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

[Roj]

Just posting these for my reference in case a certain unscrupulous person removes them. Haven't had a chance to look at the latest one very closely yet.

Lorakemath Version 2 (Did not catch the equation before the first condition was added)

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

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

Lorakemath Version 3

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

Lorakemath Version 4

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)

Lorakemath Version 5

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))

Lorakemath Version 6 (wording changed for comparison)

The probability of Y back to back chases given Anub does X chases is :

F(X,Y) = 0 when Y>X
F(X,Y) = 1 when Y = 1
F(X,Y) = .04^(Y-1) + .96 * sum(F(X-1, Y+(i-1), i = 1, i <= alpha, i++)

where
alpha = Y - 1 when (X+1)/2>Y
alpha = X - Y otherwise

Lorakemath Version 6 (original)

The probability of Y BTBs in X chases is :
Given that X,Y > 1 (It's UNPOSSIBLE to have ZERO TARGETS or 0 CHASES, OR NEGATIVE FOR THE MATTER)
F(X,Y) = 1 when Y = 1
F(X,Y) = 0 when Y>X
F(X,Y) = .04^(Y-1) + .96 * sum(F(X-1, Y+(i-1), i = 1, i <= alpha, i++)
where
alpha = Y - 1 when (X+1)/2>Y
alpha = X - Y otherwise