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In D&D 3.5, how do you add creatures with different CR which are more than 2 CR apart?

The table in the DMG 3.5, p. 49, describes how to calculate the encounter level of a mixed pair creatures which are 2 CR apart. (CR 4 + CR 2 is CR 5, etc.)

Let's say a Harpy Archer (CR 11) and a Harpy (CR 4) get into a fight with the party.

Now, how do you calculate the encounter level? It's difficult to follow the table because there is nothing written about 11+4. I tried out various encounter calculators (for example d20srd.org's calculator) but just can't get a grasp on how they get the result.

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    \$\begingroup\$ Welcome! You can take the tour as an introduction to the site and check the help center for further guidance. Good luck and happy gaming! \$\endgroup\$ – Sdjz May 15 at 12:07
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When the DM adds a CR 4 creature to an encounter with a CR 11 creature, the CR 4 creature's presence is insignificant where the Encounter Level system is concerned. In other words, the DM can add to an encounter one or more creatures that possesses CRs 3 or more lower than a creature's challenge rating to an encounter and, until a certain threshold is reached, see the encounter's encounter level remain the same.

Defeating these "bonus monsters" may still see the PCs earn XP from them, and the PCs can still take their treasure; it's just that the DM just need not include the these bonus monsters in the calculation of an encounter's Encounter Level. The lone or handful of secondary opponents are considered so worthless next to the main opponent that the lone or handful of secondary opponents are, here, considered unimportant to the encounter with the far bigger bad. (By the way, that's pretty much all that Encounter Level is—a measuring device that tells the DM how tough an encounter may be. While EL may have some minor mechanical impact on treasure, even that's what CR is usually for. Really, CR does a lot of the game's heavy lifting.)

The Dungeon Master's Guide on Mixed Pair says

In general, you can treat a group of creatures as a single creature whose CR equals the group’s EL. For example, instead of having the PCs encounter one CR 4 creature (say, a brown bear), you could substitute two CR 2 creatures (a pair of black bears), whose EL together is 4. However, creatures whose CR is far below the party’s level often provide no challenge at all, so don’t substitute hordes of low-CR creatures for a single high-CR creature. (49)

In the question's example, an encounter with a CR 11 harpy archer is an EL 11 encounter, and an encounter with a CR 11 harpy archer and a CR 4 harpy is still an EL 11 encounter. It takes 5 or 6 CR 4 harpies to equal a virtual CR 9 creature—and adding that many CR 4 harpies to the encounter with the CR 11 harpy is what will put the encounter to a mixed pair of 11+9 (i.e. the harpy archer and her 5 or 6 harpy assistants) for an EL 12 encounter.

However, creating encounters remains at least as much art as science. Even if the DMG says that an encounter's Encounter Level is unaffected by the presence of some monsters, the DM should still use his best judgment and examine any such encounter—and, ideally, every encounter—in light of his experience, his players' skills, and their PCs' abilities. See, while the Encounter Level rules can be willfully exploited by a vicious DM, but I'm not sure why a DM would want to: The Encounter Level system's goal is not to be an excuse for serial TPKs but to help the DM determine how difficult an encounter may be and, in turn, make the game more fun for everyone.

