Lightning Mace and Actual Infinity

[Jack_Simth]

Edit: Also, I can't tell yet what the original thread thought would happen at 50% odds of critting. It might come up later, but either way, I think that's really the most interesting case. Also also, either way, this formula is super neat. Gives precise answers without relying on crazy infinite series calculation.

Hmm... full formula, the 50% case...

Spoiler: works out to 0/0

Show

(((1-n)/n)^k -1)/(((1-n)/n)^x -1)
= (((1-0.5)/0.5)^k -1)/(((1-0.5)/0.5)^x -1)
= (((0.5)/0.5)^k -1)/(((0.5)/0.5)^x -1)
= ((1)^k -1)/((1)^x -1)
= (1 -1)/(1 -1) (NOTE: 1 ^ (any real) = 1, which is why k and x don't much matter - I don't think we're looking for imaginary or complex numbers of attacks, nor are we starting with them... this step is invalid outside of real x and k)
= (0)/(0)

So... to get what it should be at that point, you're going to have to play around with infinitives.

[Douglas]

At 50%, the number of attacks remaining is a simple random walk, which if I'm reading the article correctly has a 100% chance to eventually hit 0 and stop.

[Jack_Simth]

At 50%, the number of attacks remaining is a simple random walk, which if I'm reading the article correctly has a 100% chance to eventually hit 0 and stop.

In terms of infinity, sure. But that's not quite what the full formula is addressing. The full formula is supposed to be "what the odds are of hitting any number of attacks". It shows 0/0 irrespective of the target number, however (for all real values, anyway).

[Hogsy]

No, you need to actually get a critical threat to get the extra attack.



Bolded for emphasis. A natural 1 automatically misses, and you don't get a critical threat if you didn't hit.

Oh, I always assumed that simply rolling a threat was sufficient. Didn't know the exact details of what "threat" meant. Thanks for clearing that up!

[OldTrees1]

At 50%, the number of attacks remaining is a simple random walk, which if I'm reading the article correctly has a 100% chance to eventually hit 0 and stop.

The 50% -1/50% +1 random walk has 100% chance to hit 0, given infinite time. However for all finite lengths of time there is at least 1 example case where the walk has not hit 0.


A proof of termination given infinite time:
Given infinite time the probability of eventually terminating for a given number of chances remaining is given as follows:
P(K) = a*P(K-1) + (1-a)P(K+1)
//Note this formula only works in infinite time. Otherwise P(K-1, given t=0) =/= P(K-1, given t=1).
P(K) - a*P(K-1) = (1-a)P(K+1)
P(K) + (a*P(K) - a*P(K)) - a*P(K-1) = (1-a)P(K+1)
P(K) - a*P(K) + a*P(K) - a*P(K-1) = (1-a)P(K+1)
(1-a)P(K) + a*P(K) - a*P(K-1) = (1-a)P(K+1)
a * ( P(K) - P(K-1) )= (1-a) * ( P(K+1) - P(K) )
P(K) - P(K-1) = ((1-a)/a) * ( P(K+1) - P(K) )
When a = 1/2: (1-a)/a = (1-0.5)/0.5 = 0.5/0.5 = 1
P(K) - P(K-1) = P(K+1) - P(K)
//Since the difference between the probabilities is constant & probabilities are restricted to the finite range of 0 to 1, the differences must be 0 (else eventually one of the probabilities would have an invalid value).
//Since we know the difference between the probabilities is 0 and we know that P(1) > 0, then we know that the probabilities are all 1 given infinite time.

[eggynack]

Hmm... full formula, the 50% case...

Spoiler: works out to 0/0

Show

(((1-n)/n)^k -1)/(((1-n)/n)^x -1)
= (((1-0.5)/0.5)^k -1)/(((1-0.5)/0.5)^x -1)
= (((0.5)/0.5)^k -1)/(((0.5)/0.5)^x -1)
= ((1)^k -1)/((1)^x -1)
= (1 -1)/(1 -1) (NOTE: 1 ^ (any real) = 1, which is why k and x don't much matter - I don't think we're looking for imaginary or complex numbers of attacks, nor are we starting with them... this step is invalid outside of real x and k)
= (0)/(0)

So... to get what it should be at that point, you're going to have to play around with infinitives.

I think there's an easy enough way to resolve this without screwing around with math at any high level. If we reduce the (1-n)/n in the denominator at all, then the resulting odds should be strictly greater than they are now. So, reduce that denominator "(1-n)" to any extent, and run the equation from there. The new denominator fraction, with the shrunken numerator, will go to zero, so the overall denominator will go to -1, and the numerator is still going to 0, because we haven't changed that. So, you get that the old result is less than or equal to the new result, which is equal to zero, and expected values cannot be negative so the probability is zero. I think that logic holds together. It's still weird that you get that uncomfortable 0/0 in the original equation, but what I've presented seems to serve as proof of its true value.

