Lightning Mace and Actual Infinity

[eggynack]

About six months ago, I was involved in a thread (http://www.giantitp.com/forums/showthread.php?485047-Does-Lightning-Mace-grant-you-infinite-attacks)where we were arguing about whether the whole lightning mace for a pair of extra attacks about half the time trick had a theoretical chance of reaching actual infinity. I'd thought the answer was no, and the thread just kinda faded away without any real answer. Getting to the point, now I have approximately six more months of math knowledge, and by chance I think I have a definitive answer to the question, courtesy of stochastics class, because, for some reason, we covered an example that looks nearly identical.

So, it turns out that there is no chance of getting infinite attacks if you critically hit 50% or less of the time. Anything greater than that, however, will have some chance of diverging to infinity. In particular, it looks like the formula that gives the exact odds of hitting infinite attacks, where n is your extra attack chance, and k is your starting number of attacks, is 1-((1-n)/n)^k. So, for the sake of argument, assume you have a 60% chance of getting an extra pair of attacks every attack, and you're starting with ten attacks (because we've run this sequence out a bit). Then, you'd get 1-(2/3)^10, which happens to be 1-.0173, or about a 98% chance of infinity. In other words, the chances are really good if you start with enough attacks, even if you're only a bit above 50%.

You can actually calculate what the odds are of hitting any number of attacks, though that takes the full formula, which is (((1-n)/n)^k -1)/(((1-n)/n)^x -1), where x is the goal number of attacks. From all this, it should be pretty clear why 50% really isn't enough, because it'd make that stated formula return zero, as well as this new one. Anyways, I dunno if this stuff is interesting to anyone else, but I think it's kinda cool that it's possible to reach something like true infinity with a combo, assuming you optimize it well enough. I also think it's cool that this math stuff is having a real life application of any sort. If anyone cares, I could probably post a proof of the formula, though I'm not all that great at this subject at the moment so it's not gonna be perfect. Or, better yet, check out page 44 on in this stochastics textbook (https://services.math.duke.edu/~rtd/EOSP/EOSP2E.pdf), with page 48 being the source of this formula, and pages before 44 being a whole bunch of background.

[Martimus Prime]

Every time I read about Lightning Maces and the tantalizing prospect of infinite melee attacks, I get the overpowering urge to write out that build involving the choker with the feat and some way to get pounce. I just wanna charge in, drum faster than Dragon Force on some goblin's skull, then disappear around the corner with that species-specific extra action.

[Chronikoce]

I enjoy this sort of thing though I tend to disapprove of the term "reach infinity".

Actual infinite is not something you can reach because it by definition represents a concept of a limitless set. In other words you could never reach infinite because whatever count you're up to can always be added to.

However, you could calculate the odds of attacking a functionally infinite number of times. To make the numbers easy let's say it takes you 1 second to roll the attack and percentile die once. Then if we assume that you always succeed at getting a bonus attack we can use the rough time scale for the heat death of the universe to determine how many rolls could be accrued in that time.

The heat death of the universe is on the order of 10^100 years or approximately 30^107 seconds.

So if we assume 1 roll per second and calculate the odds I think that's safe to say we've basically calculated the odds of an infinite number of attacks for all intents and purposes.

Using the equation you gave with 2 starting attacks, a 0.6 chance of getting another attack and a desired result of 30*10^107 attacks the result I got was 1. I assume I've reading the rounding limit of the program I used to calculate it though.

As a note I started browsing that textbook and enjoy that they choose N=100 for (9/10)^N where N was to be infinity.

[eggynack]

I enjoy this sort of thing though I tend to disapprove of the term "reach infinity".

Actual infinite is not something you can reach because it by definition represents a concept of a limitless set. In other words you could never reach infinite because whatever count you're up to can always be added to.

However, you could calculate the odds of attacking a functionally infinite number of times. To make the numbers easy let's say it takes you 1 second to roll the attack and percentile die once. Then if we assume that you always succeed at getting a bonus attack we can use the rough time scale for the heat death of the universe to determine how many rolls could be accrued in that time.

The heat death of the universe is on the order of 10^100 years or approximately 30^107 seconds.

So if we assume 1 roll per second and calculate the odds I think that's safe to say we've basically calculated the odds of an infinite number of attacks for all intents and purposes.
My language was imprecise, perhaps, but I was talking about real infinity. The more technical way to say it, I suppose, is that that's the probability that you'll get N attacks as N approaches infinity. And that sort of limit based approach is something you can definitely do in mathematics. Sure, you can't actually roll that number of dice, but I'm talking here about the theoretical number of attacks possible through the combo, leaving aside the realistic element. Because, assuming you have as much die rolling time as you want, you could indeed consider attack quantities as its own limitless set. And then, you could assess the probability that a specific sequence of attacks might touch on every element of that set, defined inductively by considering element n, which you're assuming to have reached, and then asking whether you'd also necessarily hit element n+1.

Also, the question of functionally infinite is kinda boring. The formula I gave will also solve for any functional infinity, but if the question were whether one could hit some arbitrary number of attacks, the answer would be yes, and I wouldn't care. In fact, one could even hit those "functional infinities" for really low critical hit chances. It'd be nigh impossible, but it'd happen. Real infinity tells us an important thing here. It tells us what the lower limit is on probabilities associated with these attack strings. Once you know that you have a 98.whatever chance of never ending, you also know you have at least a 98.whatever chance of reaching your heat death number. And also any other number you care to name. It's cool. Moreover, as a side benefit to the value, it's also easier to calculate. The possibly massive denominator exponent cancels out, and you get to just consider the relatively low numerator exponent.

So, yeah. I'ma stick with this answer.

Edit:

As a note I started browsing that textbook and enjoy that they choose N=100 for (9/10)^N where N was to be infinity.
That's not actually what's happening there. The 100 is the numerator exponent, which there is the starting cash on hand, and here would be the starting number of attacks. N is the goal quantity of attacks/cash, and you're considering the limit of the equation as N approaches actual infinity. When that happens, that denominator fraction approaches zero, so the denominator as a whole becomes -1. So, you multiply the numerator by -1, and get the end result formula I mentioned at the start. That that's what happens when you use infinity is, as I mentioned above, part of why you really want to be using actual infinity, and not some crazy number. Like, you're putting stuff into programs, and you could be putting it into simple calculators, or even solving by hand if you really have nothing better to do.

[Darrin]

I think I remember someone on the WotC boards calculating something like this, but it was way over my head. As I recall, they came at it from the other end, calculating the odds of rolling enough natural "1"s to stop the combo. And I think it does terminate, but the number of attacks was so mind-bogglingly high, it couldn't be expressed in typical scientific notation.

Or was that Tleilaxu Ghola's share pain combo? I don't recall.

[Hogsy]

I have no idea how to math but I can tell you that from a rules perspective, it is possible to roll a critical threat on a chance higher than 50%, although the chance of being allowed to use such a weapon is rather slim. There is a 3rd party Swashbuckler class in Pathfinder that grants you the same effect as Improved Critical, but it stacks with it! That would make an 18-20 weapon into 12-20. Further more, a path of war stalker with the steel serpent discipline and a teamwork feat (http://www.d20pfsrd.com/path-of-war/feats#TOC-Deadly-Pairing-Teamwork-) could get that down to 11-20. In my games, I personally allow improved critical to stack with Keen, so that's down to 8-20. Add the serrated weapon property, and you have an effective critical threat range of 7-20. Isn't that good enough for you formula? Granted, it is far from legal when it comes to official settings. Honestly, the idea alone that you could somehow do something like this is the reason I love to min/max. You notice stuff like this and it just makes your day to think what if.

In pathfinder, you can actually reach a critical threat range of 13-20, and if you use the 3rd party swashbuckler you can go down to 10-20. Finally, you may be able to squeeze in the serrated property for a 9-20 and call it quasi-legal or something. Full BAB and two weapon fighting nets you a total of 7 attacks, 8 with haste. Be a monk, create a monk weapon with 18-20 critical threat range to get it down to that tasty 12-20(no swashbuckler for you) and go home. You can use the unchained flurry of blows with Ascetic Style(be a master of many styles to also use dragon ferocity for the extra swag) to get it up to 9 attacks. I am 80% you could use the feat "Lightning Fists" to gain 2 extra attacks with your unarmed strikes, which would be your monk weapons at this point, at a -5 penalty, for a total of 11 attacks. Add in the usual AoO shenanigans with karmic strike & robilar's gambit and even if you run out of attacks at your turn, you'll get to continue your streak whenever an enemy attacks you. Be right back, I gotta go vomit.

PS: I love you for this thread.

[Quertus]

Thanks for posting the proof equation. Kudos on being a fellow seeker of truth, and not limiting yourself to things that support your position!

So, could you do me a small favor? Could you baby step us through using that formula for an example or two? I see someone already misread how it works, and, honestly, for a formula I didn't personally derive, I probably would, too.

Edit: a relevant, infinite crit example, please.

[Hogsy]

This post is the reason we need a thumbs up option.

[Chronikoce]

My bad! I was trying to understand the proof at 3 am after about 5 minutes of reading and it seems I messed up.

[AvatarVecna]

Lightning Maces on its own can't realistically deliver infinite extra attacks; every time you attack, you have a 100% chance of having 1 less attack left than before you attacked, and you have a less than 100% chance of gaining an extra attack. The way the math works out, you will never make an attack and have more attacks remaining than before you made the attack, meaning the series will eventually end. Now, the higher your crit threat range, the more attacks you can expect out of this. After running through a few examples, the most basic formula to determine this is that every attack made will generate an average number of bonus attacks equal to the ratio between the {number of die results that threaten a critical} and {the number of die results that do not threaten a critical. For example, if you only crit on a 20, each attack will generate an average of ~.0523615789 attacks (1/19); if you crit on a 17-20, each attack will generate an average of .25 attacks (4/16); if you crit on 11-20, each attack will generate an average of 1 attack (10/10); if you crit on 3-20, each attack will generate an average of 9 attacks (18/2).

Now, getting basically anything higher than a 19-20 with Lightning Maces requires shenanigans with Aptitude weapons (allowing you to apply weapon-specific feats to the aptitude weapon even if the aptitude weapon isn't the weapon that feat is for; whether this legally allows you to use Lightning Maces in conjunction with, say, Keen kukris, is a matter of controversy). To get an actually infinite series of bonus attacks--that is to say, where an attack will on average generate more attacks remaining than it uses up--you need to combine this Aptitude weapon cheese with Roundabout Kick (a feat that generates an extra attack on a critical hit). If you can pump both your critical threat range and your attack bonus high enough, this will generate 2 attacks for just about every attack taken...resulting in a truly infinite series of attacks (well, until you decide to stop making attack rolls, such as when there's nothing left to attack).

Presuming that Aptitude weapons work with both Lightning Maces and Roundabout Kick in this manner, this method can be used for any weapon as long as you can crank that crit threat range into something ridiculous (getting lots of attacks helps as well). The most ridiculous example I've seen proposed was a dragonwrought kobold using such a combo with a thrown weapon that had a crit range around 8-20, and used that thrown weapon in conjunction with the epic feat Distant Shot to utterly destroy everything within line of sight in a single round. I think they used a throwing rock, purely so they could call it a martial-exclusive "rocks fall, everyone dies" option.

[eggynack]

I think I remember someone on the WotC boards calculating something like this, but it was way over my head. As I recall, they came at it from the other end, calculating the odds of rolling enough natural "1"s to stop the combo. And I think it does terminate, but the number of attacks was so mind-bogglingly high, it couldn't be expressed in typical scientific notation.

Neat benefit to this formula is that it's pretty easy.

I have no idea how to math but I can tell you that from a rules perspective, it is possible to roll a critical threat on a chance higher than 50%, although the chance of being allowed to use such a weapon is rather slim.
As I recall, most lightning mace builds terminate right at 50%, which means they fail to go infinite. However, I think there are some builds utilizing some 3.0 stuff that might be illegal now that do pass that threshold.



