if you run 1 copy 1/(30^5) & if you run 2 copies 1/(15^5)
1/30^5 is much smaller than 1/15^5. Thus, if your second calculation were correct, it would be more likely that you find no Void Contract to the last card when you put in 2 instead of 1. That doesn't make any sense (you already have to find one when you are 29 cards down).
I don't know how to calculate it out of my head (and I am too lazy to look it up on the internet), but the chance that you do not get a void contract must be smaller if you put in two than when you put in only one.