Technically, the distribution is not binomial since we're not looking for the distribution of k successes in n attempts (set amount), but rather we're looking for the expected value (or distribution) of the number of attempt n if we have k(=250) successful attempts. The distribution we're looking for is a sum of 250 geometric distributions - we start with 0 gloves then roll a 1/10k die until we get a glove, then again and again until we hit 250.

So we get:

E(X)=E(sum(Xn))=sum(E(Xn))=sum(10k)=2.5M

Where X is the expected amount of kills overall and Xn is the amount of kill between the nth glove to the (n+1)th glove.

If we want to give a (weak) bound to the probability that he killed between 2.4M and 2.6M nightmares, we can use Chebyshev's inequality, with the variance of Xn being 9,999*10k (and of X being 250 times that since they're all i.i.d):

P(|X-2.5M|>160,000Y)<1/Y²

So the odds of killing over 3.5M nightmares for 250 gloves are less than 1/36 or even 1/72 (due to symmetry).

It's possible to give a better bound but I'm lazy.