# Simple up/down vote rating but weighted by number of responses

I am trying to analyse the ratings for restaurants from a website. The rating system on the website is pretty simple: people can up-vote or down-vote.

The restaurant is then presented to website users with the number of votes it received and a percentage score (I am assuming this is simply an average of the votes: e.g. 2 up-votes and 1 down-vote will give a score of 66.7%).

Intuitively if I were to pick a restaurant to visit: I would rather pick one with a score of 90% and 100 votes than pick one with a score of 100% and 1 vote. Even though the former score is lower, I also know that the score won't be drastically affected by the next vote so I trust the score more. (Something Bayesian about this?)

How can a third metric (rating) be derived that combines score and number of votes to order restaurants not only by score but also by how trustworthy that score is?

I can think of a few empirical ways of doing this but they often result in me asking myself: What's better between (75%, 10 votes) and (80%, 8 votes)? Which isn't as obvious as the example above. Is there a more formal way to answer this question?

If I understand your question correctly, you would like to combine both the point estimate (percentage of up-votes) and its reliability (roughly represented by the total number of votes) in a single number.

As confidence intervals represent both values, a very simple approach would be to base the score on the lower limit of a (e.g. 50%) confidence interval for a binomial proportion $$\hat{p}$$. To make sure that this value is positive, you should use a confidence interval that also works with small values of $$n$$ and also with values of $$p$$ close to one ore zero, e.g. the Bayesian HPD interval, the "exact" Clopper-Pearson interval, or the WIlson interval. The latter has the attractive feature of being computable in closed form, i.e., the lower limit is: $$p_{lower} = \frac{1}{1+z^2/n}\left[\hat{p} + \frac{z^2}{2n} - z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}+\frac{z^2}{4n^2}}\right]$$ where $$z=z_{1-\alpha/2}$$ is the $$(1-\alpha/2)$$ standard normal quantile; for $$1-\alpha=50\%$$, it is $$z\approx 0.7$$.

One metric could be -- that takes into account the number of votes -- could be $$(n_+/N)\ln{N}$$ where $$n_+$$ is the number of positive votes, $$\ln{}$$ is natural log, $$N$$ is the total number of votes, $$N \geqslant1$$.