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  • \$\begingroup\$ So, in the general sense, would you say the additional low CR creatures need to reach the lower CR indicated in the table to raise the encounter level? (not necessarily the difficulty of the encounter, just talking about the mechanical encounter level) \$\endgroup\$ – Xordak May 15 at 13:14
  • \$\begingroup\$ @Xordak Yes, I think that's exactly what I'd say. In fact, I think the paragraph beginning In the question's example… strongly alludes to that, but I can make that more explicit if you want. However—and while your comment's parenthetical mentions it, I want to keep it 100% clear—, Encounter Level is largely just a ruler; the only reason I can imagine to ever game that system is in an adversarial, rules-as-written campaign (which, by the way, is not a playstyle I endorse but that I also don't claim is in any way badwrongfun.) \$\endgroup\$ – Hey I Can Chan May 15 at 13:45
  • \$\begingroup\$ While However, creatures whose CR is far below the party’s level often provide no challenge at all answers the question about low-CR creatures (and should be bolded, IMO), adding high-CR creatures is also a challenge. \$\endgroup\$ – ShadowKras May 15 at 14:42
  • \$\begingroup\$ Example: If a CR 11 creature is APL+3 to a 8th-level party, adding a CR 6 creature should still modify the encounter, as they are only 2 CR (6) lower than APL (8). But CR 11 and CR 6 are 5 levels apart. \$\endgroup\$ – ShadowKras May 15 at 14:43
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A single Monster of vastly lower CR is not enough to affect the encounter level

This is already covered by Hey I can Chan's answer in detail so this answer will not be repeating that.

Instead, this answer merely explains how the d20srd calculator is doing things because I was curious about their methods and it turns out the code for the calculator is easily found:

The d20srd calculator is mostly just using the "2 creatures of CR X-2 equals 1 creature of CR X"

In order to have a mathematical way for calculating the encounter level of an arbitrary number of a mix of different CR creatures, the way the d20srd people did it was to simply base it on the rule from the DMG:

In general, if a creature’s Challenge Rating is two lower than a given Encounter Level, then two creatures of that kind equal an encounter of that Encounter Level

The only exception they are also using is for CR below 1:

Some monsters’ CRs are fractions. For instance, a single orc (CR 1/2) is not a good challenge even for a 1st-level party. This means that you should either calculate XP as if the orc were CR 1, then divide by 2, or treat each pair of orcs encountered as a CR 1 monster.

Namely, they are just considering that:

  • 2 creatures of CR 1/2 equals 1 creature of CR 1
  • 3 creatures of CR 1/3 equals 1 creature of CR 1
  • (...)
  • 10 creatures of CR 1/10 equals 1 creature of CR 1

Using these two rules, the calculator then lumps everything together by using the great power of Mathematics to approximate a final Encounter Level (explanation may be provided below in the future). I will say though that what they do makes a lot of sense from a mathematical standpoint (in my opinion) and they are faithfully following the above given rules and simply extrapolating from it.

One thing to note is that although the displayed calculator only allows for 6 different groups of monsters and up to 20 monsters per group, their chosen method is actually generic enough for any number of groups of any size.

Step by step process of how the calculation is done:

In case for some reason you wish to follow along their calculations, I will explain what their code is doing in a step by step way:

  1. Calculate the "Power Level" (PL) of all the monsters together.

The power level is simply given by:

  • For monsters of CR 1 or less, PL is just the CR
  • For monsters of CR 2 or more, PL is:

$$ PL = 2^{\frac{CR}{2}} $$

  1. For each group of monsters, multiply the number of monsters by their corresponding individual CR
  2. Add all the power levels of groups of monsters together to get the final total PL
  3. Calculate the final encounter level (EL) of this total PL given by:

    • If the total PL is less than 2, then the EL is simply this total PL
    • Otherwise, do the following calculation to get the EL:

$$ EL = 2*log_2(PL) $$

  1. The displayed final EL is rounded to the nearest integer (Rouded half up)

One interesting thing this method provides is it allows one to obtain non-integer Encounter Levels by skipping the final (rounding) step.

For your question's example:

1 monster of CR 11 and 1 monster of CR 4:

  1. Calculate individual PL

    • 1 monster of CR 11 has approximately 45.25 PL
    • 1 monster of CR 4 has 4 PL
  2. Already done since it's just 1 monster of each CR

  3. Total PL is approximately 49.25
  4. EL is approximately 11.24
  5. Final displayed EL is rounded to 11

As you can see, when following this method adding a CR 4 to a CR 11 encounter also does almost nothing to affect the encounter level.

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