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General Tabletop Discussion
D&D Older Editions
Showing the Math: Proving that 4e’s Skill Challenge system is broken (math heavy)
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<blockquote data-quote="Eldorian" data-source="post: 4280889" data-attributes="member: 10504"><p>You mean, </p><p></p><p>P(X=x|r,p)=OVER(x+r-1, x)*p^r*(1-p)^x ?</p><p></p><p>r= successes =4, p = probably of success = .55, x= number of failures = 2?</p><p></p><p>That's what I see on wikipedia. Never used this thing, myself.</p><p></p><p></p><p>Take a look at the cumulative distribution function for your guy. According to wiki, </p><p></p><p><a href="http://en.wikipedia.org/wiki/Negative_binomial_distribution#Properties" target="_blank">http://en.wikipedia.org/wiki/Negative_binomial_distribution#Properties</a></p><p></p><p></p><p>Oh here we go. I think I've found what you're using. You did it wrong, tho.</p><p></p><p>The negative binomial random variable, denoted by X ~ nb(r, p) is a generalization of the geometric random variable. Suppose you have probability p of of succeeding on any one try. If you make independent attempts over and over, then X counts the number of attempts needed to obtain the rth success, for some designated r >=1. When r = 1, then X ~ geo(p).</p><p></p><p>We often let q = 1 - p be the probability of failure on any one attempt. Then the probability of having the rth success on the kth attempt, for k >= r, is given by</p><p></p><p>P(X = k) = C(k - 1, r - 1) * q^(k - r) * p^r .</p><p></p><p></p><p>See, you found the probability of getting the 4th success on the 6th attempt. This isn't even possible in the experiment at hand. Here k= x+r</p><p></p><p>What you want to calculate, using your method, is </p><p></p><p>P(X = 4) = C(4 - 1, 4 - 1) * (.45)^(4 - 4) * (.55)^4 = 0.09150625</p><p></p><p>plus</p><p></p><p> P(X = 5) = C(5 - 1, 4 - 1) * (.45)^(5 - 4) * (.55)^4 = 0.16471125</p><p></p><p></p><p>That is, the sum of the chance of getting the 4th success on 4th trial, and 4th success on 5th trial.</p><p></p><p>add those together and you get</p><p></p><p><span style="font-size: 26px">0.2562175</span></p><p></p><p>Easily accounts for the OPs slight round off errors.</p><p></p><p>Source: <a href="http://www.wku.edu/~david.neal/statistics/discrete/negbinom.html" target="_blank">http://www.wku.edu/~david.neal/statistics/discrete/negbinom.html</a></p></blockquote><p></p>
[QUOTE="Eldorian, post: 4280889, member: 10504"] You mean, P(X=x|r,p)=OVER(x+r-1, x)*p^r*(1-p)^x ? r= successes =4, p = probably of success = .55, x= number of failures = 2? That's what I see on wikipedia. Never used this thing, myself. Take a look at the cumulative distribution function for your guy. According to wiki, [url]http://en.wikipedia.org/wiki/Negative_binomial_distribution#Properties[/url] Oh here we go. I think I've found what you're using. You did it wrong, tho. The negative binomial random variable, denoted by X ~ nb(r, p) is a generalization of the geometric random variable. Suppose you have probability p of of succeeding on any one try. If you make independent attempts over and over, then X counts the number of attempts needed to obtain the rth success, for some designated r >=1. When r = 1, then X ~ geo(p). We often let q = 1 - p be the probability of failure on any one attempt. Then the probability of having the rth success on the kth attempt, for k >= r, is given by P(X = k) = C(k - 1, r - 1) * q^(k - r) * p^r . See, you found the probability of getting the 4th success on the 6th attempt. This isn't even possible in the experiment at hand. Here k= x+r What you want to calculate, using your method, is P(X = 4) = C(4 - 1, 4 - 1) * (.45)^(4 - 4) * (.55)^4 = 0.09150625 plus P(X = 5) = C(5 - 1, 4 - 1) * (.45)^(5 - 4) * (.55)^4 = 0.16471125 That is, the sum of the chance of getting the 4th success on 4th trial, and 4th success on 5th trial. add those together and you get [SIZE=7]0.2562175[/SIZE] Easily accounts for the OPs slight round off errors. Source: [url]http://www.wku.edu/~david.neal/statistics/discrete/negbinom.html[/url] [/QUOTE]
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