Conditional probability in dynamic model Problem text:
Now, in the previous problem we skipped over why a particular user would be interested in an article. Let us assume that each article consists of a main topic, and each user is primarily interested in one topic. We would like to find each user’s interests to better serve articles that they are interested in. Obviously, we can’t know the true interests of a user, we can only observe what she clicks on.
We can model this using a dynamic model. Each day we examine all the clicks a user has made and update the user’s Interests based on the topic she has clicked the most on that day. Let us assume we have three topics: Finance, Sport, and Tech. The conditional probability table
for P(Clicked|Interests) is:
-----------|Clicked-------------------|
Interests| Finance | Sport | Tech |
Finance | 0.50 | 0.25 | 0.25
Sport | 0.25 | 0.50 | 0.25
Tech | 0.25 | 0.25 | 0.50
All interests for new users are equally likely, as we have no prior information.
For now we assume that the user’s interests do not change over time.
Question: 
A new user arrives at our site and clicks mostly on Tech on the first day. Calculate P(Interests|Clicked=Tech). Assume a uniform prior on Interests as we have no previous information on the interests of new users. The second day, the user clicks mostly on Tech again. Calculate P(Interests|Clicked=Tech) again, but this time use the results from the first day as a prior on Interests.
I really can't make any sense of this problem, although I feel it should be rather simple. Any help appreciated.
 A: This is an application of Bayes' Theorem, which states that if $A_i$ partition the probability space, then $$ P(A_i|B) = \frac{P(B|A_i)\cdot P(A_i)}{\sum P(B|A_j) P(A_j)}. $$ I'll make use of the following abbreviations: $IF$ means interested in finance, $CT$ means clicked Tech, etc.  As the problem states, we know that since this is a new user, we have $P(IF) = P(IT) = P(IS) = \frac{1}{3}$.  Now, we have \begin{align*}
P(IF|CT) &= \frac{P(CT|IF)\cdot P(IF)}{P(CT|IF)\cdot P(IF) + P(CT|IS)\cdot P(IS) + P(CT|IT)\cdot P(IT)} \\
&= \frac{\frac{1}{4} \cdot \frac{1}{3}}{\frac{1}{4} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{1}{3} + \frac{1}{2}\cdot\frac{1}{3}}\\
&= \frac{1}{4}. 
 \end{align*}
Similarly, we get $P(IS|CT) = \frac{1}{4}$ and $P(IT|CT) = \frac{1}{2}$.  Now, since we know that the user clicked on Tech, we update $P(IF) = P(IF|CT) = \frac{1}{4}, P(IS) = P(IS|CT) = \frac{1}{4},$ and $P(IT) = P(IT|CT) = \frac{1}{2}. $  We now do the same calculation again: \begin{align*}
P(IF|CT) &= \frac{P(CT|IF)\cdot P(IF)}{P(CT|IF)\cdot P(IF) + P(CT|IS)\cdot P(IS) + P(CT|IT)\cdot P(IT)} \\
&= \frac{\frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4} \cdot \frac{1}{4} + \frac{1}{4} \cdot \frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2}}\\
&= \frac{1}{6}. 
 \end{align*}
Similarly, we get $P(IS|CT) = \frac{1}{6}$ and $P(IT|CT) = \frac{2}{3}$.
