Hey Stackexchange community! I try to get a feeling for updating probability using Bayes' theorem, but I do not really get it. I can find many examples, but I would like to consider the following one.
Imagine we have two teams $T_1$ and $T_2$ and they will play against each other on Friday. These two teams have never played against each other before, so there is no way to know the probability of one team beating the other one. Let $B$ be the event where $T_1$ is beating $T_2,$ with no data available we will assume that the probability of $B$ is $\frac{1}{2}$, that is $P(B)=\frac{1}{2}$.
Let's now assume that $T_1$ is actually beating $T_2$ that Friday. The week after these two teams will play against each other again. Now let $A$ be the event the $T_1$ is again beating $T_2,$ the question is can we use Bayes' theorem to get a estimate of the new probability?
Let me clarify where I get stock. From Bayes' theorem we have \begin{equation} P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)} . \end{equation}
We already know $P(B) = \frac{1}{2}$. I guess the same reasoning as before can be used for estimating $P(A)$, if we have no data available we have $P(A)=\frac{1}{2}$.
So if everything is correct until now the only problem is the expression $P(B \mid A)$, which states:
What is the probability of $T_1$ beating $T_2$ last Friday given $T_1$ beats $T_2$ next week?
First I thought we have to used that these two events are independent so we have $P(B \mid A) = P(B) P(A)$. From this we conclude \begin{equation} P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)} = \frac{P(B)P(A) P(A)}{P(B)} = (P(A))^2 = \frac{1}{4} \end{equation} I would not bet on this $\dots$