I think the question is ambiguous. Does "any" mean "all", or does it mean "at least one"?

I've done some MonteCarlo simulations using 10^7 trials.

(1) Assuming "any" means "at least one", I get a probability of 0.889 (to 3 significant figures).

Melody found 8/9 ≈ 0.889

(2) Assuming "any" means "all", I get a probability of 0.111 (to 3 significant figures).

1/9 ≈ 0.111.

The logic of my MonteCarlo simulation is as follows:

1. Randomly permute the integers 1 to 15.

2. Test the integer in the ninth position against those in the first eight positions.

3. If the ninth integer is less than one or more of the first eight, score 1 for the "at least" sum; if it is less than all of the first eight, score 1 for the "all" sum.

4. Repeat steps 1 to 3 10^7 times.

5. Divide the sums by 10^7 to get the probabilities.