I’ve written on this topic before, but only as part of a larger subject. But I recently had occasion to expand the EL-CR chart that I created for my own use, and decided to share it with our readers – and have written this Blog Post to explain what it is and how I use it.
This is a chart that I use all the time. It is an extremely unusual D&D game session if I don’t refer to it at least once. You can download the 3-page PDF from here. Just right-click on the icon to save the file.
Let’s start by examining the chart itself. At first glance, it’s fairly straightforward; we have CR down one side and EL across the top. The meat of the table are the cell contents, and closer study reveals some interesting (and useful) patterns. But let’s not get ahead of ourselves, here.
Each Row indexes results for a given CR or character level. The number shown in a cell is the number of creatures of that CR that are needed to achieve the EL shown at the top of the column.
This gives a completely different structure to that of the table on page 49 of the DMG, even though they are talking about the same thing. The official table has Number Of Creatures across the top, EL down the side, and CR in the cell contents. It looks like this:
Both tables are talking about the same thing, but I find the layout that I have used to be more useful, for reasons that will become clear as this article proceeds, and I have extended the table much further than the official one.
The Relationship between columns
The columns index results for a given EL. The DMG states that if a certain number of creatures give a particular EL, twice as many creatures will give an EL that is two higher than the first. In other words,
If EL(N) = N creatures at CR(X), then EL(N+2) = 2 x N creatures at CR(X).
Mathematically, that in turn means that
EL(N+1) = Sqr Root (2) x N creatures at CR(X) i.e. = 1.414 x N creatures at CR(X).
That is the foundation apon which the table has been built.
The Cell Contents
When I created the table, I started with the values given in the DMG’s much smaller table. I used the values given and the relationship defined above to extrapolate the table both across and down.
When you examine the actual table, though, there are a couple of aparrant anomalies, and those bear further examination.
What’s with the three “one’s” in a row?
If you examine the table on page 49 of the DMG, you’ll find the same thing, at least at higher EL entries. The first one is actually 1/1.4142 or 0.707, rounded off; the second is a legitimate “1″; and the third is actually 1.4142 rounded. Most of the time, the middle “1″ is the “1″ that you will use, but there are times when all three of them are valid.
The Extreme Right
The relationship is made even more explicit on the far right-hand side of the table. When the numbers started growing too large to fit conveniently into the table, I started stating them as a multiple of the last two numbers explicitly stated. So, looking at the top line (CR1 creatures) we have an entry that reads “2 x EL17″, followed by “2 x EL18″, “4 x EL17″, “4 x EL18″, and then “8 x EL17″, and so on.
Looking down the table shows some interesting patterns in this area as well – there are two entries that say the same thing, followed by two more at +2 EL. Look down the EL24 column, for example. When I first noticed this, it was so unexpected that I had to double-check and triple-check it, but it is correct.
The Pattern Of Repetition
Of course, the most obvious pattern is that each line is exactly the same as the line above it, but shifted one space to the right.
These patterns make it easy to extend the table as necessary. Cell 21, 21 is the same as Cell 1,1. Nevertheless, the table that I have prepared anticipates future needs by going all the way up to EL75 – I am running campaigns at the moment that are expected to head into Epic territory!
Definition of a fair fight
The objective of the table is to enable the GM to construct encounters that represent a fight that is fair, and that permits the characters to receive rewards commensurate with the difficulty of the fight. If the characters are outmatched, they receive more rewards for success; if the enemy is outmatched, the characters should recieve a smaller reward. The success or failure of this system hinges on a GM being able to recognise fights that are fair, and judging the encounters he creates relative to that standard. That’s what this table is there to do.
So just what is a “fair fight”? Once that is defined, the relationships within the table do the rest, so it’s important to get this right. In my book, 4 characters of level X verses four more characters of identical level qualifies; which means that a fair fight is one in which there is (in theory) a 50/50 chance of the characters losing.
