# Looking for a function that gives more weight to more “data points”, i.e. 30/60 > 1/2

I'm looking to calculate "conversion rates" of some items, as follows:

(number of times the item was clicked on) / (number of times the item was presented to the user)

I want to give a higher "score" to items that were acted upon more, even if the rate is the same (so an item with a ratio of 6/12 will get a higher score than an item with a ratio of 3/6).

Any suggestions?

I thought about this once. One thing I thought was the following. Let $k, l\in\mathbb{N}$. Given $\frac{a}{b}$ consider $T_{k, l}\left(\frac{a}{b}\right)=\frac{ka-l}{kb}$. Let us look at what this does for some specific values of $k$ and $l$.
$T_{2, 1}\left(\frac{1}{2}\right)=\frac{1}{4}<\frac{3}{8}=T_{2, 1}\left(\frac{2}{4}\right)$.
By varying $k$ and $l$ you can get many different function that will do what you want. You will always have that $T_{k, l}\left(\frac{a}{b}\right)<T_{k, l}\left(\frac{na}{nb}\right)$. Then there is the issue of how you want say $T_{k, l}\left(\frac{1}{1}\right)$ to be related to $T_{k, l}\left(\frac{999}{1000}\right)$. Changing the $k$ and $l$ will give you some flexibility when dealing with questions such as this.
You could pretend each conversion had a fixed probability of happening, and then treat the rate as a binomial proportion. Then you could look at some lower one-sided confidence interval. For example: $$\hat{p} - z_{1-\alpha} \sqrt{ \frac{1}{n} \hat{p}(1-\hat{p})}$$ where $\hat{p}$ is the conversion rate estimate, $n$ is the number of presentations, and $z_{1-\alpha}$ is a standard normal quantile equal to $1.644854$ if $\alpha = 0.05$. The terms subtracted gets smaller as the sample size increases.
I say "[f]or example" because you'd probably want to use a better interval like some of those listed at the wikipedia page for binomial proportion confidence intervals. The naive normal one does absolutely horribly for small $p$, which I imagine is very possible in your case.