[martixy]

Your answer was wrong, or at least incomplete. In the first part of your post, you claimed that a sufficiently high starting number of attacks would decrease the chance of termination to something like zero. However, for any crit chance of 50% or lower, you will always terminate, no matter what the roll count is. In the second part of your post, you did seem to get to the result that 50% is a necessarily crossed threshold, but you had only some simulations as evidence. There wasn't any underlying proof, or formula, and it wasn't definitive in any sense. It didn't even look like there was real evidence of things going absolutely infinite and not eventually trailing away, because the simulation has to stop at some point. The answer I have here is absolutely complete, with the exact probabilities for every scenario, and the underlying math really isn't all that difficult. Yes, the math underlying that underlying math is very difficult, but anyone can use the formula once they know how. Way easier, as far as I'm concerned, than running a bunch of simulations every time.

This post is not wrong:
http://www.giantitp.com/forums/shows...8&postcount=19

Not sure what you were looking at.

[eggynack]

This post is not wrong:
http://www.giantitp.com/forums/showt...finite-attacks

Not sure what you were looking at.

Your first post in that thread simply says what the end result is, that you need to be pulling at least two attacks per threat, and over 50% odds of threat, but you had no proof for the claim, or estimation of the actual odds of going infinite. Your second post is the one I was addressing, and it has the stuff I took issue with. A proof by simulation, where infinite simulations would be as necessary as they are impossible, really isn't sufficient. Proof is important, and without it, I don't think that thread was resolved.

Edit: I was referring to your second post in that thread, but that's because the first had even less to point at. You can say infinite attacks happen only under specific conditions all you like, but it's kinda meaningless without evidence backing it.

Double-edit: Also, gotta say, what this thread really means is that I had doubts about your results, due to that lack of proof, but now I agree with said results. So, in other words, I didn't agree then, and do agree now. Don't really see the problem you're having with the whole thing.

[Kelb_Panthera]

I really like that the first thing he does in a comparison of critical hits is is throw out strength modifier, the largest contributor to critical hit bonus damage.

Why did WotC or Paizo hire him for anything but lore-crafting again? That article is just made of utter fail.

[Jack_Simth]

I think there's an easy enough way to resolve this without screwing around with math at any high level. If we reduce the (1-n)/n in the denominator at all, then the resulting odds should be strictly greater than they are now. So, reduce that denominator "(1-n)" to any extent, and run the equation from there. The new denominator fraction, with the shrunken numerator, will go to zero, so the overall denominator will go to -1, and the numerator is still going to 0, because we haven't changed that. So, you get that the old result is less than or equal to the new result, which is equal to zero, and expected values cannot be negative so the probability is zero. I think that logic holds together. It's still weird that you get that uncomfortable 0/0 in the original equation, but what I've presented seems to serve as proof of its true value.

Division by 0 doesn't work that way. It doesn't matter that you canceled out the division by 0 at some point, it still produced an undefined result.

Spoiler: If that's a valid transition, it's simple to prove that 1=0

Show

Consider two non-zero numbers x and y such that
x = y.
Then x^2 = xy.
Subtract the same thing from both sides:
x^2 - y^2 = xy - y^2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.
Source, because I'm lazy

... and once you've proved 1=0, you can prove that anything equals anything by very easy addition or multiplication, like, oh, 4=2.
1=0 : Add one to both sides.
2=1 : Double both sides.
4=2 : Done.

The exactly 50% case is interesting, because there's a hole in the formula there. If the formula is 'true' for all other cases, then you can approximate the probability with infinitives, and there is going to be a "real probability" if you actually run the experiment, but the formula doesn't directly give you an answer.

[eggynack]

Division by 0 doesn't work that way. It doesn't matter that you canceled out the division by 0 at some point, it still produced an undefined result.

What? I didn't cancel anything out. I showed that there's an upper limit on the expected value of that situation, and in particular showed that said upper limit happens to be zero. I wasn't dealing directly with the 0/0 situation at all. I was modifying it in such a way that the 0/0 situation wouldn't appear, and then establishing a clear relationship between the original result and the modified result. I think that such manipulation is legal.

Edit: To the more central claim of n/0 being necessarily undefined, as I recall, that is frequently not the case when dealing with infinite limits. Which, of course, we are.

[ExLibrisMortis]

Edit: To the more central claim of n/0 being necessarily undefined, as I recall, that is frequently not the case when dealing with infinite limits. Which, of course, we are.

As I understand it, you are right; you can divide by a limit that approaches zero, because the limit is never actually zero. The result then approaches infinity, unless the numerator approaches zero faster than the denominator.

As far as I know (not all that far, but some distance beyond very close), dividing by actual zero, that is, the multiplicative inverse of the additive identity, is always nonexistent, except in the zero ring {0}, where it is 0, just like most anything in the zero ring.