I have measurements over time which give me probabilities of the state which can be X or Y and this state is constant over time. That means it either was always in state X or always in state Y. The states are mutually exclusive.
Question
How can I get the overall probability of the state being X or Y given x-amount of probability measurements?
Here is what I have thought about so far:
Multiplying does not seem to make sense, because let's say I get 2 measurements with a probability of 0.5. That actually means I have no information about the object state as both say it could be 50/50. Yet, the probability changes to 0.5*0.5=0.25. (I think because it does not take into account that the state is constant?).
So I fiddled around a bit with the numbers. And I found that if I multiply the probabilities of being in state X and then multiply the probabilities of being in state Y (inverse of X-probabilities) and normalize it so the two probabilities equal 1, I seem to get more or less what I was looking for ($p_i$ = probability being in state X at timestamp i): $$ \frac{\prod_i p_i}{\prod_i p_i + \prod_i (1-p_i)} $$
But I have no idea if this is correct in any way.
Your solution is correct if your prior knowledge of $X/Y$ is $50:50$. To see this lets use Bayes's theorem:
$P(X|p_1,\cdots ) \propto P(p_1\cdots|X)P(x) =\prod_i P(p_i|X) P(X)$
If we consider $X=[0,1]$ we have $P(p|X)=p^X(1-p)^{(1-X)} $ .
To change the proportion to an equality, we normalize ($\sum_x P(x|p_1,\cdots ) =1$) and get your expression: $P(X|p_1,\cdots )= {\prod p_i^X(1-p_i)^{(1-X)}P(X) \over \prod p_i P(1) +\prod (1-p_i)P(0)}$