10-19-2009, 11:54 PM
Lorake Wrote: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.
.04^4 + 2 * .04^3 * .96 does not = .0000050176
.04^4 + 2 * .04^3 * .96 = .00012544
You're just making up numbers and I'm tired of correcting you. I'm going to bed. I'll look at the garbage you're posting in the morning.
