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

[Roj]

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.

Come on, use some common sense. You really think there's a 15% chance of a double chase in only 5 attempts?

Think about how many double chases we've had in all our attempts. I can only think of two. And I would guess we've had about 600 total chases. According to my equation ((1/25)^X) we should have had approximately 1 double chase. Even if we lowered your percentage to a 4% chance that means we should have had 24 double chases.

I'm kind of wondering if you are going to just say HAHA GOTCHA I'VE BEEN TROLLING ALL ALONG!!!!

Edit: So wait a second. So your "equation" says the following:

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))

So using 2 and 5 it reads:

The probability of 2 back to back chases given Anub does 5 chases (answer: 15%).

That sounds to me like 2 different people being chased back to back in 5 chases. There's no way you actually think you're correct. Even if your wording is just as retarded as your equation it's still ludicrous.