In a 1998 numerical integration by Dahlgren, 2483 Guinevere is the Hilda asteroid with the greatest chance of impacting with another asteroid.
http://adsabs.harvard.edu/abs/1998A&A...336.1056D

The collision probability is mentioned in table 6 on page 1063 as PH. But since it is using scientific notation, how do I convert that to a layman statement of once every X years?

Thank you.

Well, the chances are very low, because the probability is only based on the objects known at that time, and in the PH case, only looked at the Hilda population hitting itself, I think. The number they give is

28.7 x 10 ^ -18 in km^-2 * yr ^ -1

which, if I read it right, means that the chance depends on the cross-sectional area of the asteroid in question and its potential target. 2483 is 44 km in diameter, so has a cross sectional area of (44/2)^2 * pi = ~1520 km^2.

I suppose you could add up the cross sectional areas of all the rest of the Hildas to get a rough guess of the target cross section. In this paper I think they already accounted for that because they mention the probability with 909 Hilda asteroids.

So 1520 * 28.7E-18 = 43.6e-14

If that's the probability per year of a collision, it's really small. To find out the probabillity of a collision in 1 billion years

1 - (1 - 43.6E-14)^1,000,000,000 = 4.36e-5

Even 5 billion years gives only a probability of 2 in 10000 of a collision. Remember, this is just the probability of it hitting another Hilda asteroid, not necessarily the overall probability, or the probability of a generic asteroid impacting another. Also, the chance of any single asteroid having an impact should be slim, but over all asteroids given the age of the solar system the chances increase enough to form the collisional families we see.

The paper was written in 1998, but Wikipedia shows the current Hilda population as 1,100 so the probability isn't significantly higher due to new discoveries.