I can't initially solve this problem so I made a simpler problem and solved that to start thinking about it.
The image shows the odds for a bag of 6 balls with 3 pulls. The number in brackets is the odds of success for a given combination of 2 pulls. The T shows success after 2 pulls and the F a failure after 2 pulls.
By way of example, the yellow cell is pulling a 3 on ball one and a 5 on ball two. In that scenario, there are 4 balls remaining (1,2,4,6) and only one of them (6) is a successful set of three ascending numbers. So the odds of success after that initial combination are 1/4. The odds of any initial combination is 1/30, so we can multiply the figure in each cell by 1/30 and add the answers up to get the odds in total. That gives us 20/120 or 1/6 reduced, but I think keeping it as 20/120 is more useful for thinking about the answer.
The denominator is 6*5*4, which makes sense, and I'd expect that pattern to hold for the actual problem, so it should be 1,000,000! / (1,000,000-20)! as the denominator. Roughly the only thing I remember from taking stats/probability is that there were lots of factorial signs everywhere, so that's a good place to start.
Edited to add: I repeated this exercise with 7 balls choose 3 and 8 balls choose 3. I got 35/210 and 56/336 respectively. I think it's pretty unlikely to be a coincidence that those all reduce to 1/6.
There's a pattern there, and it would need more work to prove it, but my thought is 1/(20!), as with 3 balls no matter how many in the bag I got 1/(3!), and with 2 balls the answer is trivially 1/(2!).
Edited to add again: I repeated the problem as 6 balls choose 4, and got 15/360 = 1/24 = 1/(4!), which is pretty confirmatory to me. I've changed my mind about the denominator being relevant, and think the answer here is solely dependent on the number of balls you pull (assuming the number of balls is greater than number of pulls) and so for the problem as stated it's 1/(20!)