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

[Roj]

We had 2 double chases tonight.

Uh what...? Maybe I'm not understanding what a "double chase" is. I'm thinking of someone being chased back-to-back and I'm pretty certain that did not happen tonight. I've especially been watching for it.

[Lorake]

Pretty sure we had 3 at least, because I was making a mental note of each and wanted to say something on vent, but I've kept this 'debate' out of raid. 3/20 = 15%, which isn't bad, considering we can even not notice the times where anub chases someone before coming up, then he goes under and chases that same person again (which would be a 'valid' back to back in this logic).

At this point I really don't want to think about that creating an algorithm that does back to backs with multiple chases, separated into subsets because that would really fuck with the numbers more. More than likely it would be a probability OR statement like (3,2) OR (4,2) (F(3,2) + F(4,2) * (1-F(3,2)) by basic statistics) if we wanted to match it to our current strat (which comes out to roughly 18.5%, surprising).

Again, real performance never matches 'probability' because, after all, it's a chance.

Until I hear from you, I await for your case breaker. Just saying "LOL IT DOESN'T LOOK REASONABLE" will not float. You'll have to try for something above the 7s, as I used the 3s, 4s, 5s, 6s, and 7s as my test cases to verify the algorithm works. Right now my big cases are still running so I'll post them when they become available. Like I said I'm brute forcing the actual probability so it's taking longer since it has to determine if each possible combination has the right or wrong set up.

Finally some alpha examples, so if you do go with a program you can verify you're getting the alpha I expect.

(7,3) - Alpha = 2
(8,4) - Alpha = 3
(8,5) - Alpha = 3
(7,4) - Alpha = 3

[geng]

can you do the math on the % that this fucking matters at all?

[Firecrest]

We had 2 double chases tonight.

Uh what...? Maybe I'm not understanding what a "double chase" is. I'm thinking of someone being chased back-to-back and I'm pretty certain that did not happen tonight. I've especially been watching for it.

I was. And Aragon was. I think someone else was, but I'm sure of those two at least. It's a pain in the ass because there's no raidwarning to tell you it's still on you. All you know is if you get chased and then it doesn't pop up on anyone else, it's still on you.

[Lorake]

can you do the math on the % that this fucking matters at all?

That one's obvious, 0%. I'm still enjoying the development of the algorithm personally, and if Roj can actually provide a case breaker I'm really interested in figuring out what made it break. And if Roj actually fails to break it I get to claim an ego victory. I win either way.

[Roj]

Just posting these for my reference in case a certain unscrupulous person removes them. Haven't had a chance to look at the latest one very closely yet.

Lorakemath Version 2 (Did not catch the equation before the first condition was added)

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) = 0 when Y = 1 *Edit: broke my own equation LOL
F(X,Y) = .04^(Y-1) + .96F(X-1,Y)

Lorakemath Version 3

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)

Lorakemath Version 4

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)

Lorakemath Version 5

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))

Lorakemath Version 6 (wording changed for comparison)

The probability of Y back to back chases given Anub does X chases is :

F(X,Y) = 0 when Y>X
F(X,Y) = 1 when Y = 1
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

Lorakemath Version 6 (original)

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

[Lorake]

I didn't have the F(X,Y) = 1 when Y = 1 originally. So F(20,1) was going abeshit with the original equation.

Edit - 2 and 3 are the same equation? Maybe you copied a repost.

[Mikaelo]

so who won

[Firecrest]

We all lost.

RIP Roj.

[Hitshade]

Stumbled on this and for some reason thought about this thread.

Better late than never: