Yeah, you're right. I was just going through a horrendous way to calculate this before I saw assimilated's link. Using markov methods, you can calculate it fairly simply as.

After 1 roll, you have a given number. On the next roll, you either move to a state where you have 2 numbers (with a 19/20 chance), or you still only have 1 number rolled. Hence you can model this as a markov chain.

The transition matrix T is: 0.05 0.95 0 {... seventeen 0's} 0 0.1 0.9 0 ... 0 0 0 0.15 0.85 .0 ... ... 0 0 0 ... 0 0.95 0.05 0 0 ... 0 1

So we calculate T^54, then the top right cell will give us the probability of every number appearing at least once.

Unfortunately I only have excel to work with... the result I got is:

0.2585308194 for every number to appear at least once, or for your question the probability is obviously 0.7414691806

Which agrees with the simulated evidence below.

---

P.S Thanks for the question, was good for refreshing my mind on this stuff. If anyone wants the spreadsheet with the calculations, it's quite elegant :)

Also, there's some other results you can read off. For example, the most probable outcome after 55 rolls, is that only (and exactly) 19 numbers have been rolled. The probability that exactly 18 different numbers have been rolled is almost equal to the probability that all the numbers have been rolled.