10-21-2009, 12:45 AM
Roj Wrote:Ok here we go. This is the latest version of your "equation" after you've editted it 3 times.
Lorake Wrote: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?
If F(X,Y) = 1 as X approaches infinity, then F(X+1,Y) =
F(X+1,Y) = .04^(Y-1)+ .96( 1 + F(X,Y+1)) = .96 + .04^(Y-1) + .96 * F(X,Y+1)
Assuming F(X, Y+1) is a positive number greater than 0, then
F(X+1,Y) > 1
Which would lead truth to your claim, but the problem is the limit of Y approaching infinity is 0, so that's a problem in of itself. The Diff of (3,2) and (4,2) is .0384. The difference of (4,2) and (5,2) is .03833856. Seems slow at the low numbers but it might show the limit might approach 0 as it increases. Tell you what, I'll make a script for this one too in C++ and see if the limit will exceed 1. Could be possible, could also not be possible.
That being said I don't know it's validity for the (5,2) or the (4,2) case unless I run the C++ program to get a correct probability of (5,2) and (4,2) since double chases for individuals can overlap and thus I'll hold off for now on saying it's super right. It still holds for (5,3), (6,4) as well and (5,4) and (5,5). If I had to make a guess I think it's gonna fail badly on the (5,2) case and the equation is going to look something more demented, but I'll not hold my breath until then.
And btw Roj, I'm still waiting for you to say you were wrong for the original post, which is that the probability of an anub double chase of a random person given two chases in a raid of 25 people is 4%.
I also want to hear you say you were wrong in assuming that the original function was a linear function.
Well, what I want is not what I'm going to get in this case. I highly doubt you'd ever own up to your own fuck up. If shit hits the fan and you're right about the function going over 100% I'll own up to it, and even report it with reason in detail, which is more than what you've been doing. At this point the function argument has broken far away from the original problem. Which was the chance of an anub double chase, which my answer of 4% still remains right. You'll never accept it though, because that's just you.