So, could you do me a small favor? Could you baby step us through using that formula for an example or two? I see someone already misread how it works, and, honestly, for a formula I didn't personally derive, I probably would, too.
The version with infinity is pretty straightforward. If we assume a percent chance of success of .55, then you take the chance of failure, .45, and, divide it by that chance of success. So, .45/.55. Then, you raise that to the initial number of attacks, say five, which gives about .37. You subtract that from 1, and you get about .63.

My bad! I was trying to understand the proof at 3 am after about 5 minutes of reading and it seems I messed up.
Fair enough.

Lightning Maces on its own can't realistically deliver infinite extra attacks; every time you attack, you have a 100% chance of having 1 less attack left than before you attacked, and you have a less than 100% chance of gaining an extra attack..
Yeah, that was some imprecision again. I tend to mean the version of the combo that gets two attacks with every success when I talk about it, but yes, just maces would fail at this.

Anyways, I'm glad people like the applications of high level math in a D&D context. I'm personally happy that some of this stuff had some immediate real world application, as opposed to the usual application of supporting more math.

[Name1]

As I recall, most lightning mace builds terminate right at 50%, which means they fail to go infinite. However, I think there are some builds utilizing some 3.0 stuff that might be illegal now that do pass that threshold.

The build is created by combining these with a Stump Knife and the unupdated 3.0 Class Disciple of Dispater. Every attack after the first, the Stump Knife has a crit-range of 9-20, thus allowing for a lot of attacks.

[MHCD]

"Gun Fu" from the Optimization Showcase series by Tempest et al is relevant here. Of course the goal of that build was not to reach truly infinite attacks, but using the information here, it could likely be modified to do so. Also, ranged weapons allow for better use of arbitrarily high numbers of attacks.

[icefractal]

At first I was thinking that actually infinite vs NI was just a curiosity on this, but then I realized it does have an effect - it means that Lightning Mace can slice through any arbitrarily large but non-infinite hp (such as someone using Permanent Emanation: Consumptive Field)..

[AvatarVecna]

Also, regardless of whether you're using Lightning Maces+Aptitude Weapon or Lightning Maces+Roundabout Kick+Aptitude Weapon, you'll probably find both the "Blood In The Water" stance (continuous self-stacking +1 to attack/damage for each critical hit, resets after a minute with no crits) and the "Choose Destiny" spell (9th lvl Destiny Domain, 1 round/lvl, 2d20b1 for all d20 rolls) useful; Blood In The Water guarantees your accuracy, while Choose Destiny both makes nat 1s basically impossible as well as drastically increasing your chances of getting a critical.

[Quertus]

The version with infinity is pretty straightforward. If we assume a percent chance of success of .55, then you take the chance of failure, .55, and, divide it by that chance of success. So, .45/.55. Then, you raise that to the initial number of attacks, say five, which gives about .37. You subtract that from 1, and you get about .63.

Ah, got it. Thanks. Without parenthesis to show the order, I would have done it wrong. (I tend to use a lot of parenthesis when I code, to make sure order of operations is explicit).

[ekarney]

I know there was a 2 - 20 crit range, build that wasn't 100% legal, but only used 1st party material.
Though I feel like when we're talking about extended crit ranges, we should bring up Sean K. Reynold's rant on the topic. (http://www.seankreynolds.com/rpgfiles/rants/keenimprovedcritstacking.html)

The problem being, that the PrC used for crit builds was updated* making this much tougher.


*replaced with a much worse, totally different class solely to make crit builds unusable.

However, if we discount that WotC are jerks who hate fun and present ourselves with the hypothetical situation where the PrC wasn't updated, leaving us with a 2- 20 range. I believe (someone cite me on this please) there was either a feat for class feature that allowed one to not automatically miss on a 1. Meaning that if we can get our range from 1 - 20, and find a monster with AC = (1 + Build's attack bonus) we can "infinite lightning mace"

[Zanos]

Though I feel like when we're talking about extended crit ranges, we should bring up Sean K. Reynold's rant on the topic. (http://www.seankreynolds.com/rpgfiles/rants/keenimprovedcritstacking.html)
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.

