[numeri]

EDIT: Nevermind...

EDIT2: Hmmm... maybe the geometric distribution could apply?

Consider the special case when we are ahead first to act in round 10. We would like to have enough of a buffer to diffuse the chance that our opponent will catch up. Since our opponent will not stop at any point without catching up, we can consider his turn as following a geometric distribution. (This is the number of trials before the first failure - exactly what would happen if our opponent continued to roll until rolling a 6.)

In general, the geometric distribution is:

P(n) = p(1-p)^n, where p is the probability of a failure. (1/6 in our example.) The mean of a geometric distribution is (1-p)/p , which would be 5 in our example.

Since we expect villain to roll 5 times on average, and gain 3 points for each roll, we would expect villain to get... 15 that final roll. Hmmm... So we should attempt to get 15 points ahead, right?