How can a Bayesian analysis say A < B, when both have only 0s? I've used python to analyse data from AB tests using Bayesian analysis, and for all tests I assume no prior knowledge and so set alpha = beta = 1.
However I'm finding some odd results at low data volumes, which I thought was my code, but I'm also seeing using an online Bayesian calculator. This leads me to believe I don't understand the maths properly :).
If we take an AB test with the following parameters:
A trials: 100
A successes: 0

B trials: 10
B successes: 0

There is a 90% chance that B is better according to the analysis, however I don't understand how this can be the case with no successes recorded yet? Otherwise, the true success rate could be 0.00001% and this analysis should yield a low level of significance...
So is this the case? Assuming it is, how can I adjust parameters to ensure that there is no assumption on the success rate (or at least that I can control this assumption).
 A: I'm not a Bayesian expert, but I think this is simpler than it appears.  Your data seem to be binomial, and the uninformative prior for a binomial has an expected value of $.5$—albeit with equal likelihood for all probabilities.  The conjugate prior is beta, and you get the uniform with a beta if $\alpha = \beta = 1$, which I believe is what you are referring to.  The posterior in a Bayesian analysis can be thought of simply as a weighted average of your prior and your data.  So as you observe $0$s, your posterior will slide from $.5$ towards $0$.  More $0$s will make it slide further.  Thus, with more observations in A, the posterior will be dominated by the data, whereas with few observations in B, the posterior is dominated by the prior, and the analyst believes $Pr(Y=1|A)<Pr(Y=1|B)$.  
To think about this by analogy using frequentist techniques, consider the task of forming a confidence interval for $p$ from binomial data with all $0$s.  A simple approach to determining the upper limit of a 95% CI (the lower limit is obviously $0$) is to use the rule of 3; viz.: $3/N$.  For your data, the frequentist 95% CIs are, $[0,\ .03]$ for A, and $[0,\ .3]$ for B.  
