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Message: <blockquote data-quote="Cheiromancer" data-source="post: 3429766" data-attributes="member: 141">The contretemps with Upper_Krust about the "new" formula relating CR to EL has drawn me back to the basics; the official rules for challenge ratings and encounter levels.  Specifically I'm been trying to figure out the mathematical basis behind Table 2-6: Experience Point Awards (Single Monster) (DMG p. 38) and Table 3-1: Encounter Numbers (DMG p. 49).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 CR4 creatures in it, look up the M value of CR4 (which is 4) and multiply by 6 to get 24.  24 corresponds to a CR of 9, so this is an EL 9 encounter.  To work out experience (like in table 2-6) divide the encounter's M value with the character's M value, and multiply by 300 times the character level.  For example, if an 11th level character (M=48) in a four-member party helps defeat an encounter whose CR is 13 (M=96), divide 96 by 48 to get 2, then multiply by 300 to get 600, and multiply by 11 to get 6,600.  Which is the number given by table 2-6. I don't claim any understanding of what M means, but if you manipulate it in the indicated ways, all the tables come out right.  (Well, except for fractional CR- but in an epic game I'm not going to worry about that.  Just group the fractions until you get a complete CR.)---Alright.  Now here's a major puzzle: if four identical characters (10th level fighter/rogues, say, conveniently named One, Two, Three and Four) are attacked by Five, they'll defeat their attacker without difficulty.  So what CR (and M value) should each of these characters have?  It doesn't seem like a moderate encounter (25% resources used), does it?It's easy to see that, terrain permitting, the party will get 4 attacks in on Five for every attack that Five makes.  In other words, if One and Five would take eight rounds to reduce each other to zero hit points, the party of One to Four will defeat Five in two rounds; in this time Five will have reduced One to 75% of his starting hit points.  Five will have lost all his hit points and the party will have lost 1/16 of their total hit points.  The 4:1 advantage has become a 16:1 advantage.  (I'm assuming fine granularity in attacks and average damage, etc.  There would be much more variance in a real combat, and initiatives etc. would play a role.)A similar argument will establish that a five person party will have a 25:1 advantage over an identical attacker, and, generally, a n person party will have an n[sup]2[/sup]:1 advantage.So what PC level attacker would a four person party have a 4:1 advantage against?  One that it will lose 25% of its hit points against? I think it would have to be one that is 2 levels higher than the party- in other words, the CR of a PC is level-2.Which means that the M value for a 10th level character is 16.  Four such characters would have an M value of 64; so a 10th level party is equivalent to a CR 12 monster: EL +2 (very difficult).  By reasons of symmetry, such a challenge should have a 50/50 chance of being a TPK.Incidentally, the fraction of resources used can be worked out by dividing the M scores and squaring.  If your 10th level party has an M score of 64 and you fight a creature whose M score is 48, you should use up (48/64)[sup]2[/sup] of your resources; 56% of them.  A tough fight, but winnable.All this assumes, of course, that the tables in the DMG are correct.  Which is doubtful.  If M is not an exponential curve, but something else, something that an exponential curve approximates, then these various numbers would have to be adjusted. More later.</blockquote>

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