Given a die, what is the probability that the second roll of a die will be less than the first roll?

If you are given a die and asked to roll it twice. What is the probability that the value of the second roll will be less than the value of the first roll?

There are various ways to answer this. Here is one:

There is clearly a $1$ out of $6$ chance that the two rolls will be the same, hence a $5$ out of $6$ chance that they will be different. Further, the chance that the first roll is greater than the second must be equal to the chance that the second roll is greater than the first (e.g. switch the two dice!), so both chances must be $2.5$ out of $6$ or $5$ out of $12$.

Here another way to solve the problem $$ \text{Pr }[\textrm{second} > \textrm{first}] + \text{Pr }[\textrm{second} < \textrm{first}] + \text{Pr }[\textrm{second} = \textrm{first}] = 1 $$ Because of symmetry $\text{Pr }[\text{second} > \text{first}] = \text{Pr }[\text{second} < \text{first}]$, so $$ \text{Pr }[\text{second} > \text{first}] = \frac{1 - \text{Pr }[\text{second} = \text{first}]}{2} = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12} $$

It might help to draw a picture:

$$\begin{array}{c|cccccc} &1&2&3&4&5&6 \\ \hline 1&=&<&<&<&<&< \\ 2&>&=&<&<&<&< \\ 3&>&>&=&<&<&< \\ 4&>&>&>&=&<&< \\ 5&>&>&>&>&=&< \\ 6&>&>&>&>&>&= \\ \end{array}$$

Here, the $<$ signs mark the outcomes where the row number is less than the column number, and the $>$ signs mark those where to row number is greater than the column number. It's easy to see from the picture that the number of $<$ (or $>$) signs is $5+4+3+2+1=15$ out of $6^2 = 36$.

In fact, if you look at the picture a bit longer, you might realize that there's an even easier way to count the $<$ signs: the total number of $<$ and $>$ signs equals the total number of all signs ($6^2 = 36$) minus the number of $=$ signs ($6$), and the number of $<$ signs is half of that. Thus, there are $(36 - 6)/2 = 30/2 = 15$ out of $36$ $<$ signs in the table.

Once you've noticed that, it's easy to generalize the result: if you roll two $n$-sided dice, there are $n^2$ possible outcomes, out of which in $(n^2-n)/2$ the second roll will be less than the first. Thus, the probability of the second roll being less than the first is $$\frac{n^2-n}{2n^2} = \frac{n-1}{2n}.$$

For six-sided dice, this works out to $\frac{30}{72} = \frac{5}{12} = 0.41666\ldots$

If the first roll is n, the chance that the second roll will be less is $\frac{n-1}{6}$. Summation over all possible values of n and multiplying by the chance for each value of n gives

$$ \sum _{n=1}^6 \frac{1}{6}*\frac{(n-1)}{6}=\frac{5}{12} $$