In practice, the chance of losing is a lot less. Players have a lot of time to plan their level progressions and feat choices, to optimise their character designs, and are focussed on a single PC at a time; the GM has to create his oppositions on the fly, or close to it, and – in that ‘fair fight’ situation, has to operate and coordinate four characters at once, as well as handling all the other administrative and judgemental requirements of the position. Depending on how much prep time and experience the GM has, the player’s chance of victory in a ‘fair fight’ might be 90% or more.
But it is not fair to the players to take that into account, because it’s a moving goal post. Sometimes, the GM will get the tacticals right, and the chances of PC victory will swing in the other direction, given the advantages that he can confer. The EL-CR chart is just a starting point.
The same thing measured the same way
Okay, so if four PCs of level X are in a fair fight when opposed by four NPCs of identical level, we need some mechanism of translating everything into a character-level equivalence. That, in theory, is the CR, or the character levels plus any racial level adjustment of the enemies opposing the PCs. This is a contentious issue, as revealed in my previous post A Slippery Slope: Level Adjustments Under The Microscope. I don’t intend to go over the ground of that post again; suffice it to say that you need some mechanism by which you can take an encounter of nominated CR and convert it into a character-level equivalent. What the process of achieving that equality might be is beyond the scope of this article.
What Is An EL?
Here, once again, I have to skirt contentious territory. An EL is a measure of how effective a combined force of characters is in battle. I use the term, and calculate the measure, in the same way for both “allies” and “enemies” in a battle. The concept is that you can (theoretically) combine the effectiveness of individuals of different power levels in order to determine the overall effectiveness of the group.
In other words, equal EL values on both sides should give us our hypothetical fair fight.
The purpose of the EL-CR table is to take individual power levels (CR) and facilitate the measurement of the overall power level (EL). We can then use this as a foundation point for creating battles that should test the PCs.
A gang of four
For some reason – a subject for a completely different blog post sometime (and duly noted as such) – a standard party is generally assumed to consist of four members. So let’s start examining how the EL-CR table works by looking at the simplest such case: four characters of the same level.
To determine the EL of the resulting party, find the row that matches the levels of the characters, then cross until you reach the cell with “4″ in it (the number of characters of that level), then go up the table to read off the EL from the heading of the column.
The example shown demonstrates that for four characters of level 3 (ie CR 3), the EL is 7. And, in fact, if you examine the example more closely, you will see that each +1 to the level of the four characters also raises the EL by 1 – so the EL for four characters of level 2 is 6, and of four characters of level 4 is 8, and for four characters of level 16 it will be 20.
In fact, the EL of any group of 4 characters of equal level is always 4+level. Not that this should be a surprise, since we defined the progression as 2xL = EL+2, back in the blue box above! Obviously then, 4xL will be EL+2+2, or EL+4.
With A Ringer
The same example shows quite clealy the effect of adding a 5th character of the same level, simply by following the same procedure as above. A five characters of level 3 takes us to the cell just past the one with the “4″ in it, into the cell which reads 5-6; and going up to get the EL shows us that the result is EL8. In fact, it should be slightly below EL8, while 6 would be slightly above, but the numbers are so close that they both round to the same thing.
Gaps In The Table
Another thing that you might notice after examining the example above is that there are gaps in the table. What are 13 characters of 1st level worth as an EL, for example? The cell that gives EL 8 reads “10, 11, 12″ and the cell for EL 9 reads ’14 to 18′! And, if you can make it out, the cell for EL 10 reads “20-24″ – so you might be asking what I have against the numbers 13, and 19, and – for that matter – 25, 37, 38, 39, and 49 to 55 – to name just a few.
The answer is that I have nothing against these numbers of foes, but they fall into regions where the EL is not quite right either way. This table was derived from the official one in the DMG given above, and represents a statistical analysis of these numbers of characters – and some numbers are neither one nor the other.
