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

[Mikaelo]

FOR SOMEBODY SO INTELLIGENT WITH #s IM SURPRISED AT YOUR SPELLING OF A CERTAIN WORD

[geng]

He's actually rediculously close. It's 0.04^(X), for all X greater than 0 and for the single case. The odds of a person being followed once in a single P2 is 0.04^(1) * (1-0.04)^2 ~ 3.7%, followed twice in a row is 0.04^(2) * (1-0.04) ~ 1.5%, followed all three times is 0.04^(3) ~ lolz blizzard hates you. Assuming nobody fucks up and there's only two ice patches hit.

It gets even better if you look at two phase 2's, but I obviously hate math.

o shit you just angered the lorake.

[jackslammer]

The argument in officer chat was about context more than math.

Here is math though. :ugeek:

Set size is equal to a raid of twenty five players and the subset doesn't matter but I'll use 7 players being chased in an attempt.

For all integer n > 0, n! = n·(n-1)!

((nPr1:r7)^-1)*100
n = 25
r = 1 to 7
Being followed once = 4.00000000%
Being followed twice IN A ROW = 0.16000000%
Being followed three times IN A ROW = 0.00640000%
Being followed four times IN A ROW = 0.00025600%
Being followed five times IN A ROW = 0.00001024%
Being followed six times IN A ROW = 0.00000041%
Being followed seven times IN A ROW = 0.00000002%

Is the respective probability of being selected in a row in an attempt. Otherwise it is just 4%.

[Lorake]

according to this you have a 4% chance to get targetted a million times in a row.

No, according to this, for a single person to get targetted X amount of times in a row, given the single person it targets does not matter, and 100% = 1, a person has a .04^(X-1) chance of being targeted X times.

so.. when X = 1 ... a single person has a 100% chance to be targeted 1 time? WOO

Considering the ability has to target someone, when X = 1 then yes it's 100% obviously. It's pretty safe to say, since Anub has never NOT chased someone in P2. It'd probably be safer to say that's the chance Anub has to chase the same person constantly, given he doesn't care which person he picks to chase constantly.

He's actually rediculously close. It's 0.04^(X), for all X greater than 0 and for the single case. The odds of a person being followed once in a single P2 is 0.04^(1) * (1-0.04)^2 ~ 3.7%, followed twice in a row is 0.04^(2) * (1-0.04) ~ 1.5%, followed all three times is 0.04^(3) ~ lolz blizzard hates you. Assuming nobody fucks up and there's only two ice patches hit.

It gets even better if you look at two phase 2's, but I obviously hate math.

Where are you coming up with 1-.04 off in lala land Steke. There are only two events being factored in, being chased and not being chased. You don't need to worry about the not chase cases, since those break our conditions. Thus 1-.04 = .96 chance of not being chased, we don't care. Clarify where you derive 0.04^(1) * (1-0.04)^2 ~ 3.7% since if you sum it by 25, it's not 100% total probability, which means there's a probability he does not chase anyone in raid at all, which is an event that does not exist.

[Lorake]

Probability of someone 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


etc etc etc

Figuring out the probability of someone getting chased Y times in a row in X chases gets fishy and messy. I'm thinking on that one. If it weren't required that Y be back to back, then it would be easier, but we're speaking back to back cases only. For Y times being chased where we don't give a shit if it's back to back or not in X possible chases it's

f(X,Y) = sum(t(X,i)*.04^(X-i)*.96^i ,i=0, i<X-Y+1, i++)

Alas I digress. I like math.

OP still stands true in context unless Steke can clarify his numbers.

[Lorake]

God I'm a hidden EJ number junkie at heart.

[Nouhstekehue]

0.04 is the chance of being chased, 1-0.04 is the chance of not being chased. The sum doesn't reach 1 because the numbers provided don't cover all cases.

Supposing a person is followed once, we end up with (0.04^(1)) + (1-0.04^(1)) = 1. Pretty standard stuff there.

The odds of an event occuring multiple times is (probability)^(times occuring) * (probability)^(times not occuring). The sum of separate odds is the odds of one of the summed odds occuring. Therefore, the odds of being followed twice when three targes are chosen are:
0.04^2 * 0.96 + 0.96 * 0.04^2 = 0.003072

Being followed once is:
0.96^2 + 0.04 + 0.96 * 0.04 * 0.96 + 0.04 * 0.96^2 = 0.110592

Being followed all three times is:
0.04^3 = 0.000064

Being followed 0 times is:
0.96^3 = 0.884736

Being followed 2 times, but not in a row is:
0.04 * 0.96 * 0.04 = 0.001536

And the sum of those is : 1

I was wrong the first time, hope this clarifies. Also, the odds of not being chased an entire fight that has six chases is 0.96^6 = 0.782757789696, or greater than 75%, which suits the bads ratio pretty well .

[Masex]

I hate this thread.

[Bbriguyy]

the real question is what is the % of people that will fuck up kiting when they get targeted

[abecx]

THIS SITE DOES NOT ALLOW THEORY CRAFTING OF IRL VARIABLES