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

[Lorake]

Probability of someone getting chased Once in X chases

f(X) = sum(t(X,i)*.04^(X-i)*.96^i ,i=0, i<X, i++)

where t(X,i) refers to the (i+1) value from Pascal's triangle for the Xth row.
http://en.wikipedia.org/wiki/Pascal's_triangle

So
t(4, 2) = 6
t(3, 2) = 3
t(5, 0) = 1
t(5, 1) = 5
t(5, 2) = 10
t(5, 3) = 10

etc

so
f(2) = .04^2 + 2 *.04*.96
f(3) = .04^3 + 3* .04^2*.96 + 3 * .04 * .96^2
f(4) = .04^4 + 4* .04^3*.96 + 6 * .04^2 * .96^2 + 4 * .04 * .96^3


etc etc etc

Figuring out the probability of someone getting chased Y times in a row in X chases gets fishy and messy. I'm thinking on that one. If it weren't required that Y be back to back, then it would be easier, but we're speaking back to back cases only. For Y times being chased where we don't give a shit if it's back to back or not in X possible chases it's

f(X,Y) = sum(t(X,i)*.04^(X-i)*.96^i ,i=0, i<X-Y+1, i++)

Alas I digress. I like math.

OP still stands true in context unless Steke can clarify his numbers.