What is the probability that the second and third rolls are both larger than the first roll?

Question

A die is rolled three times.

I tried using $x_1 + x_2 + x_3 = 3*6$ but to no avail.

The answer is $\frac{55}{6^3}$

Answer 1

Let's look at the possible cases:

1. 1st roll is 1, then the other 2 should be $\in \{2,3,4,5,6\}$. Or $25$ cases altogether.
2. 1st roll is 2, then the other 2 should be $\in \{3,4,5,6\}$. Or $16$ cases altogether.
3. 1st roll is 3, then the other 2 should be $\in \{4,5,6\}$. Or $9$ cases altogether.
4. 1st roll is 4, then the other 2 should be $\in \{5,6\}$. Or $4$ cases altogether.
5. 1st roll is 5, then the other 2 should be $\in \{6\}$. Or $1$ case only.
6. 1st roll is 6, then ... $0$ cases altogether.

Total so far $25+16+9+4+1=55$ thus $P=\frac{55}{6^3}$