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Skill challenges in 5e - Math help!
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<blockquote data-quote="pemerton" data-source="post: 6364807" data-attributes="member: 42582"><p>Permutations and combinations. It's actually quite painful, at least by hand and using the resources of a high school maths students. (Those who have done higher study may have learned some shortcuts?)</p><p></p><p>Your chance of success, at +2 vs DC 15, is 2 in 5 (ie need a 13+ on d20).</p><p></p><p>The chances of 7 successes in a row is 2^7 over 78125, or 128 in 78125.</p><p></p><p>The chance of 7 successes and 1 failure is 2^7 * 3 [for the failure] * 7 [because there are 7 "slots" into which the failure might fall, before the final success], all over 5^8. Which equals 128*21 = 2688 in 390625.</p><p></p><p>The chance of 7 successes and 3 failures is 2^7 * 3^2 [for the 2 failures] * [the number of ways of allocating 2 "slots" out of 8 to failures, = 8!/6!2! = 8*7/2 = 28]. So 128 * 9 * 28 = 32256, all over 5^9 = 32256 in 1953125.</p><p></p><p>Adding all these together over a common denominator:</p><p></p><p>(128*25 + 2688*5 + 32256)/1953125</p><p></p><p>= (3200 + 13440 + 32256)/1953125</p><p></p><p>= 48896/1953125</p><p></p><p>= 0.025034752</p><p></p><p>That is, a little more than 1 in 40. That seems intuitively plausible, too, which gives me confidence in my maths.</p></blockquote><p></p>
[QUOTE="pemerton, post: 6364807, member: 42582"] Permutations and combinations. It's actually quite painful, at least by hand and using the resources of a high school maths students. (Those who have done higher study may have learned some shortcuts?) Your chance of success, at +2 vs DC 15, is 2 in 5 (ie need a 13+ on d20). The chances of 7 successes in a row is 2^7 over 78125, or 128 in 78125. The chance of 7 successes and 1 failure is 2^7 * 3 [for the failure] * 7 [because there are 7 "slots" into which the failure might fall, before the final success], all over 5^8. Which equals 128*21 = 2688 in 390625. The chance of 7 successes and 3 failures is 2^7 * 3^2 [for the 2 failures] * [the number of ways of allocating 2 "slots" out of 8 to failures, = 8!/6!2! = 8*7/2 = 28]. So 128 * 9 * 28 = 32256, all over 5^9 = 32256 in 1953125. Adding all these together over a common denominator: (128*25 + 2688*5 + 32256)/1953125 = (3200 + 13440 + 32256)/1953125 = 48896/1953125 = 0.025034752 That is, a little more than 1 in 40. That seems intuitively plausible, too, which gives me confidence in my maths. [/QUOTE]
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