Solution:
First, we claim that if the display starts at some $N$, the probability that any given number $M<N$ will appear at some point is $1/(M+1)$. We can show this by induction on $N$.

If $N = M+1$ (the base case), $M$ can only be reached if it appears after the first step, and this occurs with probability $1/N = 1/(M+1)$.

If $N > M+1$ and the claim holds for $N-1$, then there are two possibilities starting from $N$. If the first step leads to $N-1$ (this occurs with probability $1/N$), the probability of seeing $M$ subsequently is $1/(M+1)$ by the induction hypothesis. If the first step leads to something less than $N-1$ (probability $(N-1)/N$), then it leads to any of the integers $\{0,1, \ldots, N-2\}$ with equal probability. But this is exactly what the first step would have been if we had started from $N-1$; hence, the probability of seeing $M$ is again $1/(M+1)$ by induction. Thus, the overall probability of seeing $M$ is
$$\frac{1}{N} \cdot \frac{1}{M+1} + \frac{N-1}{N} \cdot \frac{1}{M+1} = \frac{1}{M+1},$$
proving the induction step and the claim.

Now let $P(N, M)$ ($M<N$) be the probability of eventually seeing the number $M$ if we start at $N$; note that this is the same as the conditional probability of seeing $M$ given that we see $N$. Hence, the desired probability is
$$P(2003,1000) \cdot P(1000,100) \cdot P(100,10) \cdot P(10,1) = \frac{1}{1001} \cdot \frac{1}{101} \cdot \frac{1}{11} \cdot \frac{1}{2} = \frac{1}{2224222}.$$