[eggynack]

However, if we discount that WotC are jerks who hate fun and present ourselves with the hypothetical situation where the PrC wasn't updated, leaving us with a 2- 20 range. I believe (someone cite me on this please) there was either a feat for class feature that allowed one to not automatically miss on a 1. Meaning that if we can get our range from 1 - 20, and find a monster with AC = (1 + Build's attack bonus) we can "infinite lightning mace"
One cool thing about the formula is that it implies that you don't need nearly that much to go infinite. My initial post was assuming you hit on a 9, and the next assumed you hit on a 10, with the first using 10 attacks and the next 5. In both scenarios, you're going to diverge to infinity over half the time. An interesting aspect of this is that you can run your formula at any point in the chain, with the new information derived from your gained attacks. That count of ten, for example, was assuming you've been going for a bit and you want to know if you're going to terminate from there. The point of that being, while your chances of termination may be higher near the beginning, eventually you're just not going to stop. You get ten extra attacks, now pushing something like 15, and the value of the exponentiated fraction goes super high, and apart from critting at lower than 11, it doesn't matter much that your range goes super wide. And, take note, even in the worst case scenario, your chance of diverging is decent. You get 1-(45/55), which assumes you crit on a 10, and that you start with a single attack, and you have an 18% chance of never ever stopping. Isn't that crazy?

In fact, it looks a lot like crit range, beyond the necessary minimum, is way less critical a factor. The crit range, after all, merely defines what's getting an exponent applied to it. The real important factor seems to be the number of initial attacks, which is probably easier to boost. A neat implication of that, meanwhile, is that the odds of going infinite are really similar to the odds of hitting, say, 100. After all, the odds of getting to infinity from 100 are going to be ludicrously high. In fact, that's the scenario described in the book itself. This, in turn, means that a decently high number is a good model for infinity, and, even more critically, given the ease of use, that infinity is a good model for any decently high number. So, load up on attacks, and things should diverge well enough.

As a side note, while the infinity formula always gives zero if you have odds of 50% or less (actually, I think it gives negatives, but whatever), the larger formula does work to calculate the odds of hitting a given number of attacks from any starting number of attacks. So, if anyone's using a less pushed lightning mace build, then that's the thing to use.

[ExLibrisMortis]

A very cool result, thanks for posting. Mathematics #1 science, yo!

That S. K. Reynold post is funny. And a bit stupid.

[MaxiDuRaritry]

I do believe the notation is slightly off.

To avoid confusion, it should probably be 1-((1-n)/n)^k, otherwise the standard order of operations leaves you with 1-(1-1)^k = 1 (since n/n = 1).

[eggynack]

I do believe the notation is slightly off.

To avoid confusion, it should probably be 1-((1-n)/n)^k, otherwise the standard order of operations leaves you with 1-(1-1)^k = 1 (since n/n = 1).
Yeah, I think I have it right now.

[Hogsy]

I know there was a 2 - 20 crit range, build that wasn't 100% legal, but only used 1st party material.
Though I feel like when we're talking about extended crit ranges, we should bring up Sean K. Reynold's rant on the topic. (http://www.seankreynolds.com/rpgfiles/rants/keenimprovedcritstacking.html)

The problem being, that the PrC used for crit builds was updated* making this much tougher.


*replaced with a much worse, totally different class solely to make crit builds unusable.

However, if we discount that WotC are jerks who hate fun and present ourselves with the hypothetical situation where the PrC wasn't updated, leaving us with a 2- 20 range. I believe (someone cite me on this please) there was either a feat for class feature that allowed one to not automatically miss on a 1. Meaning that if we can get our range from 1 - 20, and find a monster with AC = (1 + Build's attack bonus) we can "infinite lightning mace"


But it doesn't matter if you roll a natural 1, if your critical threat range is 1-20. Lightning mace merely requires you to roll a critical threat range, not succeed one. Thus , even on a nat 1, you'd get an additional attack if your critical threat range was 1.

