Roj Wrote:

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.

Lorake Wrote: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.

Roj Wrote:So with just looking at your results you're saying there's a 21.7% chance of someone being double targetted in only 7 chases? I haven't even started to plug in numbers and this is already starting to look good!!!

Lorake Wrote:Just to verify - You are going to try to find A: the probability of an Anub double target on 7 chases, not B: the probability of Anub double targetting on a specific person in raid in 7 chases. If you're trying to find B this argument is moot and you just aren't listening.

Roj Wrote:We've had 2 double chases in 200 attempts.

Thorran Wrote:We had 2 double chases tonight.