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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. - Lorake - 10-19-2009
Roj Wrote:Lorake Wrote:The probability of anub double targetting is 4%Ah the ole make up some numbers trick. What happened to
The probability of anub triple targetting is .16%
The probability of anub quadruple targetting is .0064%
I'm so good at this.
2 times with 3 chases
F(3,2) = .04 + .96F(2,2) = .0784
Making the same assumption you are, which is the number of times equals the number of chases. If that's the case don't need an overly complex equation, .04^(X-1) is sufficient to describe.
I'll post the equations here for you.
Anub chooses same target X times for X chases
.04^(X-1)
Anub chooses same target Y times for X 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)
Probability of someone specifically getting chased Once in X chases,
f(X) = sum(t(X,i)*.04^(X-i)*.96^i ,i=0, i<X, i++)
where t(X,i) refers to the (i+1) value from Pascal's triangle for the Xth row.
http://en.wikipedia.org/wiki/Pascal's_triangle
So
t(4, 2) = 6
t(3, 2) = 3
t(5, 0) = 1
t(5, 1) = 5
t(5, 2) = 10
t(5, 3) = 10
etc
so
f(2) = .04^2 + 2 *.04*.96
f(3) = .04^3 + 3* .04^2*.96 + 3 * .04 * .96^2
f(4) = .04^4 + 4* .04^3*.96 + 6 * .04^2 * .96^2 + 4 * .04 * .96^3
It can also be simplified to f(x) = 1-.96^X since you can take the probability of not being targetted and subtracting it from 100%
Finally the probability of someone getting chased Y times in X chases with no regards to order, thus not back to back:
f(X,Y) = sum(t(X,i)*.04^(X-i)*.96^i ,i=0, i<X-Y+1, i++)
Re: It's 4%, AKA the Go Fuck Yourself Post. - Meekslol - 10-19-2009
fancy math algorithms isn't going to convince anyone here since the problem is not the math it is the relative perspective Lorake and Roj are taking.
Roj is arguing this from the point of view of where we see most math problems in like textbooks and what your middle school calc teacher gave you for hw.
Before anub targets someone, what is the chance of him targeting 2 people in a row? The answer to this would be 0.04 x 0.04 which is 0.0016 or 0.16%. The chance of anub targetting the same person twice in a row is 0.16% if you are considering the odds before he targets someone.
Lorake on the other hand, is taking a more practical approach. He is looking at the problem from when anub has already targetted someone.
Since anub is going always target people, Lorake is saying we don't give a shit what the probability is for the first target (which happens to be 0.04 or 4%). He has a 100% chance of targetting someone, and we only care about that next 0.04 probability of his 2nd targetting round. 100% chance of anub targetting someone (1.00) x Anub's second round of targetting (0.04) is 0.04 or 4%. So the chance of anub targetting someone twice in a practical sense is 4%.
Anub's first targetting probability of does not matter since he is going to select somebody. It is 100% even tho it's 4% for a specific person. The final practical answer is 4% because only his second round of targetting matters.
If that makes sense
Re: It's 4%, AKA the Go Fuck Yourself Post. - Nouhstekehue - 10-19-2009
Meekslol Wrote:fancy math algorithms isn't going to convince anyone here since the problem is not the math it is the relative perspective Lorake and Roj are taking.
Roj is arguing this from the point of view of where we see most math problems in like textbooks and what your middle school calc teacher gave you for hw.
Before anub targets someone, what is the chance of him targeting 2 people in a row? The answer to this would be 0.04 x 0.04 which is 0.0016 or 0.16%. The chance of anub targetting the same person twice in a row is 0.16% if you are considering the odds before he targets someone.
Lorake on the other hand, is taking a more practical approach. He is looking at the problem from when anub has already targetted someone.
