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General Tabletop Discussion
*Pathfinder & Starfinder
Success chances for Skill Challenges
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<blockquote data-quote="Elric" data-source="post: 4989636" data-attributes="member: 1139"><p>You don't need to condition on the exact sequence of rolls for skill challenges. The key is that a skill challenge of S successes before F failures can only last S+F-1 rolls. E.g., 4 successes before 3 failures. After 6 rolls, you'll either have 4+ successes (and 0-2 failures), or you'll have <=3 successes and 3+ failures. It looks like order matters when you look at the 7th roll, but that roll isn't going to matter! </p><p></p><p>It's equivalent to a normal 4 S/3 F skill challenge to assume you make six rolls, regardless of the outcomes, and then decide on success or failure of the challenge (>=4 successes-> success, >=3 failures -> fail). If you already got to 4 successes, you can't possibly reach 3 failures in six rolls, and vice versa (this isn't true if you consider the 7th roll). So order doesn't matter in the first six rolls, and you ignore all rolls after that.</p><p></p><p>As a result, calculating the chance to succeed on a skill challenge is simple when the chance to succeed on each roll is the same. It's the chance that in (S+F-1) trials, each with a probability p of success, you get <F failures. This is an application of the <a href="http://en.wikipedia.org/wiki/Binomial_distribution" target="_blank">Binomial distribution </a></p></blockquote><p></p>
[QUOTE="Elric, post: 4989636, member: 1139"] You don't need to condition on the exact sequence of rolls for skill challenges. The key is that a skill challenge of S successes before F failures can only last S+F-1 rolls. E.g., 4 successes before 3 failures. After 6 rolls, you'll either have 4+ successes (and 0-2 failures), or you'll have <=3 successes and 3+ failures. It looks like order matters when you look at the 7th roll, but that roll isn't going to matter! It's equivalent to a normal 4 S/3 F skill challenge to assume you make six rolls, regardless of the outcomes, and then decide on success or failure of the challenge (>=4 successes-> success, >=3 failures -> fail). If you already got to 4 successes, you can't possibly reach 3 failures in six rolls, and vice versa (this isn't true if you consider the 7th roll). So order doesn't matter in the first six rolls, and you ignore all rolls after that. As a result, calculating the chance to succeed on a skill challenge is simple when the chance to succeed on each roll is the same. It's the chance that in (S+F-1) trials, each with a probability p of success, you get <F failures. This is an application of the [url=http://en.wikipedia.org/wiki/Binomial_distribution]Binomial distribution [/url] [/QUOTE]
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