Say the 'acutal' rating for an item is p. (ie, if you got an infinite number of people voting, the ratio of upvotes to total votes is p). Now say your users vo

Say the 'acutal' rating for an item is p. (ie, if you got an infinite number of people voting, the ratio of upvotes to total votes is p). Now say your users vote, and they upvote with that probability p. The number of upvotes k you get out of n votes will follow a binomial distribution B(k;n,p). The binomial distribution has mean np and stdev sqrt(n p(1-p)), and is very close to gaussian in shape. Since the stdev is a rough measure of the 'width' of the distribution, common way to describe the error is (mean +/- lambda*stdev), where you can tune lambda to your desire. If you increase lambda you get a wider confidence interval, and therefore more certainty that a measurement will be within that confidence interval. Now, say you measure k upvotes out of n votes. You can divide by n to get p0, your estimated rating based on those votes. An easy estimate for the error of this measurement is to assume that p0 is approximately correct and equal to p. Then the expected number of upvotes would be n p0 with stdev(n p0(1-p0)). Divide this by n to get the fraction of upvotes, to give a final estimate for p of p0 +- lambda sqrt(p0(1-p0)/n) Now, your estimate (p0) of p is not quite right, and therefore your estimate of the error (which depends on p0) is not quite right either. The wilson score attempts to correct for that. We don't know p, but imagine if we did, we would expect any measurement p0 to be in the range p +- lambda sqrt(p(1-p)/n). That is, we expect abs(p - p0) < lambda sqrt(p (1-p)/n) If you solve this equation for p in terms of p0, you get a formula given in the article, ie, the confidence limits for p given p0.