people out of 100,000, it means it is very unlikely that not a single person in 100,000 people clocks you, correct? (0.0045%)
99.99% is NOT the probability that you won't get clocked by anyone in 10,000 people, or that at least one person clocks you in 10,000 people! It's the probability that you pass in front of ONE person.
**WARNING: BORING MATHS DETAILS below, don't read if you don't want to!!**
If you are into some maths, by some relatively simple high school maths (binomial theorem), the probability of exactly r people out of 10,000 clocking you is
nCr * p^(n-r) * (1 - p)^r,
where p = 0.9999% is your pass rate, n = 10000 is the number of people you've met, and nCr = the binomial coefficient n!/[r! (n - r)!], and thus you have something like this
r vs. probability of being clocked in front of exactly r people out of 10,000
0 36.8%
1 36.8%
2 18.4%
3 6.1%
4 1.5%
5 0.3%
...
and so on, rapidly approaching zero as r increases.
In the case of meeting 100,000 people (i.e. with n = 100,000 instead of 10,000), you expect to pass in front of approximately 10 people out of 100,000, and the probabilities show that too:
r vs. probability of being clocked in front of exactly r people out of 100,000
0 0.0045%
1 0.045%
2 0.23%
3 0.76%
4 1.9%
5 3.8%
6 6.3%
7 9.0%
8 11.3%
9 12.5%
10 12.5%
11 11.4%
12 9.5%
...
so as expected, the probability peaks near r = 10.It may be helpful if you think in terms of tosses of a fair coin, which has a 50% chance of getting a head (1 in 2 tosses on average is a head). The probability of two tosses being a head is 1/2 * 1/2 = 25%, not 50%. The probability of 20 tosses all being heads is (1/2)^20 = 0.000095%.