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

[Meekslol]

[Roj]

Do some math for me now Roj.

Your case, 5, 5.

Find out the specific combinations of people selected in a 25 man raid, and tell me the probability that selections 1,2,3,4 and 5 are all equal.

Here's a hint - the probability of a single individual being selected 5 times in one go is .04^5

Hmm this looks suspiciously like my earlier post:

The probability of any raid member being targetted "X" times in a row is (1/25)^(X).

I.E.
1 time = (1/25)^(1) = 4%
2 times in a row = (1/25)^(2) = .16%
3 times in a row = (1/25)^(3) = .0064%
4 times in a row = (1/25)^(4) = .000256%

Get raped kid.

[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.

[Lorake]

Do some math for me now Roj.

Your case, 5, 5.

Find out the specific combinations of people selected in a 25 man raid, and tell me the probability that selections 1,2,3,4 and 5 are all equal.

Here's a hint - the probability of a single individual being selected 5 times in one go is .04^5

Hmm this looks suspiciously like my earlier post:

The probability of any raid member being targetted "X" times in a row is (1/25)^(X).

I.E.
1 time = (1/25)^(1) = 4%
2 times in a row = (1/25)^(2) = .16%
3 times in a row = (1/25)^(3) = .0064%
4 times in a row = (1/25)^(4) = .000256%

Get raped kid.

Suspicious, but if you think it's .04^5 is your final answer, you're dead wrong. After all there are 25 people and not one person.

You know, sometimes, I forget who I'm talking to. Numero Uno is all that matters, so I suppose one person is the only case that matters, and that's the Roj case.

[Lorake]

So the probability of Roj getting chased 5 times in a row is .04^5. Now the probability of someone other than Roj getting chased 5 times is... is... who gives the fuck obviously Roj is the only person in raid for whom it is possible to get chased 5 times in a row, because Roj is awesome and supergreat and everyone else is obviously terrible at math because Roj thought so. So really Anub should chase Roj 100% of the time, because if Anub knew what the fuck to do he would go after Roj, because Roj is the smartest WLPHer there ever was, and the smartest there ever will be.

^ THIS MATH IS OBVIOUSLY BETTER AND MAKES MORE SENSE THAN MY CALCULATIONS.

[Firecrest]

Are you guys even doing real math anymore or are you just randomly pressing shit on your keyboards?

[Lorake]

So my math still holds true, barring any other X Y typos caught.

GET SHIT ON ROJ. HOW DO YOU LIKE THAT DICK SANDWICH.

[Lorake]

Are you guys even doing real math anymore or are you just randomly pressing shit on your keyboards?

As far as I know I'm the only one doing any real math. Roj is still stuck doing Abe math.

[Roj]

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.

[Firecrest]

Roj, your irl friends already told us that Lorake could take you in a fight. It's over man.