Dice: Probability of rolling a number between two other dice throws I was pretty suprised about this problem when I encountered it in one of my excercise sheets and would like to ask for an approach here because I have no idea how I'm supposed to get started here:
The Problem:

Player-1 throws a 20-sided die, Player-2 throws two 20-sided dice. P1
  wins, if the number of his throw lies properly within the two numbers
  that P2 got. So P2 wins, if the number of P1 is smaller or equal to
  the lower number of P2, equal to the upper number of P2 or bigger than
  both of P2s numbers. What is the probability that P1 wins?

Since I can't know how big the intervall between both numbers of P2 is, how am I supposed to calculate the probability? 
 A: Hint $1$ :
Consider all possible events (for example : player $2$ threw $10$ and $17$, player
 $1$ threw $15$) and add the probabilities for all events leading to a win for player $1$.
Hint $2$ :
Since player $1$ can only win, if the numbers of player $2$ differ, each event leading to a win for player $1$ has probability $\frac{1}{4000}$
A: The following problem is equivalent:
Paint player $1$'s die blue, and paint player $2$ dice green and red. The players throw their dice, and player $1$ wins if the blue die’s result is between the values on the red and green dice.
If three different numbers turn up when red, green, and blue dice are rolled, the blue die is the middle result (and player $1$ wins) $1/3$ of the time. If any two of the dice results match, player $1$ loses, because no result can be strictly between the other two.
Therefore the probability that player $1$ wins is $1/3$ the probability that a single roll of three $20$-sided dice will result in three different numbers.
The probability that a roll of three $20$-sided dice will have no matches is $1\cdot\displaystyle\frac{19}{20}\cdot\frac{18}{20}=\frac{171}{200}$. (Roll them in sequence. The first result can be anything; the second must be one of the $19$ other results, and the third must be different from the first two, which themselves are different.)
Player $1$ wins this game with probability $\displaystyle\frac{1}{3}\cdot\frac{171}{200}=\frac{57}{200}$.
