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+--- Thread: It's 4%, AKA the Go Fuck Yourself Post. (/showthread.php?tid=1373)
Re: It's 4%, AKA the Go Fuck Yourself Post. - Roj - 10-19-2009
Lorake Wrote: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.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.
I'm enjoying this. It's fun making Roj look stupid. The linear equation thing was classy.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Firecrest - 10-19-2009
Aseic Wrote:you said you would get rid of it rojWhy are you posting on the forums but not in raid?
Re: It's 4%, AKA the Go Fuck Yourself Post. - Roj - 10-19-2009
Ok let's take a look at your formula:
Lorake Wrote: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.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Meekslol - 10-19-2009
this thread warrants:
Re: It's 4%, AKA the Go Fuck Yourself Post. - geng - 10-19-2009
11 fucking pages of miscommunication.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-19-2009
Roj Wrote:Lorake Wrote: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.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.
I'm enjoying this. It's fun making Roj look stupid. The linear equation thing was classy.
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.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-19-2009
Roj Wrote:Ok let's take a look at your formula:
Lorake Wrote: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.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-19-2009
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
Re: It's 4%, AKA the Go Fuck Yourself Post. - geng - 10-19-2009
NOPE WRONG NOPE TRY AGIN NOPE
Re: It's 4%, AKA the Go Fuck Yourself Post. - Roj - 10-19-2009
OMG you cannot make this statement you moron.
Lorake Wrote:F(4,5) = 0 since X < Y.Read my post, X=5 and Y=5
Lorake Wrote: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