D&D 4E The Quadratic Problem—Speculations on 4e

[GoodKingJayIII]

I would like to contribute nothing to the thread by saying that my algebra days are long behind me and most of this went over my head, but it's terribly interesting.

 

[Cheiromancer]

I can't possibly imagine that WOTC intends for every fight to be a 50/50 coin flip, so it's safe to assume that an Ogre4 isn't meant to "equal" a Character4. He'll be "exponentially" weaker, so as to make for a "typical" ("tough but reasonable") encounter. So it may well be that the Ogre4 is best compared to a Character6 or Character8.

Ooh. Good point. I worked out somewhere that a character's CR (assuming an exponential system) is two less than his level. (In this thread.) Assuming standard equipment and everything. So yeah, it should be a 6th level fighter for a 50% chance. And the 20th level fighter only has to fight 2 Ogre16s.

If the party gets to gang up on the Ogre4, they should be able to finish it pretty fast, due to Lanchester's square. If it's a face to face exchange of blows between one fighter and one ogre the result will be quite different.

 

[Wulf Ratbane]

I would like to contribute nothing to the thread by saying that my algebra days are long behind me and most of this went over my head, but it's terribly interesting.

No problem, Jay-- at the very least, interest fuels the thread.

Cheiro-- I added a second edit to my post above. Not sure if it will provoke a continuation thought on your part.

 

[mmadsen]

I can't possibly imagine that WOTC intends for every fight to be a 50/50 coin flip, so it's safe to assume that an Ogre4 isn't meant to "equal" a Character4. He'll be "exponentially" weaker, so as to make for a "typical" ("tough but reasonable") encounter. So it may well be that the Ogre4 is best compared to a Character6 or Character8.

I don't think anyone's arguing that the designers want every fight to be a coin flip; they're simply changing their nomenclature.

Currently, a creature is CR-N if it's as powerful as an Nth-level character, and one CR-N creature comprises an EL-N encounter, which is an appropriate challenge for four Nth-level characters.

If we wanted to re-normalize CR and EL so that CR-N and EL-N implied that four monsters were an appropriate challenge (and not a fair fight) for four Nth-level characters, we'd simply subtract four from all CRs and ELs, since the current system adds two to EL for each doubling in numbers.

 

[Malhost Zormaeril]

Without better equipment, a Fighter improves his to-hit probability by a factor of approximately 1.1, from level to level.

Please correct me if I'm wrong, but a Fighter improves his to-hit probability by +1 per level. That would imply a logarithmic function, wouldn't it?

d[BAB]/d[Lvl] = 1/[Lvl] ==> [BAB] = ln [Lvl] + C

(People who never took Calculus, please ignore the preceding paragraph)

Similarly, HP increase also logarithmically (assuming an average roll of 5.5 for the Fighter) as [Hp] = (1/5.5) ln [Lvl] + C

It's much harder to predict how AC and damage scale, since they're a function of equipment, but as a rule, without magic or feats, it's strictly a constant (once you are wearing full plate and wielding greatswords there's nowhere to go). So we'd need to analyse how many pluses we access by level. I'm going to try to compile a list once I have access to my DMG.

I don't really know which conclusions to draw, but I just wanted to point this out for now to see if someone spots some immediate nonsense. Once I think more about this, I may have more to post.

 

[BryonD]

See below

 

[Nonlethal Force]

d[BAB]/d[Lvl] = 1/[Lvl] ==> [BAB] = ln [Lvl] + C

(People who never took Calculus, please ignore the preceding paragraph)

Kudos on bringing Calculus into this discussion. However, the numbers don't seem to support your conclusion.

[Sblock=Stat block hid for reason of conserving space]

Level | Natural Log
1     | 0
2     | 0.7
3     | 1.1
4     | 1.3

[/Sblock]

As you can see, there is no constant C that would attribute to the difference between the actual value for a fighter's BAB and ln [level].

As an aside, once the given equation is discovered, it would be the same for cleric's progression and wizard's progression. The only thing would change is that instead of a 1, a cleric would have 3/4, and instead of 1 the wizard would have 1/2. Since they are constants, though, the calculus behind the equation would remain the same and the resulting function would be the same ... just following different constants. In other words, the graph would be an identical function, it would just have a different beginning location and a slightly adjusted slope. But the function would remain the same.

EDIT: Actually, thinking about it, I think Caluculus can be used to show the relationship. But, I think the relationship is more like this:

d[BAB] / d[Lvl] = 1

Thus, the integration equation is d[BAB] = 1(d[Lvl]. That would imply that:

BAB = LVL + C, with C being equal to zero in this case.

The reason that the equation is equal to 1 instead of 1/level is that a fighter's base attack increases a full 1 with each level.

EDIT2: Sorry, I don't mean for the prior edit to state what BryonD posts below. i got an emergency phone call and my edit took longer than I thought.

 

[BryonD]

Kudos on bringing Calculus into this discussion. However, the numbers don't seem to support your conclusion.

Good catch.

The mistake is that d[BAB]/d[L] = 1. It is NOT 1/[L]. (Which would result in a decreasing rate of BAB gain, as your numbers show.)

The "per level" part of the equation is already in the left hand side.

 

[Cheiromancer]

No problem, Jay-- at the very least, interest fuels the thread.

Cheiro-- I added a second edit to my post above. Not sure if it will provoke a continuation thought on your part.

Thanks for pointing it out. I wouldn't have noticed it, otherwise (the new page makes it easy for edits to slip by).

I think I made a mathematical error, though, when I was setting the benchmarks. (Easy to do in these kinds of calculations!). If 2 Ogre4s are going to have a 50/50 shot against an Ogre6, an Ogre6 has to be 4 times the power of an Ogre4. That follows from Lanchester's square (he has to kill twice as many opponents will withstanding two attackers). So the Ogre16 has to be 12 quadruplings of power over the Ogre4; not 12 doublings. That's quite a beast.

However, under these assumptions a 6th level fighter should have a 50/50 chance against an Ogre4. So a 20th level fighter should have a 50/50 chance against an Ogre18. I have to run, but if no-one has done it by the time I come back I'll stat out an Ogre18. It should be quite a monster.

My argument is that a character's CR is their level minus 2. Four characters will be the equivalent of a monster 4 higher (two doublings at +2 CR per doubling). And so a party of four 4th level characters should have a power of 4 times that of a CR 4 monster. In other words a CR 4 monster should use up 25% of their resources.

That's assuming that the exponential power curve is a strict law. Actual characters and monsters will vary.

 

[mmadsen]

Please correct me if I'm wrong, but a Fighter improves his to-hit probability by +1 per level. That would imply a logarithmic function, wouldn't it?

d[BAB]/d[Lvl] = 1/[Lvl] ==> [BAB] = ln [Lvl] + C

I'm afraid I don't follow. BAB increases linearly with level: +1 BAB/level.

Probability of hitting depends on the target AC, which is a moving target. If we only compare nearby levels -- e.g. Ftr1 vs. Ftr2 -- then to-hit probability increases by a factor of approximately 1.1 -- e.g. from 50 percent to 55 percent.

If we assume our target is always an Orc War1 with AC 13, then to-hit probabilities will go from, say, 40 percent, to 45, 50, 55, and so on, up to 95 percent. That's obviously linear with level, to a point. As with hit points, the ratio of one level's to-hit probability to the previous level's drops from level to level: 1.12, 1.11, 1.10, 1.09, and so on.