I could have revised the table to cover the entire field, growing the boundaries of each area enough that there are no gaps, but I decided not to do that; instead, I offer three alternative policies for the GM to use in this situation:
- use the average of ELs that fall in between, effectively defining an ad-hoc EL of “Eight-and-a-half” or whatever; or,
- use the higher number for lower-level characters and the higher one for higher level ones in the same party (this is getting into more complicated territory that will be addressed in subsequent sections of the article); or,
- use the higher of the two numbers and then drop the CR of one of the creatures by 1. This is a fudge, but it’s not far off the mark.
Or, of course, you can always contrive never to use an encounter that falls into the gap – but I don’t believe in the limits of the system confining a referee’s options to that extent.
A more complicated question
It’s actually a fairly rare situation, especially when rating encounters that the GM has created to oppose the players, for every character in a group to have exactly the same CR. In fact, it’s often not the case even for PC parties. So what’s the procedure for more complicated mixes – say, a group consisting of three 2nd-level characters, plus one 3rd level character, plus two fourth level characters, plus a fifth level character?
Now that we’re getting into more complicated situations, the first thing to address is the way that these are described. Actually spelling something out – like our example question – takes up far too much space and takes too long. I long-ago started using the “@” symbol to mean “at” and leaving out the word level; to me, and to most people, the meaning of the result is immediatly obvious. But there were people who were confused by it the last time I abbreviated things in that way, so I thought I would take a moment to explain it.
Our example problem can be stated:
3 @ 2 + 1 @ 3 + 2 @ 4 + 1 @ 5 = ?.
If you don’t follow that, remember that “@” means “at”, and just read it left to right. You will end up with something that is remarkably similar to the long form used at the end of the previous section.
Lowest Common Denomenator
Okay, so on to the actual methodology of using the EL-CR table to analyze this compound force. The first step is to identify a common denomenator. It’s usually most convenient to actually use the Lowest common denomenator, but some people panic when that term is employed.
In this case, it’s just a fancy name for the lowest CR in the compound group. In this case, the lowest CR of any member of our force is 2.
Translating to the common denomenator
All we have to do is go down the left-hand column to the CR of each item on the list, go across until we find the appropriate number of characters of that level, then go up the table until we get to the line that describes our common denomenator. Then write down the number in the cell.
Taking our example item by item:
3 @ 2: Since CR 2 is our common denomenator, we can simply write down “3″.
1 @ 3: Go down the table rows to CR3, then across until you get to a “1″. Problem: there are three of them in a row: which one do we use? If I have the time, I use all three, and write the result as a range; if I need a quick and dirty solution, use the middle one. Then go up one cell and write down the numbers in the cell belonging to the “CR 2″ row. The first of the three ones gives a “1″; the second also gives a “1″; and the third gives a “2″. So I would usually write 1-2, but if in a hurry, I would use a 1.
2 @ 4: This is actually the item that I chose to illustrate this process. Go down to CR4 and across until you find a two and then up to the row that describes CR2, as shown, and you will find that the number in the cell is “4″.
1 @ 5: Again we encounter the “three one’s”. The first gives a “2″ on the CR2 row, the second gives a “3″, and the third gives a “4″. So I would normally use 2-4, but if I was in a hurry, I would just use 3.
All this can be written far more succinctly:
3 @ 2 = 3 @ 2
1 @ 3 = 1-2 @ 2
2 @ 4 = 4 @ 2
1 @ 5 = 2-4 @ 2
Mixing the compound score
The next step is to add up the results. If any of your numbers shows a range, your result should also be a range. So:
3 +1 +4 +2 = 10; and 3 +2 +4 +4 = 13. The result is a range of 10-13 – at “CR2″.
In other words, this combined force is the equivalent of 10-13 second-level characters.
A measure of efficacy
The final step is to locate the cell on the table that best describes that result and go up the column to the EL of the compound force. While there is no entry for 10-13, there is an entry for 10-12 characters of CR 2 – so our example has an EL of 9. That means that anything in the EL 9 column would be a fair fight for this particular combination of characters or creatures – and this combination would be a fair fight for anything else in this column.
