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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 - 10-21-2009
Roj Wrote:Ok here we go. This is the latest version of your "equation" after you've editted it 3 times.
Lorake Wrote:Find the probability of Y back to back chases given Anub does X chases
F(X,Y) = 0 when Y>X
F(X,Y) = 1 when Y = 1
F(X,Y) = .04^(Y-1) + .96(F(X-1,Y) +F(X-1, Y+1))
First of all, I need to post a disclaimer. I'm going to calculate this out using the Lorakefailsatmath method. The equation is broken in the first place but even using your own methods it doesn't compute. Here's the first 3 sets I tried:
F(3,2) = .04^(2-1) + .96(F(3-1,2) +F(3-1, 2+1))
F(3,2) = .04^(1) + .96(F(2 ,2) +F(2 , 3 ))
F(3,2) = .04 + .96( ( .04 ) + ( 0 ))
F(3,2) = .04 + .0384
F(3,2) = .0784
F(4,2) = .04^(2-1) + .96(F(4-1,2) +F(4-1, 2+1))
F(4,2) = .04^(1) + .96(F(3 ,2) +F(3 , 3 ))
F(4,2) = .04 + .96( (.0784) + ( .0016 ))
F(4,2) = .04 + .0768
F(4,2) = .1168
F(5,2) = .04^(2-1) + .96(F(5-1,2) +F(5-1, 2+1))
F(5,2) = .04^(1) + .96(F(4 ,2) +F(4 , 3 ))
F(5,2) = .04 + .96( (.1168) + (.003136 ))
F(5,2) = .04 + .11513856
F(5,2) = .15513856
So this "equation" says that the probability of 2 back to back chases given Anub does 5 chases is 15.5%!!!
Obviously that's retarded and it won't be long before the probability is greater than 100% of back to back chases. That's strike 3 on your nonsense "equations". Are you going to start from scratch this time or keep adding on to this epic fail of stupidity?
If F(X,Y) = 1 as X approaches infinity, then F(X+1,Y) =
F(X+1,Y) = .04^(Y-1)+ .96( 1 + F(X,Y+1)) = .96 + .04^(Y-1) + .96 * F(X,Y+1)
Assuming F(X, Y+1) is a positive number greater than 0, then
F(X+1,Y) > 1
Which would lead truth to your claim, but the problem is the limit of Y approaching infinity is 0, so that's a problem in of itself. The Diff of (3,2) and (4,2) is .0384. The difference of (4,2) and (5,2) is .03833856. Seems slow at the low numbers but it might show the limit might approach 0 as it increases. Tell you what, I'll make a script for this one too in C++ and see if the limit will exceed 1. Could be possible, could also not be possible.
That being said I don't know it's validity for the (5,2) or the (4,2) case unless I run the C++ program to get a correct probability of (5,2) and (4,2) since double chases for individuals can overlap and thus I'll hold off for now on saying it's super right. It still holds for (5,3), (6,4) as well and (5,4) and (5,5). If I had to make a guess I think it's gonna fail badly on the (5,2) case and the equation is going to look something more demented, but I'll not hold my breath until then.
And btw Roj, I'm still waiting for you to say you were wrong for the original post, which is that the probability of an anub double chase of a random person given two chases in a raid of 25 people is 4%.
I also want to hear you say you were wrong in assuming that the original function was a linear function.
Well, what I want is not what I'm going to get in this case. I highly doubt you'd ever own up to your own fuck up. If shit hits the fan and you're right about the function going over 100% I'll own up to it, and even report it with reason in detail, which is more than what you've been doing. At this point the function argument has broken far away from the original problem. Which was the chance of an anub double chase, which my answer of 4% still remains right. You'll never accept it though, because that's just you.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Crazy Diana - 10-21-2009
gentlemen, let me through, im a scientist
Re: It's 4%, AKA the Go Fuck Yourself Post. - Sanderz - 10-21-2009
As an engineer, i say this is a waste of time.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Crazy Diana - 10-21-2009
im pretty sure who anub charges is determined the same way loot is, by who zoned in first
Re: It's 4%, AKA the Go Fuck Yourself Post. - Sanderz - 10-21-2009
cr4zyd Wrote:im pretty sure who anub charges is determined the same way loot is, by who zoned in first
They're both based on what abe gets from Taco Bell.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Roj - 10-21-2009
Lorake Wrote:If F(X,Y) = 1 as X approaches infinity, then F(X+1,Y) =Come on, use some common sense. You really think there's a 15% chance of a double chase in only 5 attempts?
