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<blockquote data-quote="Pseudopsyche" data-source="post: 5943148" data-attributes="member: 54600">If n kobolds each have probability p of failing a save, then the number of kobolds who fail the save is drawn from a binomial distribution with parameters n and p. The average value for this random variable is n*p, so yes, if you had to guess how many of 40 kobolds fail if they each need a 14 to save, then 26 failures is generally the best guess.Personally, I wouldn't want to tell the player of the wizard that he or she will always do &quot;average damage&quot; against a large number of enemies. Just for kicks I wrote a quick Perl script that estimates the inverse cumulative distribution function for this distribution (using Monte Carlo sampling), to give you a sense of the variance. You could use the following table to simulate all 40 die rolls using a single d20 roll, assuming that each kobold fails on a 13 or lower. Here, the roll is interpreted as the player's &quot;attack roll&quot;, so higher die rolls lead to more kobolds failing their saves.1: 122: 213: 224: 235: 236: 247: 248: 259: 2510: 2611: 2612: 2613: 2714: 2715: 2816: 2817: 2918: 2919: 3020: 31[sblock][code]#!/usr/bin/perlmy $n = 40;&nbsp; &nbsp; # Number of saving throwsmy $p = 13/20; # Probability of FAILING the savemy $k = 100000; # Number of simulations of $n rolls eachsub simulate {&nbsp; my $num_saves_failed = 0;&nbsp; for (1..$n) {&nbsp; &nbsp; ++$num_saves_failed if $p &gt; rand;&nbsp; }&nbsp; return $num_saves_failed;}my [MENTION=6904]res[/MENTION]ults = sort {$a &lt;=&gt; $b} map simulate(), 1..$k;print &quot;$_: $results[($_ - 1) * $k / 20]\n&quot; for 1..20;[/code][/sblock]</blockquote>

[QUOTE="Pseudopsyche, post: 5943148, member: 54600"] If n kobolds each have probability p of failing a save, then the number of kobolds who fail the save is drawn from a binomial distribution with parameters n and p. The average value for this random variable is n*p, so yes, if you had to guess how many of 40 kobolds fail if they each need a 14 to save, then 26 failures is generally the best guess. Personally, I wouldn't want to tell the player of the wizard that he or she will always do "average damage" against a large number of enemies. Just for kicks I wrote a quick Perl script that estimates the inverse cumulative distribution function for this distribution (using Monte Carlo sampling), to give you a sense of the variance. You could use the following table to simulate all 40 die rolls using a single d20 roll, assuming that each kobold fails on a 13 or lower. Here, the roll is interpreted as the player's "attack roll", so higher die rolls lead to more kobolds failing their saves. 1: 12 2: 21 3: 22 4: 23 5: 23 6: 24 7: 24 8: 25 9: 25 10: 26 11: 26 12: 26 13: 27 14: 27 15: 28 16: 28 17: 29 18: 29 19: 30 20: 31 [sblock][code]#!/usr/bin/perl my $n = 40; # Number of saving throws my $p = 13/20; # Probability of FAILING the save my $k = 100000; # Number of simulations of $n rolls each sub simulate { my $num_saves_failed = 0; for (1..$n) { ++$num_saves_failed if $p > rand; } return $num_saves_failed; } my [MENTION=6904]res[/MENTION]ults = sort {$a <=> $b} map simulate(), 1..$k; print "$_: $results[($_ - 1) * $k / 20]\n" for 1..20;[/code][/sblock] [/QUOTE]

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