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

[Lorake]

Now for the fun part:

Probability of a random person in raid getting targetted Y times in a row with X amount of chases

F(X,Y) = 0 when Y>X
F(X,Y) = 1 when Y = 1 *Edit: broke my own equation LOL
F(X,Y) = .04^(Y-1) + .96F(X-1,Y)

So targetted 2 times with 2 chases, F(2,2)

F(2,2) = .04 + .96F(1,2)
F(1,2) = 0 thus F(2,2) = .04


2 times with 3 chases
F(3,2) = .04 + .96F(2,2) = .0784

and it goes from there. If you think I'm wrong break it with a case, otherwise your arguements are invalid and retarded.

Edited, fixed a typo. F(X,Y) = 1 when Y = 1 obviously, since it's a 100% chance for anub to follow someone.

This simply does not work. How did you magically come up with this part?

F(1,2) = 0

Your equation is broken. Copy/paste something better.

Given that X is the amount of times he's going to chase someone, and Y is the number of back to back chases he needs to do, tell me what's the probability he's gonna chase someone Y times when X is less than Y.

0 obviously, he doesn't have enough chases left.

When Y = 1, he just has to chase someone, so it's obvious F(X,1) = 1 since 1 = 100%