That might be a single foe of CR 8 to 10, or two foes of CR 7, or three foes of CR 6, or four of CR 5, or 5-6 of CR 4, or 7-9 of CR 3, or 10-12 of CR 2 (by definition), or 14-18 of CR1.
Or even a different compound that converts to a lowest denomenator matching one of the EL9 entries.
More Difficult Conversions – Small CRs
Some creatures have CRs of less than 1 – there are examples with CR 1/2, CR 1/4, and CR 1/10th. To use these low CRs, simply multiply the numbers in the “1″ column by the number under the “/”.
An EL 9 encounter – to continue with the example – consists of 14-18 CR1 creatures, as stated above; or 28-36 creatures of CR 1/2; or 56-72 creatures of CR 1/4; or 140-180 creatures of CR 1/10th.
More Difficult Conversions – bridging Totals
There will be occasions when the total you come up with will bridge a gap in the table.
The combination of 4 @ CR 6 + 1 @ CR 9 gives a total of 6-8 characters at CR 6, for example. That bridges two EL columns – there is an entry for 5-6, and an entry for 7-9, giving EL 11 or 12.
There are a couple of different ways to handle this.
- One is to reduce the size of the span by “1″ on each side of the range until you get a single answer. So if we have a range of 6-8, adding 1 to the 6 and subtracting 1 from the 8 would give us a smaller range of 7 – which puts the result firmly in the EL 12 column.
- Another is to simply find the mid-point of the range by averaging the two extremes – the results are the same in this case.
- I don’t like either of these approaches; there is too much risk that the average will land in a gap in the table. I have found that the range of ELs result is actually an accurate reflection of something more important: the sensitivity of the result to the power of the single CR 9 party member. It’s his combat effectiveness that is actually clouding the issue, and that reflects the fact that character prowess equivalence at a given class level is only an aproximation. If the CR9 in our example is ever-so-slightly better in combat than most, the party combination is equivalent to 8 characters at 6th level; if the character is ever-so-slightly less than normal at level 9, the combination is equivalent to 6 characters of CR6, and there is a middle ground.
The best approach is to take the result literally – hit the opposition with a force composed of 6 characters of CR 6, and have two more (or equivalents) “in reserve” that can show up in the middle of the fight if the opposition seem to be winning too easily.
This not only preserves the accuracy of the results, it makes the battle feel more real; instead of being a set challenge, with fixed force strengths, it makes combat more dynamic.
In effect, this is actually a consequence of the rounding errors that produced the “three one’s” anomaly that was discussed earlier. If the CR9 participant is slightly less combat-effective, he only counts as 0.7 of a typical CR9; and if he is slightly more combat-effective, he actually counts as 1.4 of a typical CR9.
What’s more, this variability is not fixed; it will vary from encounter to encounter, depending on the specifics. If the targets have a resistance of some sort to the best attack of the CR9, he will be equivalent to the “0.7 of a CR9″ character; if they are more vulnerable to it, he will be equivalent to the “1.4 of a CR9″. An example of the latter is Undead vs a CR9 cleric.
Judging this ahead of time is where the artistry of encounter design remains significant. It might seem that this table reduces encounter design to a science – it doesn’t.
More Difficult Conversions – Spanning Totals
It’s also possible to achieve totals that span three EL columns.
This usually only occurs in one of three circumstances:
- There is too big a difference between the common denomenator and the most powerful combatant;
- There are low numbers of characters, so that the “three ones” have scope to have a big impact;
- There are a lot of creatures, so that the spread of ranged results becomes very broad.
The best solution to this situation depends on which of these causes is responsible, so the first step in resolving this situation in identifying the cause.
Spanning Totals due to CR difference
I’m including this cause as it is one that many people who have seen the table seem to think is real, even though I will prove that it doesn’t actually exist. To illustrate this problem, let’s say that you are designing an encounter group that consists of 1@CR2 + 2@CR6 + 1@CR9 + 1@CR11.
1 @ CR 2 = 1 @ CR 2
2 @ CR 6 = 7-9 @ CR 2
1 @ CR 9 = 7-18 @ CR 2
1 @ CR 11 = 14-36 @ CR 2
totals = 29-64 @ CR 2 = EL 12-14.
If you actually work through the process of getting the above results using the table, you will find that the real problem isn’t the CR difference; it’s the number of “1 @” in the encounter group. To prove this, let’s ignore the lone CR 2 character and rework the problem:
2 @ CR 6 = 2 @ CR 6
1 @ CR 9 = 2-4 @ CR 6
1 @ CR 11 = 4-9 @ CR 6
total = 8-15 @ CR 6 = EL 12-14.
Heck, we can even ignore the CR 6 duo and get almost the same answer:
1 @ CR 9 = 1 @ CR 9
1 @ CR 11 = 1-4 @ CR 9
total = 2-5 @ CR 9 = EL 11-14.
These two top-heavy (in CR terms) characters are causing the spread, for the reasons previously described. The presence of the lower-level party members is actually helping to stabalise the result.
Cause #1 of the spanning problem doesn’t actually exist – it’s an illusion, another case of Cause #2. Whatever solution works for that cause will also work here.
Spanning Totals due to the “three ones”
This is, as demonstrated above, a real issue. It’s also one that has been addressed previously, in the discussion on “bridging totals”.
Statistically, the more characters you have, the more these little variations will even out. That’s why there would be no issue if the compound was 5@CR6 + 3@CR9. But when we’re dealing with small numbers, like we are in a 2@CR6 + 1@CR9 + 1@CR11 – which is effectively the example mooted in the previous section – individual variations have too much scope. And there is too wide a gap to be handled by using “reserves” in the way I proposed in the section on “bridging totals” – the “reserves” would be greater in number than the main encounter. How many of them do you use at once? Far from creating additional realism within the combat, it just starts to get tiresome and dreary. “Another patrol of 15 Kobolds springs from the side tunnel.” “Again? This is the 5th patrol of reinforcements, enough already! Besides, I thought Dave blocked that side tunnel last round.”
The solution in this case is to strip away the rounding errors and go with your best guess as to where the two high-level characters will fall on the 0.7-1.4 continuum. You have three choices: 0.7, 1, and 1.4.
Once you know that, you will cut down the variability of the results enormously, and get back to a practical result.
Let’s say, for example, that the CR 9 member of the encounter group is a combat monster, easily able to trounce characters of his own level, while the CR 11 member is more typical of his class level. In this case, you would use the highest of the three “one’s” for the CR 9 and the middle one for the CR 11.
Instead of (reproduced from the earlier section),
2 @ CR 6 = 2 @ CR 6
1 @ CR 9 = 2-4 @ CR 6
1 @ CR 11 = 4-9 @ CR 6
total = 8-15 @ CR 6 = EL 12-14,
2 @ CR 6 = 2 @ CR 6
>1 @ CR 9 = 4 @ CR 6
=1 @ CR 11 = 5-6 @ CR 6
total = 11-12 @ CR 6 = EL 13.
Note the use of the equals sign to indicate the middle choice and the > sign to indicate the high choice in this example. Or, if it was the CR 11 who was the combat monster, and the CR 9 who was typical:
2 @ CR 6 = 2 @ CR 6
=1 @ CR 9 = 3 @ CR 6
>1 @ CR 11 = 7-9 @ CR 6
total = 12-14 @ CR 6 = CR 13-14, a result which bridges a gap in the table. In accordance with the procedure outlined previously, I would use CR 13, but hold an additional 2 CR 6 opponants in reserve to us in the event that the fight proved too easy.
This last example shows why it is not a good approach to simply “pick the middle one” when you get a result spanning three (or more, theoretically) ELs.
Okay, so let’s say we have a party of 4 characters, all of CR 6. That gives them an EL of 10, which defines a whole column of opponants that would provide a fair fight. We can go up an EL for a harder fight, or down one for an easier fight.