[AvatarVecna]

But it doesn't matter if you roll a natural 1, if your critical threat range is 1-20. Lightning mace merely requires you to roll a critical threat range, not succeed one. Thus , even on a nat 1, you'd get an additional attack if your critical threat range was 1.


Whenever you roll a threat on an attack roll while using a light mace in each hand, you gain an additional attack at the same attack bonus.

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


Increased Threat Range
Sometimes your threat range is greater than 20. That is, you can score a threat on a lower number. In such cases, a roll of lower than 20 is not an automatic hit. Any attack roll that doesn’t result in a hit is not a threat.

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

[Quertus]

However, if we discount that WotC are jerks who hate fun and present ourselves with the hypothetical situation where the PrC wasn't updated, leaving us with a 2- 20 range. I believe (someone cite me on this please) there was either a feat for class feature that allowed one to not automatically miss on a 1. Meaning that if we can get our range from 1 - 20, and find a monster with AC = (1 + Build's attack bonus) we can "infinite lightning mace"

Did you miss the math that indicated that we don't need anywhere near these numbers to go infinite?

[fallensavior]

I'm pretty sure you will either run out of targets or get kicked for wasting time doing zero damage before getting anywhere close to infinity.

[martixy]

the thread just kinda faded away without any real answer.

I'm sorry, but what?

I thought I gave a pretty clear and definite answer.
With way simpler math to boot.

[eggynack]

I'm pretty sure you will either run out of targets or get kicked for wasting time doing zero damage before getting anywhere close to infinity.
Perhaps, but as I've noted, the probability of this combo reaching infinity has implications on the combo's behavior at non-infinite regions. And the overall formula happens to tell you exactly how often you're going to reach a non-infinite goal, if you'd rather know an exact chance rather than a lower bound. Also, things going infinite is intrinsically cool. We're always seeking out those abilities that loop upon themselves, and we can ask of those abilities whether they reach a true infinity or only give ridiculously large results. At first glance, one might assume that lightning maces falls in the latter category, but it turns out that you're getting that true infinity surprisingly often. And that's neat. Sure, this kinda thing hangs out in the more theoretical domain, but it's cool to think about. And, even if you mostly lack targets, you could always aim downwards and start exploding the earth. That's the kinda target where you really do want true infinity in reach.

Edit:
I'm sorry, but what?

I thought I gave a pretty clear and definite answer.
With way simpler math to boot.
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.

[Heliomance]

This (http://www.giantitp.com/forums/showthread.php?132804-Highest-Possible-Crit-Range) is the original thread on the subject, back in 2009. I believe the build that we ended up with there had a 5/9 chance to go infinite, which was proven amidst much arguing about the definition and nature of infinity, and debates over whether it was possible to have a chance of going infinite that is not either 0 or 1 (spoiler: the answer was yes).

[eggynack]

This (http://www.giantitp.com/forums/showthread.php?132804-Highest-Possible-Crit-Range) is the original thread on the subject, back in 2009. I believe the build that we ended up with there had a 5/9 chance to go infinite, which was proven amidst much arguing about the definition and nature of infinity, and debates over whether it was possible to have a chance of going infinite that is not either 0 or 1 (spoiler: the answer was yes).
I'ma read that, cause it seems interesting, but it looks like that poster was wrong if I'm right (and I think I am right, because it's not me that's saying it but a textbook about this exact problem). The build in question is critting on a 6 or higher, which means 75% chance of two attacks, and it's starting with a single attack. By my formula, that means 1-(.25/.75), which is clearly 2/3, or, y'know, 6/9, higher than the claimed chance.

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.

[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... (((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 (https://en.wikipedia.org/wiki/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 (https://en.wikipedia.org/wiki/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 (https://en.wikipedia.org/wiki/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... (((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/showsinglepost.php?p=20674428&postcount=19

Not sure what you were looking at.

[eggynack]

This post is not wrong:
http://www.giantitp.com/forums/showthread.php?485047-Does-Lightning-Mace-grant-you-infinite-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.
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 (https://www.math.hmc.edu/funfacts/ffiles/10001.1-8.shtml)


... 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.