A verification request upon a combinatorial problem An urn contains $100$ balls numbered from $1$ to $100$. Four are removed at random without being replaced. Find the probability that the number on the last ball is smaller than the number on the first
ball.
MY ATTEMPT
If the first removed ball is $2$, then there is $1$ possibility for the last ball and $98\cdot 97$ for the other two. Analogously, if the first removed ball is $3$, then there are $2$ possibilities for the last ball and $98\cdot 97$ for the other two. According to this reasoning, there are
\begin{align*}
1\cdot 1\cdot 98\cdot 97 + 1\cdot 2\cdot 98\cdot 97 + \ldots + 1\cdot 99\cdot 98\cdot 97 = \left[\frac{(1+99)\cdot99}{2}\right]\cdot98\cdot 97
\end{align*}
possible results which satisfy the given restriction. Consequently, the sought probability is
\begin{align*}
\textbf{P}(E) = \frac{50\cdot 99\cdot 98\cdot 97}{100\cdot 99\cdot 98 \cdot 97} = \frac{1}{2}
\end{align*}
Could someone please tell me if I am right? Thanks in advance.

Here is a simple argument:
Either the number on the last ball is greater than the number on the last ball or less than the first ball's number. Hence the required probability is $\frac12$.
