D&D 101: A lesson in fun

[Faerl'Elghinn]

Umm, about the math ... you've already been proven wrong. A mathematician named Kenneth Arrow managed to come up with this theorem that's called Arrow's Theorem (what a coincidence!), you can argue with it if ya want, but I think he got the math equivalent of a Nobel Prize for it, so don't expect the professors to listen to you. Basicly, given that all units in a conflict are equivalent in effective strength, and it's combat until one side is eliminated, you can calculate the most likely number of survivors of the larger side by taking the square root of the difference of the squares of the number of the units. Notice it's multiplicative, rather than additive, in nature. So if 5 orcs take on 4 orcs with no strategic advantage for either side, 25 - 16 = 9, and the square root of 9 is 3, so the larger force will probably have 3 survivors. Double the 5 to 10, and it's 100-16=84, so instead of losing 2 the larger force will now probably not even lose one. A more direct and intuitive example for you: You have a gun. So does the other guy. You can shoot him at a 1/1 ratio. Now, his friend with a gun walks up. You now have one shot, and they have two, but they have one target and you have two. So you're four times as screwed, not in twice as much trouble.

See, the problem with this theory is that you're automatically assuming that the party is the larger force. According to the DMG, the EL of a group of monsters remains the same regardless of the number of party members. EL doesn't take into account the number of members included in an opposing force. Also, the theory you've presented doesn't take into account how skilled each of the arriving shooters is. If you're twice as quick and twice as accurate, then you're possibly no less screwed either way. Therefore, this theory is technically not applicable to the game without a complete rewrite of the rules.

 

[Wulf Ratbane]

Your complaints about EL are simply due to a misunderstanding of the EL system, which is exponential (technically, logarithmic), not linear. Doubling the number of enemies does not "double" the EL because the EL values are not "mapped" linearly 1 to 1.

If numbers confuse you, don't use them.

An EL that is equal to the party is a "Moderate Challenge." That's your starting point.

If you double the number of enemies, it becomes a "Difficult Challenge." (Not EL +2, not EL x2, just "Difficult.")

If you quadruple the number of enemies, it becomes a "Very Difficult Challenge."

Wulf

 

[EricNoah]

EL is meant to be a DM aid in preparing challenging encounters. It has no other bearing on the game. So if you think, from your experience, that the method of "calculating" EL is flawed then by all means ignore it or calculate some other way.

In my experience, twice as many monsters of a given CR doesn't mean twice as difficult. It often means "about half again as difficult." This, I think, has something to do with the all-or-nothing nature of the combat system (the creatures either hit you or they don't -- if one ogre is having a hard time hitting my AC, then 2 ogres will likewise still have trouble hitting my AC).

I must add that I certainly haven't sat down and calculated the odds of this or that happening.

 

[Faerl'Elghinn]

Your complaints about EL are simply due to a misunderstanding of the EL system, which is exponential (technically, logarithmic), not linear. Doubling the number of enemies does not "double" the EL because the EL values are not "mapped" linearly 1 to 1.

If numbers confuse you, don't use them.

An EL that is equal to the party is a "Moderate Challenge." That's your starting point.

If you double the number of enemies, it becomes a "Difficult Challenge." (Not EL +2, not EL x2, just "Difficult.")

If you quadruple the number of enemies, it becomes a "Very Difficult Challenge."

Wulf

And your argument is a misunderstanding of my argument. My argument is that the system used is not the correct one, regardless of whether or not it involves logarhithms. Your argument makes absolutely no sense whatsoever. How can it be that 3 Ogres are only twice as challenging to one character as one Ogre is to one character? Therein lies the flaw.

 

[Thotas]

Important part of the theory: equivalent in effective strength. And we all know how rare that is. The number of units comparison is only one factor in what actually happens on the battlefield, as we all know. Obviously, if the 100 are 1st level kobold warriors and the 3 are Dire Bears, some kobolds will probably die. How many depends on terrain, who gets surprise, whether the kobolds try to retreat, do the kobolds have missile weapons, have they set pit traps, etc. Throw in sorcerers with area effects into an example, and things get even more complicated. There's more involved than numbers on each side, obvious without being a professional mathematician. But Arrow's Theorem does handle that single aspect of the issue accurately. So 2 kobolds, 2 dire lions, 2 medusa, 2 spectres are 4 times as deadly as a single specimen in any case (assuming they have the same number of hit points and the same equipment, feats, whatever). And for the fellow who said that those of us addressing your math errors (which you directly challenged us to do) are missing the point about your roleplaying comments, I'm in 100% agreement with you on those.