F(X+1,Y) = .04^(Y-1)+ .96( 1 + F(X,Y+1)) = .96 + .04^(Y-1) + .96 * F(X,Y+1)
Assuming F(X, Y+1) is a positive number greater than 0, then
F(X+1,Y) > 1
Which would lead truth to your claim, but the problem is the limit of Y approaching infinity is 0, so that's a problem in of itself. The Diff of (3,2) and (4,2) is .0384. The difference of (4,2) and (5,2) is .03833856. Seems slow at the low numbers but it might show the limit might approach 0 as it increases. Tell you what, I'll make a script for this one too in C++ and see if the limit will exceed 1. Could be possible, could also not be possible.
That being said I don't know it's validity for the (5,2) or the (4,2) case unless I run the C++ program to get a correct probability of (5,2) and (4,2) since double chases for individuals can overlap and thus I'll hold off for now on saying it's super right. It still holds for (5,3), (6,4) as well and (5,4) and (5,5). If I had to make a guess I think it's gonna fail badly on the (5,2) case and the equation is going to look something more demented, but I'll not hold my breath until then.
And btw Roj, I'm still waiting for you to say you were wrong for the original post, which is that the probability of an anub double chase of a random person given two chases in a raid of 25 people is 4%.
I also want to hear you say you were wrong in assuming that the original function was a linear function.
Well, what I want is not what I'm going to get in this case. I highly doubt you'd ever own up to your own fuck up. If shit hits the fan and you're right about the function going over 100% I'll own up to it, and even report it with reason in detail, which is more than what you've been doing. At this point the function argument has broken far away from the original problem. Which was the chance of an anub double chase, which my answer of 4% still remains right. You'll never accept it though, because that's just you.
Think about how many double chases we've had in all our attempts. I can only think of two. And I would guess we've had about 600 total chases. According to my equation ((1/25)^X) we should have had approximately 1 double chase. Even if we lowered your percentage to a 4% chance that means we should have had 24 double chases.
I'm kind of wondering if you are going to just say HAHA GOTCHA I'VE BEEN TROLLING ALL ALONG!!!!
Edit: So wait a second. So your "equation" says the following:
Lorake Wrote:Find the probability of Y back to back chases given Anub does X chases
F(X,Y) = 0 when Y>X
F(X,Y) = 1 when Y = 1
F(X,Y) = .04^(Y-1) + .96(F(X-1,Y) +F(X-1, Y+1))
So using 2 and 5 it reads:
The probability of 2 back to back chases given Anub does 5 chases (answer: 15%).
That sounds to me like 2 different people being chased back to back in 5 chases. There's no way you actually think you're correct. Even if your wording is just as retarded as your equation it's still ludicrous.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Roj - 10-21-2009
Lorake Wrote:And btw Roj, I'm still waiting for you to say you were wrong for the original post, which is that the probability of an anub double chase of a random person given two chases in a raid of 25 people is 4%.So let's expand on this because I think even though you now realize your "equation" is incorrect all you really care about is demonstrating that the chance to be chased twice in a row is 4% not .16%. You have failed to prove this mathematically so let's look at it practically from all of the attempts we have done. We've done 150+ attempts. Anub does approximately 7 chases per full attempt, but we sometimes wipe early so let's say 4 chases per attempt which gives us 600 total chases. .16% of 600 is .96. 4% of 600 is 24. In all of our attempts I've only seen two double chases. Maybe I missed one, but regardless it is not even CLOSE to 24 double chases.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-21-2009
Roj Wrote:Lorake Wrote:And btw Roj, I'm still waiting for you to say you were wrong for the original post, which is that the probability of an anub double chase of a random person given two chases in a raid of 25 people is 4%.So let's expand on this because I think even though you now realize your "equation" is incorrect all you really care about is demonstrating that the chance to be chased twice in a row is 4% not .16%. You have failed to prove this mathematically so let's look at it practically from all of the attempts we have done. We've done 150+ attempts. Anub does approximately 7 chases per full attempt, but we sometimes wipe early so let's say 4 chases per attempt which gives us 600 total chases. .16% of 600 is .96. 4% of 600 is 24. In all of our attempts I've only seen two double chases. Maybe I missed one, but regardless it is not even CLOSE to 24 double chases.
And when you flip a coin 100 times, because it's a 1/2 chance for it to be heads (roughly), it's suppose to be 50 heads and 50 tails. So obviously what you said above is true.
NOTE: PROBABILITY AND ACTUAL PERFORMANCE DON'T HAVE TO MATCH. PROBABILITY IS THE CHANCE THAT SOMETHING IS LIKELY TO HAPPEN, BUT DOESN'T HAPPEN. COMPARING IT TO "REAL" PERFORMANCE IS HILARIOUS AND IS A HUGE FAILURE ON YOUR PART.
