Define a configuration to be a state of the array $A[]$ after receiving some of the numbers. Define a configuration to be obviously wrong if there exists indices $i<j$ such that $A[i],A[j]$ have already been filled in with numbers such that $A[i]>A[j]$.
A randomized algorithm
Here is a natural algorithm: guess where to put each number uniformly at random, conditioned on the one constraint that you never place a number in an obviously wrong position.
The analysis of this approach seems pretty tricky. Based on some back-of-the-envelope estimates I conjecture that this attains a $\sim e^{-\Theta(n)}$ probability of placing all numbers in the correct position, which is better than the naive $1/n!$ algorithm. However I have no proof, so this could be wrong.
A deterministic algorithm
Here's what I can show. Let's consider a variation on the above algorithm: when you receive each number, find the range of array indices that wouldn't be obviously wrong, and then place that number in the exact middle (median) of that range. This algorithm is now deterministic.
For this algorithm, I can show that the probability of placing all numbers in the correct position is asymptotically larger than $1/n!$. In particular, it is at least something like $1/2^{\Theta(n)}$. Here's a hand-wavy analysis. For simplicity, let's assume $n$ is one less than a power of two. Place all $n$ numbers in a complete binary search tree. Now consider all sequences where you enumerate all the numbers at depth $1$ in some order, then all the numbers at depth $2$ in some order, then all the numbers at depth $3$ in some order, and so on. A hand-wavy estimate suggests that there are something in the vicinity of $n!/2^{\Theta(n)}$ such sequences, i.e., a random permutation has very roughly a $1/2^{\Theta(n)}$ probability of having this form. Moreover, the second algorithm above always succeeds in placing every number in the correct position. Therefore, the second algorithm above achieves a success rate of something in the vicinity of $1/2^{\Theta(n)}$. In particular, the success rate is significantly larger than $1/n!$.
For values of $n = 2^i-1$, we can characterize the exact probability of being correct: letting $P(n)$ denote the probability that the second algorithm succeeds in placing every number in the correct position, we find
$$P(n) = {1 \over n} \times P(\lfloor n/2 \rfloor)^2,$$
because the first element needs to be the median, and then you have two problems of half the size (the subsequence of numbers smaller than the median has to be a sequence with the same property, and the same for the subsequence of numbers larger than the median). We have the base cases $P(1)=1$ and $P(3)=1/3$. This recurrence relation grows like $P(n) \sim 2^{-e/2 \cdot (n+1)}$. Here $e/2 \approx 1.359\ldots$, so $P(n)$ grows roughly like $2^{-1.36 n}$, and in particular, much faster than $1/n!$.
In particular, here's one way to analyze this recurrence relation. Let $Q(n) = -\lg P(n)$. Then we find
$$Q(n) = 2 Q(\lfloor n/2 \rfloor) + \lg n,$$
where $Q(1)=0$ and $Q(3) = \lg 3$. Letting $R(i) = Q(2^i-1)$, we find
$$R(i) = 2 R(i-1) + \lg(2^i - 1),$$
with base case $R(1)=0$. Expanding, we find
$$R(i) = 2^{i-1} \times \left(\lg(1) + {\lg 3 \over 2} + {\lg 7 \over 4} + {\lg 15 \over 8} + {\lg 31 \over 16} + \dots + {\lg (2^i-1) \over 2^{i-1}}\right).$$
Summing the series, we find that to an excellent approximation
$$R(i) \approx 2^{i-1} \times e.$$
Therefore, $Q(n) \approx e/2 \cdot (n+1)$ and $P(n) \approx 2^{-e/2 \cdot (n+1)}$.
Credits: My thanks to @Algorithms with Attitude for the recurrence relation and the idea of analyzing the probability in this way.
