Everybody up!

There are $10!$ ways the cards could be arranged at first.

$9!$ of those result in a pattern where you flip over all the cards.

The $9!$ comes from $9$ ways to pick the first card (not an Ace), $8$ ways to pick the one in the position it points to, etc. etc. ....

$\dfrac{9!}{10!} = \boxed{0.1}$

The arrangement of the cards represents a permutation $\pi \in S_{10}$ . You turn over the cards numbered $1$ , $\pi1$ , $\pi^21$ , $\pi^31$ and so on. Thus you will fail to turn over all the cards if the constituent cycle of $\pi$ that contains $1$ has length less than $10$ . Thus the only way that you will turn over all cards before stopping is if $\pi$ is a $10$ -cycle.

There are $9!$ $10$ -cycles and $10!$ permutations altogether. Thus the probability of turning over all the cards is $\boxed{\tfrac{1}{10}}$ .

Everybody else who has posted a solution here has written a better solution than this one. I wrote some quick and dirty python code to solve this problem. My reasoning was to search through every permutation (about 3 million possibilities which is small enough to brute force search through each and every possibility, although, would not scale well if given more cards) and go through the process of "turning them over" for each one. It is easy to see that when the ace is reached, a cycle forms. We are looking only for cycles of length 10.

from itertools import permutations

def valid(x):
    met = [x[0]]
    i = 0
    while x[i] != 1:
        i = x[i] - 1
        met.append(x[i])
    return len(met) == 10

valids = 0
for p in permutations([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], r = 10):
    if valid(p): valids += 1
print(valids / 3628800)

Which outputs $\boxed{0.1}$ .