I'm not double checking this, but I think 1 more on the magnitude scale is 32 times the energy, so a M8 has 32,768 times a 5 (not really that precise). There is something else that is 10 times per magnitude increment: it may be the amplitude of motion on the Wood Anderson instrument for Richter scale, or it may be fault rupture area....

I think the 2 cents comments on this invalidating recurrence intervals are incorrect: stress release by small earthquakes is not how recurrence intervals are done. Instead, the long term average slip on a fault is determined (by offset dateable features). Then, the time and amount of slip in the last earthquake is determined. Then, simple math to determine how much strain has accumulated: enough for 1 m of slip, or 2, or what. The problem is whether or not the next earthquake will have as much slip as the previous: an earthquake may have 1 m of slip, and 1 m may have accumulated, but the fault does not fail until 2 m have built up (when it fails in a larger earthquake). As 2 cents mentions, the rupture area of a previous M8 may fail in a couple of M7.5s instead-and no reason why this can't be in the same day or year. In fact, there was a poster by Toppazada (sp?) at AGU last December where he suggested that the 2 large earthquakes 13 days apart in 1812 were both on the San Andreas (it is known from tree rings that one of these had to be on the San Andreas). (Dolan and Rockwell suggested that the second one was on the San Cayetano fault in Dec 2001 BSSA. It had previously thought to be in or near Santa Barbara Channel (and may have been there regardless). I tend to distrust the probabilities for the San Andreas fault: they are very low for Carrizo Plain and San Franciso-North Coast, and very high for Parkfield. I don't think this is understood as well as they once thought, so the very low areas (2 or 3 % in 30 years) should be increased, and the very high areas (90+%) should be decreased. Just to take into account that things aren't as predictable as one might like. Chris