 

[Wulf Ratbane]

And your argument is a misunderstanding of my argument. My argument is that the system used is not the correct one, regardless of whether or not it involves logahithms. Your argument makes absolutely no sense whatsoever. How can it be that 3 Ogres are only twice as challenging to one character as one Ogre is to one character? Therein lies the flaw.

As I said, if the concept of doubling confounds you, then excise the concept of doubling from your mind and start over.

If you can accept the premise that one Ogre is a Moderate challenge, you should be able to accept the premise that two Ogres are a Difficult challenge, and four Ogres are a Very Difficult challenge.

Is this approach flawed? The language is different but the underlying premise is the same.

You'll also note that one Ogre is worth 1x XP, two Ogres are worth 2x XP, and four Ogres are worth 4x XP.

So... How is this a problem?

Wulf

 

[Faerl'Elghinn]

As I said, if the concept of doubling confounds you, then excise the concept of doubling from your mind and start over.

If you can accept the premise that one Ogre is a Moderate challenge, you should be able to accept the premise that two Ogres are a Difficult challenge, and four Ogres are a Very Difficult challenge.

Is this approach flawed? The language is different but the underlying premise is the same.

You'll also note that one Ogre is worth 1x XP, two Ogres are worth 2x XP, and four Ogres are worth 4x XP.

So... How is this a problem?

Wulf

The problem really comes in at higher levels. Say a party of four 10th-level PCs comes across 2 Tendriculi. Round 1 goes like this: party is surprised due to the difficulty of recognizing the nature of the creature, party member 1 is swallowed whole, party member 2 is swallowed whole. Round 2: either the remaining members flee, or 2 hits are scored, party member 3 is swallowed whole, and party member 4 is swallowed whole. TPK... It's not slightly more difficult here (of course, the circumstances would raise the EL, but even so...). Another example: 4 7th-level characters come across 2 Hill Giants. Is it slightly more difficult to defeat 2 than 1? How are they not going to take twice as long to kill? How will the giants not get to attack twice as many times? They will. It is linear. That's my point.

 

[Wulf Ratbane]

It is linear. That's my point.

One Hill Giant is a Moderate Encounter.

Two Hill Giants is a Difficult Encounter.

Whether you believe it's linear, whether it is linear, doesn't matter, because the assessment isn't linear.

One is Moderate.

Two is Difficult.

Four is Very Difficult.

Your problem is that you feel that because the EL system applies "numbers" to relative difficulty, that you think there is an obvious (read: linear) numerical relationship between them.

You can call it EL+0. You can call it Moderate. You can call it "Code Green." You can call it whatever you want, but this encounter is worth x1 XP.

Now double the number of opponents. You can call it EL +2. You can call it Difficult. You can call it "Code Yellow." Doesn't matter whether this encounter is exactly "twice as difficult" or not, all that matters is that there are twice as many opponents, that the GM knows it is DIFFICULT, and it's worth twice as many XP.

Double the number again (x4 from the original). You're at EL+4. Very Difficult. Code Red. The GM knows this encounter will be Very Difficult, and it's worth 4x the XP.

Wulf

 

[Faerl'Elghinn]

Important part of the theory: equivalent in effective strength. And we all know how rare that is. The number of units comparison is only one factor in what actually happens on the battlefield, as we all know. Obviously, if the 100 are 1st level kobold warriors and the 3 are Dire Bears, some kobolds will probably die. How many depends on terrain, who gets surprise, whether the kobolds try to retreat, do the kobolds have missile weapons, have they set pit traps, etc. Throw in sorcerers with area effects into an example, and things get even more complicated. There's more involved than numbers on each side, obvious without being a professional mathematician. But Arrow's Theorem does handle that single aspect of the issue accurately. So 2 kobolds, 2 dire lions, 2 medusa, 2 spectres are 4 times as deadly as a single specimen in any case (assuming they have the same number of hit points and the same equipment, feats, whatever). And for the fellow who said that those of us addressing your math errors (which you directly challenged us to do) are missing the point about your roleplaying comments, I'm in 100% agreement with you on those.

So actually what you're arguing here is that 2 Ogres should be EL 12, as they are four times as deadly to 1 character as 1 Ogre. Yet the book says that 2 Ogres are only 1.66 times as deadly as one Ogre. How, then, does that theory disprove my argument that the EL system is flawed?

 

[Wulf Ratbane]

So actually what you're arguing here is that 2 Ogres should be EL 12, as they are four times as deadly to 1 character as 1 Ogre. Yet the book says that 2 Ogres are only 1.66 times as deadly as one Ogre. How, then, does that theory disprove my argument that the EL system is flawed?

You have GOT to stop looking at the EL system as a linear system.

An Ogre is CR3.

If there are four Ogres, that does NOT make them EL12. It makes them EL7.

Your inability to grasp the application of EL and what the specific numbers mean does not make it flawed.