Let $ \{ X_i \}_{i=1}^{50}$ be a path of the markov chain.
The probability of observing the path then reads as
\begin{align} P_{\theta}(X_1, ..., X_{50}) &= P_{\theta}(X_{50} | X_{49}, ..., X_1) \cdot P_{\theta}(X_1, ..., X_{49}) \\ &= P_{\theta}(X_{50} | X_{49}) \cdot P_{\theta}(X_1, ..., X_{49}) \\ &= \prod_{i=1}^{50} P_{\theta}(X_i | X_{i-1} ) \end{align}
However, I am not interested in the probability of observing the whole path, but in observing $\{X_2, X_{25}, X_{50}\}$, meaning I want to neglect some observations.
My intuition is:
\begin{align} P_{\theta}(X_2, X_{25}, X_{50}) &= P_{\theta}(X_2 | X_{1}) \cdot P_{\theta}(X_{25} | X_{24}, ..., X_1) \cdot P_{\theta}(X_{50} | X_{49}, ..., X_1)\\ &= P_{\theta}(X_2 | X_{1}) \cdot P_{\theta}(X_{25} | X_{24}) \cdot P_{\theta}(X_{50} | X_{49}) \end{align}
Can I do it like this?
No. One way is \begin{align} P(X_2, X_{25}, X_{50}) &= P(X_2) \cdot P(X_{25} \mid X_2) \cdot P(X_{50} \mid X_{25}, X_2) \\ &= P(X_2) \cdot P(X_{25} \mid X_2) \cdot P(X_{50} \mid X_{25}), \end{align} where the first equality holds for any joint distribution (not necessarily Markov chain), and the second line uses the Markov chain memoryless property.
Alternatively, you can write it as \begin{align} P(X_2,X_{25}, X_{50}) &= \sum_{X_i : i \notin\{2,25,50\}} P(X_1,X_2,X_3,\ldots,X_{25},\ldots,X_{49}, X_{50}) \\ &= \sum_{X_i : i \notin\{2,25,50\}} P(X_1) \prod_{j=2}^{50} P(X_j \mid X_{j-1}), \end{align} where the sum is actually $47$ sums, one for each variable $X_i$ over all the values it can take.