Open a single random VLC .xspf from folder? (Python)

Question

I have a folder containing:

A_Playlist1.xspf
A_Playlist2.xspf
A_Playlist3.xspf
B_Playlist1.xspf
B_Playlist2.xspf
(etc etc, there are ~50 numbered files each under "A" and "B")

I have two buttons (Button 1 = A_Playlist, Button 2 = B_Playlist). How would I get these to open a random Playlist from my directory in VLC?

When using command="vlc /home/user/Desktop/Playlist_folder/A_Playlist*.xspf", it only opens the first file (Playlist1.xspf), and queues up the remaining .xspf files. I would like to open a single one of them at random if possible?

Apologies if this is a stupid question. Pretty new here!

Thanks in advance :)