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

[Lorake]

OMG you cannot make this statement you moron.

F(4,5) = 0 since X < Y.

Read my post, X=5 and Y=5

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 (OBVIOUSLY IF THE AMOUNT OF CHASES LEFT IS LESS THAN THE AMOUNT OF BACK TO BACKS NEEDED, WE WILL FAIL DUE TO LACK OF CHASES LEFT)
F(X,Y) = 1 when Y = 1 *Edit: broke my own equation LOL
F(X,Y) = .04^(Y-1) + .96F(X-1,Y)

But because you're retarded I'll use different numbers:

1. Probability of a random person in raid getting targetted 5 times in a row with 4 amount of chases

Oh wait, how can you be targetted 5 times in 4 chases? Impossible.

2. Probability of a random person in raid getting targetted 4 times in a row with 5 amount of chases

Oh wait according to your formula F(X,Y) = 0 when Y>X this equals 0.

NICE FORMULA JACKASS

Typo in the copy paste job I did. Let's re-adjust to what I meant. That's my bad on the typo, and fixed for the catch. Equation still holds, so let's continue with your case.

Let's take your probability of a random person in raid getting targetted 4 times in a row with 5 chases.


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)
F(5,4) = .04^4 + .96F(4,4) = .04^4 + .96*(.04^4 + .96F(3,4))
F(3,4) = 0 since X<Y thus F(5,4) = .04^4 + .96 * .04^4 = .0000050176



T T T T T = .04^5
T T T T F = .04^4 * .96
F T T T T = .04^4 * .96
Are the only cases where 4 back to backs happen for an individual.
Individual = .04^5 + 2 * .04^4 * .96
25 People = 25 * (.04^5 + 2 * .04^4 * .96) = .04^4 + 2 * .04^3 * .96 = .0000050176

I APOLOGIZE FOR THE TYPO IN THE COPYPASTING JOB I DID. EQUATION STILL HOLDS YOU TWAT.