Well, let's see...
(nCx)(.1^x)*(.9^(n-x)) x being number of times silence works and n being the number of mandrakes you have.
Since you only need it to work once, all we have to do is find the probability that it won't work, at x=0, and subtract that from 1.
For 10 mandrakes...
Let x=0 and n=10, so that's 1*1*.9^10= 0.3486784401, or 34.9%.
So There's a 65.1 % chance of it working, with 10 on the field...
Okay, you're right. Probably not OP, especially since it would cost 20
quanta for 10, assuming the first 9 failed.