Two dice (d6) were being rolled. We wish to calculate the probability that the outcome $X$ of the first die will be greater than the outcome $Y$ of the second. According to the textbook, the probability of this event ($A$) is:
$P(A) \ = \ \displaystyle \sum_{i=1}^{6} P(Y < X \ | \ X = i) \ P(X=i)$
$= \ \displaystyle \sum_{i=1}^{6} P(Y < i) \ \frac{1}{6} $
$= \ \displaystyle \sum_{i=1}^{6} \sum_{j=1}^{i-1} P(Y=j) \ \frac{1}{6}\\$
$= \ \displaystyle \frac{1}{6} \ \sum_{i=2}^{6} \frac{i-1}{6} \ = \ \frac{5}{12}$
I was really confused by the summation tricks from line 2 to line 3 and from line 3 to line 4. Intuitively, I can understand that those summations are equivalent by reasoning under the context of rolling dice of 6 faces. But is there a formal way to show their equivalence mathematically?
How can I show $\displaystyle \sum_{i=1}^{6} P(Y < i) = \sum_{i=1}^{6} \sum_{j=1}^{i-1} P(Y=j) = \sum_{i=2}^{6} \frac{i-1}{6}$?
$$P(Y<i)=\sum_{j: j < i} P(Y=j)=\sum_{j=1}^{i-1} P(Y=j)$$ because $Y$ is a discrete random variable with integer values $\ge 1$.
$$P(Y=j)=\frac{1}{6}$$ because $Y$ is a discrete uniform random variable over $\{1,\dots,6\}$.
$$\sum_{j=1}^{i-1} \frac{1}{6} = (i-1) \frac{1}{6}$$ because there are $i-1$ terms, each with value $1/6$, in the sum.
$$\sum_{i=1}^6 \frac{i-1}{6} = \sum_{i=2}^6 \frac{i-1}{6}$$ because the $i=1$ term is $0$.