I'm developing a website at the moment.

The website allows users to "rate" a post from 0 to 5.

Posts can then be displayed in order of popularity.

At the moment, my method of calculation is pretty primitive:

average_rating = total_rating/ratings

the problem is that a story with 1 rating of 5 is more popular than a story with 99 ratings of 5 and 1 of 4.

(5/1) > (499/100)

Could someone suggest a more accurate way to calculate popularity both on the number of votes and the quality of each vote?


4 Answers 4


A standard procedure (frequently -and loosely- called 'bayesian average') is to make a weighted average between the individual rating and the 'a priori' rating:

$R_a = W \; R + (1 - W ) \; R_0$


$R_a = $ averaged ('bayesian') rating

$R = $ individual rating: average rating for this item.

$R_0 = $ a priori rating: global average rating, for all items in your database.

$W = $ weight factor: it should tend to $0$ if this items has few votes, and it should tend to $1$ if it has many.

Some choices: $W = \frac{n}{N_{max}}$, or $W = max( \alpha \frac{n}{N_{av}},1)$ , etc ($n=$ number of votes for this item, $N_{max}=$ maximum number of votes for all items, $N_{av}=$average, $\alpha=$ some number between 0.5 and 1... ) Also, frequently one discards items that have very low/big values when computing the statistics.

See some examples

Added: for another approach, specially for yes/no like/diskike votes, see here.

  • $\begingroup$ +1 Excellent response! $\endgroup$ May 26, 2011 at 15:24
  • $\begingroup$ It's not clear to me whether this procedure can really be called Bayesian or not. (What's the prior? What posterior loss function are we using?) But I agree with the principle of using Bayesian estimation. $\endgroup$
    – Zhen Lin
    May 26, 2011 at 18:02
  • $\begingroup$ @Zhen Agreed. It's true that in some scenarios the formula agrees with the real bayesian estimator (eg. estimate the media of a gaussian with another gaussian as prior), but I feel that this is just a loose generalization of that result. $\endgroup$
    – leonbloy
    May 26, 2011 at 18:07
  • $\begingroup$ as per the equation, if you have more votes, (1-W) will tend to be a lower quantity and hence lower weight on prior. Is this what is intended? $\endgroup$ Feb 24, 2018 at 7:28

You could consider each post to have a 'true' average rating which you zero in on as more users vote on it. If we consider that the votes on each post come from a set of all possible votes that could have been cast that have a 'true' mean $\mu$ and standard deviation $\sigma$, then your average of the votes actually cast can be considered as an estimator of the true value. This estimator can be more or less accurate, depending on the number of votes cast. So your question is "how do I take account of the inaccuracies of my measured averages in the ranking of posts?"

We can solve this with a little math. Let vote $i$ be denoted $x_i=1,\dots,5$. Start by giving each post a rating of 3, so $x_1=3$. Then for each subsequent rating calculate the mean

$$\mu_{(n)} = \frac{1}{n}\sum_{i=1}^n x_i$$

and the sample standard deviation

$$\sigma_{(n)} = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_{(n)})^2$$

Then the standard deviation of your measured mean $\mu_{(n)}$ (which is a measure of its accuracy) is given by

$$\hat{\sigma}_{(n)} = \frac{\sigma_{(n)}}{\sqrt{n}}$$

i.e. the quality of your estimate of the true mean increases as the number of votes increases, which is what you'd expect.

To translate this into a ranking, note that knowing the standard deviation of the mean gives you an approximate confidence interval for the mean. In particular, you can calculate a minimum value for the true mean, with a certain level of confidence. A 95% confident estimate for the minimum value of the mean is given by $\mu_{(n)} - 1.64\hat{\sigma}_{(n)}$. Loosely translated, you can be 95% sure that the true mean is above $\mu_{(n)} - 1.64\hat{\sigma}_{(n)}$.

Ranking your posts by this quantity rather than by the measured mean naturally takes into account your uncertainty for posts with only a small number of votes.