The big problem is that all the opposition combatants have to be the same level – if there are 4 of them giving an EL of 9 (easy fight), each has a CR of 5 – you can find that out by finding the EL column that you want, going down until you get to 4 enemies, then tracking across to get the CR, as shown in the example above.
But how do you work backwards from an EL to get a complex compound of opposition that still represents a fair fight?
There are two approaches: the high CR approach and the low-CR approach. I’ll demonstrate and explain each one seperately; I use both of them regularly.
A third approach is the brute-force approach, where you start with something that sounds about right and tweak it until it adds up to the EL that you want. That’s a lot of work, which is is what these procedures aim to avoid, so I don’t recommend it.
Working Backwards using the High-CR Approach
In this approach, the GM defines equivalences, splitting the encounter into equal shares of difficulty and then determining what CRs and how many enemies at those CRs are needed to achieve the total EL. This approach is somewhat easier but less flexible than the Low-CR approach that follows.
How many shares?
The first step is to decide how many equal shares the required EL is going to be split into. The most common values are 2, 3, or 4. Another good choice is to make the division by the number of characters on the side of the “allies”, ie the group that this opposition will be fighting – that usually means PCs and their allies. In theory, that means that each PC will have an equal opportunity to contribute; using a lower number means that one or more members of the opposition will be fighting multiple members of the allies. This creates an opportunity for the PCs to employ tactics to their benefit. A higher number means that each member of the allies will be faced with multiple enemies, possibly giving the opposition a tactical advantage at least initially.
For an example, I’ll use three, and work with a goal of EL10.
An even split
The next step is to break out the EL-CR chart, locate the EL10 column, and find the entry for 3. That row gives the maximum CR of a single member of the opposition force.
EL10 = 3 @ CR 7.
The first equivalence
So the first part of our compound of enemies is 1 @ CR 7. The others are all at a lower CR than this. The next step is to define the first equivalence – that is, a CR that is less than the maximum. It’s easiest to drop the CR by a multiple of two, so for this example I’ll drop it by 3, and say that our second teir of opposition will be a number of CR 7-3=4 enemies that are equal in power to a single CR 7.
Since a single CR 7, by definition, is the same thing as EL 7, all I have to do is go to the EL 7 column and go down until I get to the row that matches CR 4. The value in the cell will give us our second line of the compound. The answer to this is three, so the second line of our compound reads “3 @ CR 4″.
In the case of this example, the next equivalence is also the last equivalence. If I were dividing the total EL into 4 or more pieces, there would be another one to follow. I use a different procedure for the final equivalence, so the information in this section will be theoretical.
We now have a maximum CR of 3, because this subgroup of opposition are defined as having less personal power than those of the previous tier, which we set at CR 4. For the sake of example, I’ll choose CR 2.
Again, all I have to do is work down the EL 7 column until I get to the CR 2 row and read off the number of enemies. The table shows 5-6, so I can choose either 5 or 6 enemies at CR 2 to be the third tier. If I had chosen CR 1, the answer would have been 7-9, and I would probably have chosen 8 as a nice intermediate number, or I could have shaded the difficulty one way or the other ever-so-slightly by using one of the more extreme values. Another option would have been to choose 7 as the number of characters and keep two CR2′s in reserve; they probably wouldn’t make that big a difference to the battle, but coming in once the party are weakened by the battle, they could have an impact.
The final equivalence
There is an inherant fuzzyness about the results that are achieved during the equivalence stages. In theory, that fuzzyness gets smaller with each step, but the final equivalence is obviously the last opportunity to get exactly the total effectiveness that is desired.
That calls for a slightly different approach. Instead of looking at achieving an equivalence to the other elements of the compound, the technique is to determine the total effectiveness needed at our minimum CR to get the total effectiveness we need.
Let’s say that the minimum CR that we’re talking about is CR 1. According to the process detailed for intermediate equivalences, above, which does not allow for the fuzzyness of results, we would get a total of 7-9 characters @ CR 1. Let’s see what we get when use the final equivalence approach.
