Let $X_1 .... X_K$ represent $K$ independent random variables. These var (of unknown distribution). I am trying to understand how to simplify the probability that $X_1 $ is the greatest of all. That is:
$$Pr(X_1 > X_2, X_1 > X_3 ... X_1 > X_K)$$
Notice that I am not excluding the chance that the variables are equal. They could be equal, so there is a non-zero probability of that.
My understanding is that, in general, the events of interest (the logical comparisons) are not independent. That being the case, I am trying to decompose the formula above into simple probabilities (not joint or conditional).
With the simple case of $K = 3$, and using the formula of conditional probabilities, I get at:
\begin{align}Pr(X_1 > X_2, X_1 > X_3) = Pr(X_1 > X_3 | X_1 > X_2) \cdot Pr(X_1 > X_2) \end{align}
Which I think can be simplified like:
\begin{align}Pr(X_1 > X_2, X_1 > X_3) =& Pr(X_1 > X_2)\cdot[Pr(X_3 \leq X_2)+Pr(X_3>X_2)Pr(X_1>X_3)]\\ & \cdot Pr(X_1 > X_2) \end{align}
What I am trying to understand is, first, if the above simplification makes sense, and second, how that could be extended to $K > 3$ in a generic way.
EDIT: I can impose the assumption that $X_1,...,X_K$ follow the same distribution family, but I cannot assume they follow exactly the same distributin, that is, with same distribution parameters.
One way to compute a generic answer is to condition on $X_1$. Given $X_1$, the comparisons do become independent, and we obtain $$ \mathbb P(X_1>X_2,\ldots,X_1>X_{K}\mid X_1)=\mathbb P(X_1>X_2\mid X_1)^{K-1}. $$ Even though you didn't say it explicitly, I am assuming that $X_1,\ldots,X_K$ all have the same distribution. Thus, $$ \mathbb P(X_1>X_2,\ldots,X_1>X_{K})=\mathbb E \left[\mathbb P(X_1>X_2\mid X_1)^{K-1}\right]. $$ Note, I am using the notation of conditional expectation as described here, and if you aren't familiar with this concept then the meaning of the expression I have written may be lost. It has another equivalent expression that may be clearer to you. Let $f(x)$ be (a variant of) the CDF of $X_2$, defined so that $f(x)=\mathbb P(X_2<x)$. Then the quantity $\mathbb P(X_1>X_2\mid X_1)$ is equal to $f(X_1)$. Note that it is random, obtained by substituting $X_1$ into the (deterministic) function $f(x)$. Using this notation, the formula can be equivalently written as follows: $$ \mathbb P(X_1>X_2,\ldots,X_1>X_{K})=\mathbb E \left[f(X_1)^{k-1}\right]. $$