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

[Thorran]

Yeah hit up like page 5 or 6 where I explain this is all irrelevant because it isn't actually a solely based on probability.

[Ethax]

einstein meets gandalf came to mind

Priceless.

[Crazy Diana]

dont think i didnt see you edit that

[Kaistine]

Is there anything worth reading in this thread?

No

[Roj]

This is probably the closest you will ever get to admitting you're wrong.

Check my numbers and verify I did the sum of the individual and F(X,Y) right. If this is true Roj is right about my formula, and my formula and I have to eat crow on this one, but it still stands Anub has a .04^(X-1) chance of X selecting people in X tries, as my formula has held and matched. My formula holds true for X chases in a row with X chances, and X-1 chases in a row with X chances, which were the two original cases I checked myself before putting this monstrosity up.

And there you go Roj. If you had only given a simple case and not just said NO NO LOL YOU'RE RETARDED and make yourself look like shit, then you could have walked away from this one clean and not looking like a retard bantering nonsense for 8 pages. The first few cases you gave me were shot down, only because you didn't even bother thinking it out. I'm sure if you actually had tried to make sense you could have came up with the case to break it easily, but thanks to your random number generation YOU LUCKED OUT.

But don't worry I'll keep checking your math for you my student. Let's start with formula 8 v2.5

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))

F(5,3) = .04^(2) + .96(F(4,3)+F(4,4)) = .04^(2) + .96(.04^(2)+ .96F(3,3) + .04^3)

We'll be simple. How the fuck can you explain .04^(2) + .96(F(4,3)+F(4,4)) equalling .04^(2) + .96(.04^(2)+ .96F(3,3) + .04^3) with your "equation"?

[Ethax]

I quoted the wrong post.

[Lorake]

This is probably the closest you will ever get to admitting you're wrong.

Check my numbers and verify I did the sum of the individual and F(X,Y) right. If this is true Roj is right about my formula, and my formula and I have to eat crow on this one, but it still stands Anub has a .04^(X-1) chance of X selecting people in X tries, as my formula has held and matched. My formula holds true for X chases in a row with X chances, and X-1 chases in a row with X chances, which were the two original cases I checked myself before putting this monstrosity up.

And there you go Roj. If you had only given a simple case and not just said NO NO LOL YOU'RE RETARDED and make yourself look like shit, then you could have walked away from this one clean and not looking like a retard bantering nonsense for 8 pages. The first few cases you gave me were shot down, only because you didn't even bother thinking it out. I'm sure if you actually had tried to make sense you could have came up with the case to break it easily, but thanks to your random number generation YOU LUCKED OUT.

But don't worry I'll keep checking your math for you my student. Let's start with formula 8 v2.5

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))

F(5,3) = .04^(2) + .96(F(4,3)+F(4,4)) = .04^(2) + .96(.04^(2)+ .96F(3,3) + .04^3)

We'll be simple. How the fuck can you explain .04^(2) + .96(F(4,3)+F(4,4)) equalling .04^(2) + .96(.04^(2)+ .96F(3,3) + .04^3) with your "equation"?

F(4,3) = .04^2 + .96(F(3,3) + F(3,4)) O BTW F(3,4) = 0
F(4,4) = .04^3 + .96(F(3,4) + F(3,5)) O BTW F(3,5) = 0
So
F(4,3) = .04^2 + .96F(3,3)
F(4,4) = .04^3

Teacher's obviously retarded if he couldn't do that simple math on his own.

[Zoia]

Lets start a poll who the fuck do you think is going to give up first roj or lorake

[Crazy Diana]

theyre both ignorant, i wager roj for 20 coppers

[Roj]

Ok here we go. This is the latest version of your "equation" after you've editted it 3 times.

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?