How to calculate prior probabilities I have two possible events $A$ and $B$ that could lead to $n$ possible consequences $X_1, X_2, \ldots , X_n$, $P(A) + P(B) = 1$, $P(X_1) + P(X_2) + \ldots + P(X_n) = 1$. I know all conditional probabilities $P(X_1 \mid A), P(X_1 \mid B), \ldots , P(X_n \mid A), P(X_n \mid B)$, and I need to find $P(A \mid X_1), P(B \mid X_1), \ldots , P(A \mid X_n), P(B \mid X_n)$.
According to the Bayes' rule,
$$
P(A \mid X_1) = \frac{ P(A) P(X_1 \mid A)}{ P(A) P(X_1 \mid A) + P(B) P(X_1 \mid B)} .
$$
The question is what is the proper way to calculate priors $P(A)$ and $P(B)$. Are they equal to each other? Are they equal to the proportions of their appearances in my dataset? In the latter case, should I calculate priors separately for each participant?
UPD. My data is basically the sequences of $X$'s for different participants.
 A: Since you have the conditional probabilities (i.e. $P(X_i | A)$ and $P(X_i | B)$), and that $P(B) = 1 - P(A)$, you need to fix the value of only one variable, namely $P(A)$; let's denote it by $p$. 
You have to pick $p$ that fits your data best. To do so, you can look at the log likelihood function and set $p$ so that the (log) likelihood function, denoted by $L(p)$, is maximized. In other words, you are looking for $p^*$ defined as:
$$
p^* = \arg\max_{0 \leq p \leq 1} L(p)
$$
Let $D = \left\{ X^{(1)}, \dots, X^{(N)} \right\}$ be $N$ samples that you observed. Also, let $M$ be a $n \times 2$ matrix representing the conditional probabilities; i.e. $M[k, 1] = P(X_k | A)$ and $M[k, 2] = P(X_k | B)$. The log likelihood function on $D$ is defined as:
$$
\begin{align}
L(p) &= \sum_{i=1}^N \log \left( P(A) P(X^{(i)} | A) + P(B) P(X^{(i)} | B) \right) \\
&= \sum_{i=1}^N \log \left( p \times M[X^{(i)}, 1] + (1 - p) \times M[X^{(i)}, 2] \right). \\
\frac{\partial L}{\partial p} &= \sum_{i=1}^n \frac{M[X^{(i)}, 1] - M[X^{(i)}, 2]}{p \times M[X^{(i)}, 1] + (1 - p) \times M[X^{(i)}, 2]}
\end{align}
$$
Note that the log likelihood function is concave, because the second derivative is negative (this is easy to show). Assuming that $\nexists k, M[k, 1] = M[k, 2] = 0$, we have:
$$
\begin{align}
p^* &= 
\begin{cases}
1 & \text{if } \frac{\partial L}{\partial p} > 0, \text{for } p \in (0, 1) \\
0 & \text{otherwise}
\end{cases}
\end{align}
$$
