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Message: <blockquote data-quote="Wulf Ratbane" data-source="post: 2240893" data-attributes="member: 94">If you have a 20% chance to hit, you will hit 20% of the time, on average. If you take a d20 and roll 1000 times, an impartial, scientific observer will reasonably expect 200 of those rolls (20%) to be 17 or better.Can you guarantee it? Of course not. Is it possible you might roll a natural-1 a thousand times in a row? Sure. Is it even possible you could, as you say, "roll forever and never hit that 20%"? Uhh, yeah, it's possible.But it's not remotely probable. The most likely result is that you will roll 17 or better almost exactly 20% of the time (approaching exactly 20% as the number of iterations approaches infinity).An intelligent player will play the law of averages in order to maximize his odds.If you have a 20% chance to hit, it means your damage over multiple iterations (ie, on average) is 20% of your expected damage. So if your damage is, for example, X, then your average damage on any given attack where you have a 20% chance to hit is (.20)(X).Why? 80% of the time my damage is zero (because I missed). 20% of the time my damage is X (because I hit). For any single attack roll, when you add up all the expected damage for every possible number I could roll between 1 and 20, it's:0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0+X+X+X+X.So the average is 4X/20, or (.2)(X), or 20% of X. (Remember, X is my weapon damage.)(This is no different than determining that the average on a d6 is 3.5-- you add up all the possible results from 1 to 6, then divide by the total number of possibilities: (1+2+3+4+5+6)/6 = 3.5. Just trying to keep you on the same page.)I'll continue and hope I haven't lost you yet:And your expected, average damage when you have a 10% chance to hit is (.10)(X).So, your expected, average damage when you have TWO rolls with a 10% chance to hit is (.10)(X) + (.10)(X).(Just as the average of 2d6 =  3.5 + 3.5 = 7.)Now let's try to show you how this would work using a very specific example. Let's say you have a magic bow that always does 5 points of damage per hit. Let's furthermore say that you hit on a 2 or better. (Your archer is awesome.) That's 95% of the time.If you roll a single attack, your average damage is 4.75 (.95*5).If you roll two attacks at -2, now needing a 4 or better to hit, your average damage is 8.5 (.85*5 + .85*5).If you were to continue comparing results all the way up to "I need a 17 or better to hit," then rolling two attacks is ALWAYS better, on the average.At the point where you need a 17 to hit, the results areSingle Attack: (.20)(5) = average 1 damage.Double Attack: (.10)(5) + (.10)(5) = average 1 damage.So if you need a 17 to hit, it's statistically a dead heat whether you are better off with one attack or two.At the point where you need an 18 to hit, the results areSingle Attack: (.15)(5) = average .75 damage.Double Attack: (.05)(5) + (.05)(5) = average .5 damage.This is the point where it's no longer mathematically wise to make two attacks, because one attack has a better AVERAGE damage output than two attacks.As a gut reaction, I can't fault, "I'd rather have one good solid hit than a dozen maybes." I'm down with that sentiment. But that's just not the case here.I'm trying to show you that the -2 penalty does NOT result in "a dozen maybes." The drop in probability, and the commensurate drop in average damage output, is not as bad as you believe it is. It's just ignorant to continue to hold onto that fallacy.</blockquote>

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