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

[Lorake]

The probability of anub double targetting is 4%
The probability of anub triple targetting is .16%
The probability of anub quadruple targetting is .0064%

I'm so good at this.

Ah the ole make up some numbers trick. What happened to

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

Making the same assumption you are, which is the number of times equals the number of chases. If that's the case don't need an overly complex equation, .04^(X-1) is sufficient to describe.

I'll post the equations here for you.

Anub chooses same target X times for X chases
.04^(X-1)

Anub chooses same target Y times for X 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)

Probability of someone specifically 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

It can also be simplified to f(x) = 1-.96^X since you can take the probability of not being targetted and subtracting it from 100%

Finally the probability of someone getting chased Y times in X chases with no regards to order, thus not back to back:
f(X,Y) = sum(t(X,i)*.04^(X-i)*.96^i ,i=0, i<X-Y+1, i++)