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

[Roj]

Yea I know. Half of it is Roj trying to restate the context of the problem so he's right, or just saying I'm wrong with no basis of understanding. He doesn't want to listen either.

I'm enjoying this. It's fun making Roj look stupid. The linear equation thing was classy.

Classy? Are you trying to say classic? You still haven't plugged numbers into your nonsense formula and I know you won't because it won't work.

[Firecrest]

you said you would get rid of it roj

Why are you posting on the forums but not in raid?

[Roj]

Ok let's take a look at your formula:

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)

We'll try 5 for both.

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

F(5,5) = .04^(5-1) + .96F(5-1,5) which equals

F(5,5) = .04^(4) + .96F(4,5) which equals

F(5,5) = .00000256 + .96F(4,5)

Now go ahead and finish the equation for me Lorake.

[Meekslol]

this thread warrants:

[geng]

11 fucking pages of miscommunication.

[Lorake]

Yea I know. Half of it is Roj trying to restate the context of the problem so he's right, or just saying I'm wrong with no basis of understanding. He doesn't want to listen either.

I'm enjoying this. It's fun making Roj look stupid. The linear equation thing was classy.

Classy? Are you trying to say classic? You still haven't plugged numbers into your nonsense formula and I know you won't because it won't work.

No I mean classy. Because first you said I was making up linear and quadratic equations when I wasn't, and then you said I called it a linear equation when I didn't. And no, just because you can enter points into an equation, it does not suddenly, magically become linear.

Ok so let's plug and chug some numbers for you.

3 chases, chance of an individual being chased at least twice.

f(3,2) = sum(t(X,i)*.04^(X-i)*.96^i ,i=0, i<X-Y+1, i++)
t(3,0) = 1, t(3,1) = 3 (based on pascal's triangle and my use of it)
= .04^3*.96^0 + 3*.04^2*.96 = .04^3 + 3*.04^2*.96

All possible cases.
T T T = .04^3
T T F = .04^2 * .96
T F T = .04^2 * .96
T F F = Invalid case, no double
F T T = .04^2 * .96
F T F = Invalid case, no double
F F T = Invalid case, no double
F F F = Invalid case, no double

.04^3 + 3 * .04^2 * .96

Equation holds for individual cases.


Now for Anub's back to back specifics with 3 chases. So no more target preference, just needs a back to back event.
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(3,2) = .04 + .96F(2,2) = .04 + .96(.04 + .96F(1,2)) = .04 + .04*.96 = .0784

All possible cases.
T T T = .04^3
T T F = .04^2 * .96
T F T = Invalid case, no btb
T F F = Invalid case, no btb
F T T = .04^2 * .96
F T F = Invalid case, no btb
F F T = Invalid case, no btb
F F F = Invalid case, no btb

.04^3 + 2 * .04^2 * .96

But that's for one person. We unfortunately have 25 people, so

25 * (.04^3 + 2 * .04^2 * .96) = .04^2 + 2 * .04 * .96 = .0784%


This has got me thinking, what is the probability of him chasing someone twice in general, and not individually. I'll think on that one.

[Lorake]

Ok let's take a look at your formula:

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)

We'll try 5 for both.

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

F(5,5) = .04^(5-1) + .96F(5-1,5) which equals

F(5,5) = .04^(4) + .96F(4,5) which equals

F(5,5) = .00000256 + .96F(4,5)

Now go ahead and finish the equation for me Lorake.

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

Thus F(5,5) = .00000256 = .04^4 = I'm right, bitch.

[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

[geng]

NOPE WRONG NOPE TRY AGIN NOPE

[Roj]

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
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