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<blockquote data-quote="Crazy Jerome" data-source="post: 5942752" data-attributes="member: 54877">You could do the math each time, to get realistic variance. However, what I do is eyeball it, divy up the monsters in convenient group sizes, divide most of those up as keterys said by chance of failure or success, then roll for the rest.&nbsp;This is particularly handy because you are seldom going to get monsters in even units of 20. So say you have 36 kobolds. If they fail on a 1-13, say that 30 of them are so divided. That's about 20 that fail and 10 that succeed (e.g. 65% being so close to 2/3 as not to matter). Then roll the last 6 kobolds. &nbsp;For even less math, when you have an odd percentage (e.g. 55%, 65%, etc.), divide the monsters into groups of 10, plus leftovers, and roll for one in each group, plus roll for leftovers. Say 23 kobolds. They have this 65% chance of failure. So in each group of 10, 6 fail, 3 make it, and 1 rolls. Then you also roll for the 3 leftovers.&nbsp;That's 12 auto fails, 6 auto succeeds, and 5 that could go either way. So you'll have 6-11 kobolds succeed. Pick up 23 unbiased dice, roll them in 20 sets (i.e. roll all 23 and record number of successes, do that 20 times), and you'll stay within those bounds on most rolls--and most of the sets outside the bounds won't be far outside it. &nbsp;Once you get up over about 15-20 creatures, the chances that you get outside the bounds of the possible results rolling every last one of them, is usually fairly small. (It will vary, of course, with the exact divisions you use and how accurate your eyeballing is.)</blockquote>

[QUOTE="Crazy Jerome, post: 5942752, member: 54877"] You could do the math each time, to get realistic variance. However, what I do is eyeball it, divy up the monsters in convenient group sizes, divide [B]most[/B] of those up as keterys said by chance of failure or success, [B]then[/B] roll for the rest. This is particularly handy because you are seldom going to get monsters in even units of 20. So say you have 36 kobolds. If they fail on a 1-13, say that 30 of them are so divided. That's about 20 that fail and 10 that succeed (e.g. 65% being so close to 2/3 as not to matter). Then roll the last 6 kobolds. For even less math, when you have an odd percentage (e.g. 55%, 65%, etc.), divide the monsters into groups of 10, plus leftovers, and roll for [B]one[/B] in each group, plus roll for leftovers. Say 23 kobolds. They have this 65% chance of failure. So in each group of 10, 6 fail, 3 make it, and 1 rolls. Then you also roll for the 3 leftovers. That's 12 auto fails, 6 auto succeeds, and 5 that could go either way. So you'll have 6-11 kobolds succeed. Pick up 23 unbiased dice, roll them in 20 sets (i.e. roll all 23 and record number of successes, do that 20 times), and you'll stay within those bounds on most rolls--and most of the sets outside the bounds won't be far outside it. Once you get up over about 15-20 creatures, the chances that you get outside the bounds of the possible results rolling every last one of them, is usually fairly small. (It will vary, of course, with the exact divisions you use and how accurate your eyeballing is.) [/QUOTE]

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