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Message: <blockquote data-quote="Cheiromancer" data-source="post: 3442328" data-attributes="member: 141">I'm been trying to figure out the mathematical basis behind Table 3-1: Encounter Numbers (DMG p. 49).  Here's what I have so far:It seems that there is a variable that can be calculated by figuring out how many creatures of various CRs there are in an encounter, and that this number can be used to determine the EL, and also average experience.  Call this number M (for monster).  It is defined as 2^(CR/2).  Or you can use the following table (the M values for odd numbered CRs are about 6% too high):[code]CR     M---------1       1.52       23       34       45       66       87      128      169      2410     3211     4812     6413     9614    12815    19216    25617    38418    51219    76820   1024[/code]Given an encounter with monsters of various CRs, add up the M values, and find a CR whose M value is close; that is the EL of the encounter.For example, if your encounter has 6 CR 4 creatures in it, look up the M value of CR 4 (which is 4) and multiply by 6 to get 24.  24 corresponds to a CR of 9, so this is an EL 9 encounter.  It should provide a moderate challenge (25% of resources used) for a 9th level party.  A CR 4 and a CR 7 is a CR 8 encounter (4 and 12 is 16, which is the M value for CR 8), and so on.To figure out the amount of resources used, divide the M values and square the result.  So if one CR 8 creature fights a group of four CR8 creatures, it will use up (16/64)[sup]2[/sup] = 1/16 (about 6.25%) of the larger group's resources.  This assumes no scissor-paper-stone is going on.  If one monster can fly and has ranged attacks, and the other can't and doesn't, then the results will probably be quite different!  Similarly for creatures with special attacks and/or immunities, etc.. But generally speaking it should be true.[sblock=reasoning]Think of two identical abstract monsters fighting another such monster.  Assume average damage per round, average hit points and no initiative- everything happens simultaneously.  The two do twice as much damage to the one as they get back, so by the time that the single monster dies, the monster it was hitting is down to half hit points, and the other monster is at full; considered as a whole, they have 75% of the hit points they started with.  In other words, 1/4 the resources are used up when it is a 2:1 matchup.  A similar reasoning indicates 1/16 for a 4:1 matchup, 1/9 for a 3:1 matchup and so on; generally the resources used are the square of the numbers.  Since M values are proportional to the number of monsters, the resources used are proportional to the square of the M values.[/sblock]This means that if a CR 10 monster uses up 25% of the resources of a 10th level party (with 4 characters), then the 10th level party must be a CR 12. (Whose M value is twice that of a CR 10 monster, and so the resources used are 1/2 squared = 1/4)  Which means that each party member is CR 8 (since four CR 8s is equivalent to a CR 12).  Similar results hold for other levels; CR is equal to character level (ECL, actually) minus 2.Which is kind of interesting.  You can verify that this method agrees with table 3-1, and it also agrees with the intuition that a PC of a given level is weaker than a monster of the same CR, but it doesn't agree with the rule (or shall we say guideline?) that the CR of a PC is given by character level.  I say it is character level -2.  (Or, rather, the M value is half what you would expect if CR = ECL)If we are talking about a character with NPC wealth (but PC classes) they'd do even worth.  And if they had NPC classes (Warrior, Adept, etc.) then it would be worse still.  The rule I'm referring to is for PCs with the recommended wealth.Thoughts?</blockquote>

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