It's 4%, AKA the Go Fuck Yourself Post.

[Lorake]

0.04 is the chance of being chased, 1-0.04 is the chance of not being chased. The sum doesn't reach 1 because the numbers provided don't cover all cases.

Supposing a person is followed once, we end up with (0.04^(1)) + (1-0.04^(1)) = 1. Pretty standard stuff there.

The odds of an event occuring multiple times is (probability)^(times occuring) * (probability)^(times not occuring). The sum of separate odds is the odds of one of the summed odds occuring. Therefore, the odds of being followed twice when three targes are chosen are:
0.04^2 * 0.96 + 0.96 * 0.04^2 = 0.003072

Being followed once is:
0.96^2 + 0.04 + 0.96 * 0.04 * 0.96 + 0.04 * 0.96^2 = 0.110592

Being followed all three times is:
0.04^3 = 0.000064

Being followed 0 times is:
0.96^3 = 0.884736

Being followed 2 times, but not in a row is:
0.04 * 0.96 * 0.04 = 0.001536

And the sum of those is : 1

I was wrong the first time, hope this clarifies. Also, the odds of not being chased an entire fight that has six chases is 0.96^6 = 0.782757789696, or greater than 75%, which suits the bads ratio pretty well .

http://www.mathgoodies.com/lessons/vol6 ... vents.html

Ok I get where you're coming from. But the original argument is back to back chases occuring to someone, with the total chases being equal to back to back chases being assumed. Bad assumption to presume I suppose but that's the assumption. So .04^(X-1) still holds true for when the number or back to back chases equal the number of total chases, since we still hold that the first target selected does not matter. So if it's 2 back to back chases out of two chases, it's still .04, 3 back to back chases out of 3 chases is .0016, etc etc. Things get messy when the total chases don't equal back to back chases.