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

[Lorake]

So go fuck yourself Roj, have fun with this one.

So problem fixed for now new equation holds under (5,3) as well as the previous cases of (5,4) and (5,5), but hold your breath until I can really test out the case of F(5,2)
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))

F(5,3) = .04^(2) + .96(F(4,3)+F(4,4)) = .04^(2) + .96(.04^(2)+ .96F(3,3) + .04^3) = .04^(2) + .96(.04^(2) + .96(.04^(2) + F(2,3) + .04^3))
= .04^(2) + .96(.04^(2) + .96(.04^(2)) + .04^3 ) = .04^2 + .04^2 * .96+ .04^2 * .96^2 + .04^3 * .96
Individual's chance (use previous post)
Sum = .04^5 + 4 * .04^4 * . 96 + 3 * .04^3 * .96 ^ 2
^ Individual case. But we have 25, so multiple
Sum(25) = .04^4 + 4 * .04^3 * . 96 + 3 * .04^2 * .96 ^ 2 =0.004672
F(5,3) = .04^2 + .04^2 * .96 + .04^2 * .96^2 + .04^3 * .96 = 0.004672
Eliminating the difference.

The forumula still holds for F(5,4) since F(5,4) = .04^3 + .96(F(4,4) + F(4,5)) = .04^3 + .96F(4,4) = Old algorithm.


I have a problem now. Because the cases of a triple chase of this case are independent, it’s possible to have a double chase occur twice in 5 chases (ROJ, ROJ, LORAKE, LORAKE, ABEC) thus it’s not logical to multiple by 25, since each individual’s chance of a overlap with each other in certain cases. I’ll see if I can get my hands on a C++ program or something to have it calculate the probability of a double chase through raw power. This problem hasn’t been apparent in previous ones since it’s unpossible for two people to have the triple chase occur in 5 chases (UNPOSSIBLE).