I am working on a tabletop roleplaying game in which damage is calculated by rolling a pool of 10-sided dice. If more than one of the dice in this pool show a result of "10", the character in question takes an automatic Wound.
I have been trying to calculate the probability of an attack inflicting an automatic Wound based on the size of the damage pool, and I'm not sure what that formula would look like. I know that if there are 2 dice in the pool the odds are 1%, as both dice need to show a result of 10. How would I calculate the probability of at least two 10s appearing when three or more 10-sided dice or rolled?
You can find the probability of zero 10's and one 10, and subtract that from one.
P(zero 10's) $= (9/10)^n$
P(one 10) $=n (1/10)(9/10)^{n-1}$
You get the $n$ in the second one because there are $n$ orders that result in one 10 and the rest not 10's.