what is the difference here ?

Hello,

I found the solution to exercise 2 of the 99 ocaml problems.
The solution I found was this :

let rec last_two list =
    match list with
      | [] | [_] -> None
      | [x;y] -> Some (x,y)
      | _::xs -> last_two xs ;;


Now I wonder something.
For me it looks like [x;y]  and _::xs are the same.
If I understand it right you can write [x;y] as x::y and _::xs looks 
then the same to me.
But if I change [x;y] to x::y then I see a message that _:xs is not used.

Can someone explain in simple words what is the difference between those 
two ?

Roelof