I start by looking up the value of our total EL at the target CR. I then subtract the values of each of the intermediate stages already defined, also at the CR 1 value. The result is the precise number of CR 1 characters needed to achieve the target.
EL 10 gives 20-24 CR1 characters.
1 @ CR 7 is equivalent to 5-12 characters at CR 1. This is a spanning total, which means that we need to refine the CR 7 to just one of the three “ones” available. In this case, let’s assume that we’re talking about a typical CR 7; that means that we choose the middle “one” and get:
=1 @ CR 7 = 7-9 @ CR1.
Next, we have 3 @ CR 4. So we go across the CR 4 line until we get to the cell with a “three” in it, then go up to get the CR 1 equivalent.
3 @ CR 4 = 7-9 @ CR1 – which we should have expected.
So, subtracting 7-9, twice, from 20-24, will give us our answer – well, almost. Do we subtract low from low and high from high, or high from low and low from high? Is the low value of our results range 20-7-7 or 20-9-9? The first gives a smaller range, the second gives a broader one.
Ultimately, it doesn’t matter; one (20-9-9 & 24-7-7) will reduce the size of the initial force, but increase the number of reinforcements; the other (20-7-7 & 24-9-9) will reduce the size of the reinforcements, but the encounter will start with more low-level participants present. I use both, jusging the results by the specific circumstances – if those circumstances make it more appropriate for few low-level flunkies to be present at the start of combat, I use the second choice, otherwise I use the first. And if I can’t make up my mind, I split the difference: 20-8-8 and 24-8-8.
Just to complete the example, let’s assume that I chose the first of the options listed; the range of results is, therefore, 20-9-9 to 24-7-7, or 2 to 10 @ CR 1. That’s a very different number to the result that was derived from the intermediate methodology, i.e. 7-9. And its all due to evening out the fuzzyness.
At the end of the high-CR approach, we end up with:
total = EL 10 = 1 @ CR 7 + 3 @ CR 4 + 2 @ CR 1, + reinforcements 8 @ CR 1.
This would work well to describe a villain, three lieutenants, and a couple of bodyguards/servants, plus eight enforcers in the next room that will join in the fight part-way through the fight.
Working Backwards using the Low CR Approach
The other way to go about it is to work from the bottom up. In effect, to use the circumstances described previously, this methodology asks “how many flunkies?”, then “how much is left for lieutenants?”, and finally, “how much is left for the boss?”.
The first step to employing this approach is to decide how low the CR of the flunkies is going to be. The combination of our EL and that CR will give the basic parameters against which everything else will be measured.
EL 10 = 20-24 @ CR 1.
If CR 1 is our minimum EL, then 20-24 is our fundamental. We then get to split that up into the ratios that best express our goal, using the standard ranges given on the CR1 line.
Let’s say that we dedicate 14 of that 20-24 to flunkies. That leaves 6-10 for everything else. Of that 6-10, we can put 2 of them toward a lieutenant – just one of them – leaving 4-8 for our head villain of the encounter.
In this step, the allocations are converted into actual encounter elements. I have reproduced the table extract again so that you can follow along.
We don’t have to convert the flunkies, since they are automatically at the level we defined.
So we start with the Lieutenants. Going across the CR1 row (because that’s what our fundamental was measured as), we find the cell that contains a “2″ – it’s the EL3 column. We then go down the column until we find an entry with a “1″ in it – there will be three of them in a row. We can choose any one of the three that we want, but whichever one we choose, we next have to choose the opposite. If there is no “next time” to even things out, we have to choose the middle.
Let’s choose the first one (ie the lower CR) for the lieutenant, so that we’ve got a stronger main bad guy. The CR that is the first to show a “1″ in the EL3 column is CR2.
Next, we repeat the process with the head villain. Immediatly, we are confronted with a problem: a range of 4-to-8 spans three columns, there’s a four, a five-to-six, and a seven-eight-nine. Because we chose a low value for our lieutenant, we need to even that out by choosing the higher value on our next choice. so we choose the seven-eight-nine cell, which is in the EL7 column. We then drop down until we get to the cells with “ones” in them. Since we have no choices to even out, we have to pick the middle option – and that gets us CR 7.