One valid criticism of this scheme is that it is expensive to recalculate the mean and variance every time someone submits a vote. You may find the following on-line algorithms for calculating mean and variance useful: http://en.wikipedia.org/wiki/Algorithms_for_calculating_variance#On-line_algorithm

  • 1
    $\begingroup$ +1 This is the kind of statistics they should teach in school! $\endgroup$
    – Zhen Lin
    May 26, 2011 at 11:03
  • 1
    $\begingroup$ I'd like to make it clear that I think leonbloy's answer is superior to this one, and that's what I'd do if I was coding this up :) $\endgroup$ May 27, 2011 at 6:19

While Chris Taylor's answer is already good, I feel I should say something about statistical estimation. The idea is that your data are observations of a random variable $X$, and you want to say something about the distribution of $X$ using this data. There are two main types of models in statistics, namely parametric and non-parametric models. For example, plotting your data as a histogram is an example of non-parametric method. While non-parametric methods are robust, it is sometimes more difficult to obtain information about $X$ using them, so we often make assumptions about the distribution of $X$ and use a parametric model instead. An example of parametric modelling would be to assume that the users' votes are normally distributed with some unknown mean $\mu$ and variance $\sigma^2$ and try to estimate $\mu$ and $\sigma^2$ from the sample data. One advantage of making such an assumption is that we can give good confidence intervals for our estimates.

Indeed, if we have samples $X_1, \ldots, X_n$ of $X \sim N(\mu, \sigma^2)$, then, the probability that the sample mean $\overline{X} = \frac{1}{n} (X_1 + \cdots + X_n)$ is contained in the interval $(\mu - a, \mu + a)$ is $$\Phi \left(\frac{a}{\sigma \sqrt{n}}\right) - \Phi \left(-\frac{a}{\sigma \sqrt{n}}\right)$$ where $\Phi$ is the cumulative distribution function of the standard normal distribution. If we set $a = 1.64 \sigma \sqrt{n}$ then we see that there is a $90\%$ chance that $\overline{X} \in (\mu - a, \mu + a)$. Equivalently, there is a $90\%$ chance that $\mu \in (\overline{X} - a, \overline{X} + a)$. Unfortunately, I've cheated here a little — we don't know what $\sigma^2$ is, so we can't calculate $a$. So now we have to estimate $\sigma^2$. It's a known fact that $$S = \sum_{i=1}^{n} \frac{(X_i - \overline{X})^2}{\sigma^2}$$ is a $\chi^2$ random variable with $n - 1$ degrees of freedom, and is independent of $\overline{X}$. Hence, $$\mathbb{E} \left[ \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \overline{X})^2 \right] = \sigma^2$$ i.e. $\tilde{\sigma}^2 = \frac{1}{n - 1} S$ is an unbiased estimator of $\sigma^2$. We could, in principle, compute a confidence interval for $\sigma^2$ as well, but we only need a point estimate, and we can just substitute $\tilde{\sigma}$ for $\sigma$ in the formula above for the confidence interval for $\mu$ to obtain a concrete answer. (But note that it isn't a $90\%$ confidence interval!)

Of course, one can iterate this game, and (loosely speaking) Bayesian statistics is what you get when you assume that your parameters themselves are random variables distributed according to some prior distribution.

  • $\begingroup$ OK Thanks for your advice... I'm not a mathematician; could someone please explain this formula in a syntax I could understand? $\endgroup$
    – bodacious
    May 26, 2011 at 13:11

Yes, give every post an initial 3 to start with, so adjust your formula to

$${\rm average\,\ rating} = ({\rm total\ \,rating} + 3) / ({\rm ratings} + 1)$$

Alternatively, you may also amplify the posts with many ratings, e.g. by

$${\rm score} = \log ({\rm ratings}+1) ({\rm total\ \,rating} + 3) / ({\rm ratings} + 1) $$

This score may grow by a "degree" every time the number of ratings gets multiplied by $e$, but this total degree will be reduced by the average rating.

  • 3
    $\begingroup$ Your amplification method has the slightly bizarre result that a post with five 5* ratings would be ranked below a post with one hundred 2* ratings. $\endgroup$ May 26, 2011 at 10:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .