Solution:
Consider the more general problem for $n+1$ numbers instead of $100$ and denote by $p_{k, n+1}$ the probability that $a_{m}$ be equal to $1$. If $1$ is not the last number of a permutation of the positive integers $1, 2, \ldots, n+1$, then $a_{m} = 1$ with probability $p_{k, n}$ (we assume that $p_{n, n} = 0$). Otherwise, $a_{m} = 1$ only if $2$ and $3$ are among the first $k$ numbers of the permutation. Then
$$p_{k, n+1} = \frac{n}{n+1} p_{k, n} + \frac{1}{n+1} \cdot \frac{k}{n} \cdot \frac{k-1}{n-1}$$
and hence
$$\begin{aligned} & (n+1) p_{k, n+1} = \sum_{j=k}^{n} \left((j+1) p_{k, j+1} - j p_{k, j}\right) \\ & = k(k-1) \sum_{j=k}^{n} \frac{1}{j(j-1)} = \frac{k(n+1-k)}{n} \end{aligned}$$
It follows that $p_{k, n} = \frac{k(n-k)}{n(n-1)}$ and therefore $p_{k, 100} = \frac{1}{4}$ for $k = 45$ and $k = 55$.