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

[Lorake]

Now for the fun part:

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)

So targetted 2 times with 2 chases, F(2,2)

F(2,2) = .04 + .96F(1,2)
F(1,2) = 0 thus F(2,2) = .04


2 times with 3 chases
F(3,2) = .04 + .96F(2,2) = .0784

and it goes from there. If you think I'm wrong break it with a case, otherwise your arguements are invalid and retarded.

Edited, fixed a typo. F(X,Y) = 1 when Y = 1 obviously, since it's a 100% chance for anub to follow someone.

[Sharfalma]

Now for the fun part:

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)

So targetted 2 times with 2 chases, F(2,2)

F(2,2) = .04 + .96F(1,2)
F(1,2) = 0 thus F(2,2) = .04


2 times with 3 chases
F(3,2) = .04 + .96F(2,2) = .0784

and it goes from there. If you think I'm wrong break it with a case, otherwise your arguements are invalid and retarded.

...obviously

[ArchmageFurly in three days]

How have I missed out on this thread.

Anyways, Lorake is correct in saying that the chance of A person being followed twice in a row is 4%.

The reason that this is so, is because, as he said, who he targets first does not matter. There is a 100% chance for SOMEONE to get targeted the first time. The odds of that same person getting targeted again is 4%.

I'll explain it in a non-mathematical way. Imagine you have a bag of 25 marbles, all with a different name on them. Anub'Arak picks out a marble from that bag, there is a 100% chance of this happening. He then throws that marble back into the bag and picks out another one. The chance that it will be that same marble is 1/25 or 4%. As lorkae said, if we had been asking about a specific marble, let's say, Lorake. The odds of him picking Lorake the first time are 4% and then again the second time is .16%. But that's not what we're asking. The question is whether or not he will follow ANYONE twice. Because that is what causes the problem.

Gemg, what yo said before was wrong. He didn't say there was a 4% chance for someone to get followed the first and second time. There is in fact a 100% chance that someone will get followed once, and what that means is that there is a 100% chance that Anub will follow a member of the raid. Someone, refers to one member of the 25 man raid, so if it still doesn't make sense, think of it as there is a 100% chance for 1 of the 25 people to get followed by Anub.

[Roj]

Now for the fun part:

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)

So targetted 2 times with 2 chases, F(2,2)

F(2,2) = .04 + .96F(1,2)
F(1,2) = 0 thus F(2,2) = .04


2 times with 3 chases
F(3,2) = .04 + .96F(2,2) = .0784

and it goes from there. If you think I'm wrong break it with a case, otherwise your arguements are invalid and retarded.

Edited, fixed a typo. F(X,Y) = 1 when Y = 1 obviously, since it's a 100% chance for anub to follow someone.

This simply does not work. How did you magically come up with this part?

F(1,2) = 0

Your equation is broken. Copy/paste something better.

[Lorake]

Now for the fun part:

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)

So targetted 2 times with 2 chases, F(2,2)

F(2,2) = .04 + .96F(1,2)
F(1,2) = 0 thus F(2,2) = .04


2 times with 3 chases
F(3,2) = .04 + .96F(2,2) = .0784

and it goes from there. If you think I'm wrong break it with a case, otherwise your arguements are invalid and retarded.

Edited, fixed a typo. F(X,Y) = 1 when Y = 1 obviously, since it's a 100% chance for anub to follow someone.

This simply does not work. How did you magically come up with this part?

F(1,2) = 0

Your equation is broken. Copy/paste something better.

Given that X is the amount of times he's going to chase someone, and Y is the number of back to back chases he needs to do, tell me what's the probability he's gonna chase someone Y times when X is less than Y.

0 obviously, he doesn't have enough chases left.

When Y = 1, he just has to chase someone, so it's obvious F(X,1) = 1 since 1 = 100%

[Roj]

We both know you're full of shit. That does not make sense. You copy/pasted a linear equation when you should have googled an exponential equation. Think about it. Give it up already, you fail.

[Lorake]

We both know you're full of shit. That does not make sense. You copy/pasted a linear equation when you should have googled an exponential equation. Think about it. Give it up already, you fail.

http://en.wikipedia.org/wiki/Linear_equation

If X and Y confuse the shit out of you, I can just use (X1) and (X2) notation for variables. Sorry I come from a level of math where you can have a ton of variables and you describe functions of variables as F(variables).

Wonder how long it took you to look up linear and exponential equations. Did you Wiki a strat?

[Roj]

We both know you're full of shit. That does not make sense. You copy/pasted a linear equation when you should have googled an exponential equation. Think about it. Give it up already, you fail.

http://en.wikipedia.org/wiki/Linear_equation

If X and Y confuse the shit out of you, I can just use (X1) and (X2) notation for variables. Sorry I come from a level of math where you can have a ton of variables and you describe functions of variables as F(variables).

Wonder how long it took you to look up linear and exponential equations. Did you Wiki a strat?

Explain to me why your equation is linear. You don't even know what that means do you?

[geng]

roj you are being too obvious

[Aseic]

so 1+1 is 2 then?