I watched a video of a guy playing russian roulette by himself. He pulled the trigger twice and the second time, it blew his head open. I'm assuming it was a 6 chamber revolver and he loaded 1 round.

What is the chance of him shooting himself?

for it to be 33.33%, each chamber would need a 16.67% chance of randomly being loaded completely independently of the other chambers, meaning that 0 rounds and 6 rounds are a possibility.

I think the probability is 1/6 + 1/5 so 36.67% but I think it might also depend on how the question is framed.
I'm kind of a dumbass and I haven't taken any statistics courses and it's really bothering me now.