Now for the equation. Get ready to shit on yourself in the end.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Roj - 10-21-2009
Lorake Wrote:NOTE: PROBABILITY AND ACTUAL PERFORMANCE DON'T HAVE TO MATCH. PROBABILITY IS THE CHANCE THAT SOMETHING IS LIKELY TO HAPPEN, BUT DOESN'T HAPPEN. COMPARING IT TO "REAL" PERFORMANCE IS HILARIOUS AND IS A HUGE FAILURE ON YOUR PART.You are a joke. A complete fucking retard. You would expect some deviation based on sample size, but with this amount of data you would not expect to be off by over 1000%. Or at least an intelligent person wouldn't.
Lorake Wrote:Now for the equation. Get ready to shit on yourself in the end.You've come up with so many failures you're bound to be successfully eventually because of sheer luck.
Re: It's 4%, AKA the Go Fuck Yourself Post. - Lorake - 10-21-2009
So to address the situation I'll go through the steps completely, because this is the last post on the equation, because, I believe, I now have the actual equation in question, thanks to several tests via a C++ program to emulate my formula and compare it to the brute force count of the number of Y BTBs in X attempts.
First off, my 'new' equation is wrong for (5,2) and (4,2) just like I suspected. Funny thing is, my original equation is 100% correct for (4,2) and (5,2) ( 11.5264% and 15.0653% probability, respectively. Yes (5,2) is ~15% for a 2 BTB, brute force assured me it). In addition, my 'new' equation was right for (6,4) and (6,3), so I went ahead and tested the equation on the 7 series ((7,2) to (7,6)). My old equation held on both (7,2) and (7,6), the new one held on (7,3) and (7,5), both failed on (7,4). However, injecting F(X-1, Y+2) into the mix of functions suddenly fixed (7,4).
Good thing my 'new' equation was wrong, because Roj's thought that it would exceed 100% was right, as it broke through the barrier at (27,2) on the 'new' equation. At (27,2) it was roughly 102% so yea the equation was wrong. Now this is where error correction starts coming in.
LORAKE'S FINAL (FOR NOW?), BRAND SPANKIN NEW EQUATION.
The probability of Y BTBs in X chases is :
Given that X,Y > 1 (It's UNPOSSIBLE to have ZERO TARGETS or 0 CHASES, OR NEGATIVE FOR THE MATTER)
F(X,Y) = 1 when Y = 1
F(X,Y) = 0 when Y>X
F(X,Y) = .04^(Y-1) + .96 * sum(F(X-1, Y+(i-1), i = 1, i <= alpha, i++)
where
alpha = Y - 1 when (X+1)/2>Y
alpha = X - Y otherwise
Well now Lorake, looks like that's even more garbage than before. Not only are you using two new variables (i and alpha) it's gonna still be over 100% in the end so obviously this will fail.
Especially since with your new function,
F(3,2) = .0784
F(4,2) = .115264
F(5,2) = .150653
F(6,2) = .184627
F(7,2) = .217242
O you are WRONG good sir. This equation will never break 100%, thanks to the magic I will show you next.
Given F(X,2) < 1 show that F(X+1, 2) < 1
So F(X+1,2) = .04 + .96F(X,2) (alpha = 1, thus the sum is equal to F(X,2))
If F(X+1,2) < 1, then
1 > .04 + .96F(X,2)
.96 > .96F(X,2)
1 > F(X,2) which is true based on our assumption.
In addition, the limit as Y approaches infinity is 0 since a finite number X will always be less than infinity.
Now for the DOUBLE LIMIT
lim F(N,N) = .04^(N-1) + 0 (.96F(N-1,N) = 0 since N > N-1)
N->INF
= .04^(N-1) = 0 since N approaches INFINITY.
I was confident with the original equation because I tested it a few times and it looked good (4,3) (4,2) (3,2) and (3,3) all worked out. When Roj broke it, I was less confident about the new equation, since a simple fix like F(X, Y+1) seemed fishy and unreasonable. Now with the several test cases I ran, including some really high ones, and thanks to the comparison to a brute force, I'm 98.6% confident in the above algorithm. The last little bit of percentage is to make up for the fact that I'm currently running (100, 30) in my algorithm, and because of the brute force, it's taking quite a bit of time. We'll see how it pans out.
SORRY ROJ NO JK LOL TROLLING, THIS EQUATION LOOKS LEGIT. GOOD THING I GOT THE EQUATION RIGHT BEFORE STRIKE 3, UNLESS YOU REALLY WANNA COUNT A TYPO IN THE WORDING OF THE PROBLEM AND NOT THE FORMULA A STRIKE FOR SOME REASON.