Since anub is going always target people, Lorake is saying we don't give a shit what the probability is for the first target (which happens to be 0.04 or 4%). He has a 100% chance of targetting someone, and we only care about that next 0.04 probability of his 2nd targetting round. 100% chance of anub targetting someone (1.00) x Anub's second round of targetting (0.04) is 0.04 or 4%. So the chance of anub targetting someone twice in a practical sense is 4%.
Anub's first targetting probability of does not matter since he is going to select somebody. It is 100% even tho it's 4% for a specific person. The final practical answer is 4% because only his second round of targetting matters.
If that makes sense
Good post.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-19-2009
Meeks understands what I'm trying to say. Steke understands what I'm trying to say. Furly understands what I'm trying to say. I'll treat it as a valid attempt to be clear and not just me being a fuckwad and saying you're wrong.
You are wrong, Roj, as long as you keep saying I'm wrong and full of shit.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-19-2009
Meekslol Wrote:fancy math algorithms isn't going to convince anyone here since the problem is not the math it is the relative perspective Lorake and Roj are taking.
Roj is arguing this from the point of view of where we see most math problems in like textbooks and what your middle school calc teacher gave you for hw.
Before anub targets someone, what is the chance of him targeting 2 people in a row? The answer to this would be 0.04 x 0.04 which is 0.0016 or 0.16%. The chance of anub targetting the same person twice in a row is 0.16% if you are considering the odds before he targets someone.
Lorake on the other hand, is taking a more practical approach. He is looking at the problem from when anub has already targetted someone.
Since anub is going always target people, Lorake is saying we don't give a shit what the probability is for the first target (which happens to be 0.04 or 4%). He has a 100% chance of targetting someone, and we only care about that next 0.04 probability of his 2nd targetting round. 100% chance of anub targetting someone (1.00) x Anub's second round of targetting (0.04) is 0.04 or 4%. So the chance of anub targetting someone twice in a practical sense is 4%.
Anub's first targetting probability of does not matter since he is going to select somebody. It is 100% even tho it's 4% for a specific person. The final practical answer is 4% because only his second round of targetting matters.
If that makes sense
Meeks, if I'm talking about the chance anub chooses the same target twice, you'd take the sum of the individual person's being double targetted to determine the probability of Anub double targetting.
So 25 x .16% (single person double target probability) = 4%
Re: It's 4%, AKA the Go Fuck Yourself Post. - Roj - 10-19-2009
Lorake Wrote:Meeks understands what I'm trying to say. Steke understands what I'm trying to say. Furly understands what I'm trying to say. I'll treat it as a valid attempt to be clear and not just me being a fuckwad and saying you're wrong.I understand what you're trying to say, you're just wrong. Meeks understands also and appears to be the only person who has any grasp of math in this thread. Even though you're trying to make a different point, your equation is wrong for what you're trying to say. Regardless, the question was:
You are wrong, Roj, as long as you keep saying I'm wrong and full of shit.
What is the probability of being targetted twice in a row?
Only someone who is mentally handicapped would say the answer is 4%.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-19-2009
Roj Wrote:Lorake Wrote:Meeks understands what I'm trying to say. Steke understands what I'm trying to say. Furly understands what I'm trying to say. I'll treat it as a valid attempt to be clear and not just me being a fuckwad and saying you're wrong.I understand what you're trying to say, you're just wrong. Meeks understands also and appears to be the only person who has any grasp of math in this thread. Even though you're trying to make a different point, your equation is wrong for what you're trying to say. Regardless, the question was:
You are wrong, Roj, as long as you keep saying I'm wrong and full of shit.
What is the probability of being targetted twice in a row?
Only someone who is mentally handicapped would say the answer is 4%.
What is the probability of Anub doing a double target? 4%
And the question was what is the probability of Anub doing a double target.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Ethax - 10-19-2009
10 pages this is fucking incredible.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-19-2009
Ethax Wrote:10 pages this is fucking incredible.
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.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Roj - 10-19-2009
Lorake Wrote:What is the probability of Anub doing a double target? 4%Mind boggling
And the question was what is the probability of Anub doing a double target.
hock: