$5$ distinct numbers are randomly distributed to players numbered $1$ to $5$. Whenever two players compare their numbers, the one with the higher one is the winner. Initially, players 1 and 2 compare their numbers; the winner then compares with player 3 and so on. Let $X$ denote the number of times player 1 is a winner. Find $P(X=i), i \in\,(0,1,2,3,4)$
Attempt: $P(X=0) = P(\text{player 1 loses}) = P(\text{player 2 has higher number than player 1}) = 1/2$ I reasoned a half because we know know nothing about the distribution of numbers, so either he has a number lower or higher (not the same, because it is given that they are distinct), and both are equally likely.
$P(X=1) = P(\text{player 1 has higher than player 2 and player 1 has lower than player 3}) $ To find this I am pretty sure I need to condition on player 1 beating player 2 first. Also, is it correct that these events are not independent, since the event that player 1 is lower than player 3 is only valid provided that player 1 beats player 2, that is requires knowledge about whether player 1 beat player 2?
So I said $P(\text{p.1 > p.2 and p.1 < p.3}) = P(\text{p1 < p3|p1 > p2})P(p1 > p2)$ I am struggling to compute the first term, but I think the second term is just 1/2 (this is the same 1/2 I computed for the first part) I tried working with a reduced sample space and listing possible outcomes but I didn't manage to attain the right answer this way.