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

[Lorake]

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 .

http://www.mathgoodies.com/lessons/vol6 ... vents.html

Ok I get where you're coming from. But the original argument is back to back chases occuring to someone, with the total chases being equal to back to back chases being assumed. Bad assumption to presume I suppose but that's the assumption. So .04^(X-1) still holds true for when the number or back to back chases equal the number of total chases, since we still hold that the first target selected does not matter. So if it's 2 back to back chases out of two chases, it's still .04, 3 back to back chases out of 3 chases is .0016, etc etc. Things get messy when the total chases don't equal back to back chases.

[sophronius]

THIS SITE DOES NOT ALLOW THEORY CRAFTING OF IRL VARIABLES

yeah NERDS

[Crazy Diana]

im so happy that jack and lorake went to such lengths to answer my 3 times in a row question

[Roj]

Let's establish context first. The argument is simple, and warrants a clean, clear explanation.

The question: What is the probability of Anub targetting the same person twice in a row.

Some things to understand -
A single person has a 4% chance to be targetted.
There is no preferential treatment for back to back selection - IE if someone gets selected once, it's still a 4% chance to be selected again.

Scope: The person Anub targets twice does not have to be a specific person - It can target Roj twice or Lorake twice, it just has to be a back to back selection.

Answer: 4% This is justified in two different methods.

Method A: Given that the first person it selects does not matter, it just needs to select one person. Thus the 4% probability of selection does not matter and does not factor in, since we don't give a shit who it selects. Thus it's first person, target A, is selected. After a stun, a new target is selected, and target A has a 4% chance of being selected, thus there is a 4% chance of a back to back targetting. Still not good enough?

Method B: So you don't buy into the fact that the first selection can be ignored completely. Ok, so you're retarded, but I can work with that. Given, that there is 25 people in the raid. For a single person in the raid, he has a 4% chance of getting it the first time, and 4% the second time to be targetted, thus a .16% percent of being targetted twice. So each person in a 25 man raid has a .16% chance of being double targetted, but since we're talking about a raid member in general and not a specific member being targetted twice, we need to take the sum of those probabilities. Thus 25 * .16% = 4%.

TLR It's 4%, go fuck yourself while eating dicks.

Wrong Wrong Wrong

I don't think you understand the difference between these two statements:

1. The probability of a person being targetted 2 times in a row.
2. After being targetted once, the probability of that person being targetted again.

Either you're really stupid or you're terrible at math. The probability of any raid member being targetted "X" times in a row is (1/25)^(X).

I.E.
1x = 4%
2x = .16%
3x = .0064%
4x = .000256%

[Nouhstekehue]

Angry Roj :<

[Roj]

Angry Roj :<

Lol I'm always angry. And I love being called retarded.

[Sanderz]

ITT: Idiots that didn't take statistics

[geng]

50% chance to be followed.
25% chance to be followed 2 times in a row.
12.5% chance to be followed 3 times in a row.
100% chance this is thread is fully retarded.

[Lorake]

Let's establish context first. The argument is simple, and warrants a clean, clear explanation.

The question: What is the probability of Anub targetting the same person twice in a row.

Some things to understand -
A single person has a 4% chance to be targetted.
There is no preferential treatment for back to back selection - IE if someone gets selected once, it's still a 4% chance to be selected again.

Scope: The person Anub targets twice does not have to be a specific person - It can target Roj twice or Lorake twice, it just has to be a back to back selection.

Answer: 4% This is justified in two different methods.

Method A: Given that the first person it selects does not matter, it just needs to select one person. Thus the 4% probability of selection does not matter and does not factor in, since we don't give a shit who it selects. Thus it's first person, target A, is selected. After a stun, a new target is selected, and target A has a 4% chance of being selected, thus there is a 4% chance of a back to back targetting. Still not good enough?

Method B: So you don't buy into the fact that the first selection can be ignored completely. Ok, so you're retarded, but I can work with that. Given, that there is 25 people in the raid. For a single person in the raid, he has a 4% chance of getting it the first time, and 4% the second time to be targetted, thus a .16% percent of being targetted twice. So each person in a 25 man raid has a .16% chance of being double targetted, but since we're talking about a raid member in general and not a specific member being targetted twice, we need to take the sum of those probabilities. Thus 25 * .16% = 4%.

TLR It's 4%, go fuck yourself while eating dicks.

Wrong Wrong Wrong

I don't think you understand the difference between these two statements:

1. The probability of a person being targetted 2 times in a row.
2. After being targetted once, the probability of that person being targetted again.

Either you're really stupid or you're terrible at math. The probability of any raid member being targetted "X" times in a row is (1/25)^(X).

I.E.
1x = 4%
2x = .16%
3x = .0064%
4x = .000256%

Ok so maybe my point isn't clear. Maybe I should restate the point as the probability of a RANDOM person being targetted twice. In which case it's just the same as a person being targetted again, which was the crux of the debate, since the debate was what's the probability that someone in raid (IE RANDOM PERSON) would get double targetted. And it's 4%. If you still insist that I'm wrong, you are retarded.

If we were talking, say, the probability of LORAKE being targetted twice in a row, then it's .16%, since we've set the conditional statement of who the first target must be. In my complete explanation of method B I completely explain how the sum of individual probabilities equals 4%. Statistics 101 brah.

Somehow I just doubt you read the OP in completeness, and just said I was wrong in your old man ignorance. I am right. You are wrong. Stop doing bad math in your head.

[geng]

targeted once = 4%
targetted twice in a row = 4%

does not make any fucking sense