So our total using this approach is: 14@CR1 + 1@CR2 + 1@CR7.
The strength of our main bad guy hasn’t changed, but we’ve lost one lieutenant and weakened the other in return for more flunkies.
What if we had wanted more than 1 lieutenant? If we had wanted two, then we would have dropped down to the cell that contained a “2″ instead of one of the cells that contained a 1 when converting the lieutenant – and that would have given us a CR of 1 for the lieutenants, because we hadn’t allocated enough power to them, we had it all in the flunkies and the big bad guy.
Why is this more flexible?
The first system is simpler because it equates every subgroup within the group to the same effective power level. This approach is more flexible, and often less work, but requires the GM to make decisions about the makeup of the opposing force without being entirely sure of what the consequences will be; the consequence of the choices made in the example is a lieutenant that is not much better than the flunkies.
Of course, if you don’t like a result, you can always change your allocations with the benefit of hindsight. Dropping the head bad guy’s allocation to 5-6 would have left more room for the lieutenants to be reasonably powerful. Then you select the number of lieutenants so that individually, they are weaker than the head villain, even if the combination is more powerful.
You can alter the combination any way that you want to. That makes it more flexible – and more decision-driven. If it’s an easy answer you want, stick with the High CR approach.
It is possible to further refine the encounters produced by the system. Giving one side a tactical advantage could be reflected by increasing the EL of that group of combatants. A character whose abilities are more at home in an urban environment may get an effective CR increase in such an environment and a reduction in a more natural setting – and vice versa. These refinements are appropriate because the objective is always to define “a fair fight”, and these circumstances alter the combat effectiveness of the group.
Not every combat needs to be a “fair fight”. Some should be easier, some more difficult. Knowing what constitutes an even fight gives an invaluable baseline from which to work, nothing more. So how far should GMs generally range from the “fair fight” value?
To some extent, this will vary with circumstances. Hitting the PCs with even a moderately difficult encounter may be extremely difficult for them if they have been weakened by prior battles and given no chance to recover. A succession of moderately-difficult encounters leading up to a major encounter can add to the difficulty of that major encounter by consuming resources. There remains an element of artistry in the way one encounter is connected to another which no analysis tool can replace.
I consider it quite routine to hit the PCs with an encounter that is 1 EL higher or lower than the “fair fight”.
It is less common but still not exceptional to employ encounters that are 3 ELs higher or lower than the “fair fight” value.
It is quite rare but still not out of the question to employ encounters that are 5 ELs removed from the “fair fight” standard. However, I would not do so without building a substantial advantage into the weaker side or a substantial disadvantage into the stronger; an anarchic situation in which there is conflict or even actual battle amongst members of the opposition, so that the PCs are not expected to win the fight outright but only overcome whatever is left after the major enemies have fought it out, for example.
Circumstances would have to be extraordinary before I went further than this; and there would be many warnings delivered along the way. Nevertheless, overconfidence may lead to really extreme situations, in which case, let the chips fall where they may!
Let’s be clear about what these ranges actually mean: if a “fair fight” is EL 10, consisting of 10-12 foes at CR 3:
- EL 9-11 ( = 7-18 foes at CR 3) is routine;
- EL 7-8 ( = 4-6 foes at CR 3) or EL 12-13 ( = 20-36 foes at CR 3) are acceptable;
- EL 5-6 ( = 2-3 foes at CR 3) or EL 14-15 ( = 40-72 foes at CR 3) are rare but also acceptable, if the weaker side is strengthened or vice-versa.
The extremes range from roughly 18% of a fair fight to 6 times a fair fight. That’s a fairly wide span!
Even when considering such extreme ranges, the information given by the EL-CR chart is useful, giving some indication of how much of an advantage should be given to bring the results back to an acceptable challenge.
So that’s how I use the EL-CR Chart. And now you can, too.