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

[Roj]

Ok here we go. This is the latest version of your "equation" after you've editted it 3 times.

Find the probability of Y back to back chases given Anub does X chases
F(X,Y) = 0 when Y>X
F(X,Y) = 1 when Y = 1
F(X,Y) = .04^(Y-1) + .96(F(X-1,Y) +F(X-1, Y+1))

First of all, I need to post a disclaimer. I'm going to calculate this out using the Lorakefailsatmath method. The equation is broken in the first place but even using your own methods it doesn't compute. Here's the first 3 sets I tried:

F(3,2) = .04^(2-1) + .96(F(3-1,2) +F(3-1, 2+1))
F(3,2) = .04^(1) + .96(F(2 ,2) +F(2 , 3 ))
F(3,2) = .04 + .96( ( .04 ) + ( 0 ))
F(3,2) = .04 + .0384
F(3,2) = .0784

F(4,2) = .04^(2-1) + .96(F(4-1,2) +F(4-1, 2+1))
F(4,2) = .04^(1) + .96(F(3 ,2) +F(3 , 3 ))
F(4,2) = .04 + .96( (.0784) + ( .0016 ))
F(4,2) = .04 + .0768
F(4,2) = .1168

F(5,2) = .04^(2-1) + .96(F(5-1,2) +F(5-1, 2+1))
F(5,2) = .04^(1) + .96(F(4 ,2) +F(4 , 3 ))
F(5,2) = .04 + .96( (.1168) + (.003136 ))
F(5,2) = .04 + .11513856
F(5,2) = .15513856

So this "equation" says that the probability of 2 back to back chases given Anub does 5 chases is 15.5%!!!

Obviously that's retarded and it won't be long before the probability is greater than 100% of back to back chases. That's strike 3 on your nonsense "equations". Are you going to start from scratch this time or keep adding on to this epic fail of stupidity?