Updating a probability with additional knowledge. Bayes Theorem I am quite confused with using Bayes theorem for the following problem. And I am not sure it can be applied at all.
I have a football website data with user views. Each view corresponds to a specific team. Therefore, for each user I can calculate it's frequency and determine which football team he prefers. 
Also, I have some geo statistics corresponding to football fans in different cites. With this I can calculate a probability of being a fan of specific team in some city.
Then with this additional information I want to update probability that a user likes some specific team.
I tried using Bayes Theorem for it but I am not sure if it is right.
P =  P(user likes the team) * P(Probability of liking this team in this city) /P(Probability to be a website user from this city)
Andy advice will be appreciated.
 A: I can't quite make sense of your question. For one thing it seems that the geo
data would provide prior information and information from your website would 'update' that. I have tried to put your ideas into the framework of a typical application of elementary Bayesian inference.
Suppose that City A has a football team and several nearby cities also have football teams.
Your prior information (from geo statistics) is that about 40% of football fans in City A with favor the local 'A-team'. In Bayesian statistics the probability $\theta$ of preferring the A-team is provided as a probability distribution, often a beta distribution because beta distributions have $(0,1)$ as their support.
Perhaps you choose as your PRIOR distribution $\theta \sim \mathsf{Beta}(4,6)$ which has $E(\theta) = \frac{12}{12+18} = 0.4$ and puts over 70% of its probability in the interval $(.4,.6)$ and about 95% of its probability in
$(0.234, .0577),$ according to computations in R:
diff(pbeta(c(.3,.5), 12, 18))
[1] 0.7381535
qbeta(c(.025, .975), 12, 18)
[1] 0.2352402 0.5773954


Then from your website you find that $x = 1283$ of $n = 2500$ visitors to
the site from City A, show interest in the A-team. That information based on data
is represented in terms of a LIKELIHOOD function.
Bayes' Theorem in this setting is expressed as:
$$\text{POSTERIOR} \propto \text{PRIOR} \times \text{LIKELIHOOD},$$
where the symbol $\propto$ (read 'proportional to') indicates that
we are using 'kernels' of density functions, omitting their norming constants.
In particular, your posterior distribution is
$$p(\theta|x) = p(\theta) \times p(x|\theta)\\
\propto \theta^{12-1}(1-\theta)^{18-1} \times
    \theta^{1283}(1-\theta)^{2500-1283}\\
\propto \theta^{1295-1}(1-\theta)^{1235-1},$$ 
where we recognize the final expression as the kernel of the posterior
distribution $\mathsf{Beta}(1295, 1235).$
Thus, the posterior mean is $E(\theta|x) = \frac{1295}{1295+1235} =  0.512$
and the 95% Bayesian posterior probability interval is $(0.492, 0.531)$
qbeta(c(.025, .975), 1295, 1235)
[1] 0.4923767 0.5313210

Thus your prior information that $\theta$ is roughly $0.4$ has been
updated by website data to show that $\theta$ is very nearly $0.51.$
