Nobody can predict the day of the next big earthquake. What scientists can do is estimate chances, using slip rates, records of past quakes and compound probability. A forecast such as "a 72% chance of a magnitude 6.7 or larger in the San Francisco Bay Area within 30 years" comes from this kind of reasoning.
Why it works
Suppose the chance of a large quake in any one year is $p$, independent from year to year. The chance of no quake in a year is $1 - p$. For $n$ quiet years in a row, multiply: $(1 - p)^n$ (an exponential function). So the chance of at least one is $P = 1 - (1 - p)^n$. You can't simply add yearly chances: adding counts the years with two or more quakes twice, and for long spans it can even pass 100%. (Real forecasts also let $p$ grow as strain builds, but the "at least one" logic is the same.)
Key idea
Chance of at least one event in $n$ years: $1 - (1 - p)^n$, where $p$ is the yearly chance. For a recurrence interval $T$, $p \approx \tfrac{1}{T}$.
Worked examples
Worked example: Two percent a year
A fault has a 2% chance of a large quake each year. What is the chance of at least one in 30 years?
- No quake in a year: $0.98$.
- No quake in 30 years: $0.98^{30} \approx 0.545$.
- At least one: $1 - 0.545 = 0.455$, about 45%, not 60%.
Answer: About 45%.
Worked example: From a forecast back to a year
A 30-year forecast gives a 50% chance. What yearly chance does that mean?
- $(1 - p)^{30} = 0.5$, so $1 - p = 0.5^{1/30} \approx 0.9772$.
- $p \approx 0.0228$, about 2.3% per year.
Answer: About 2.3% per year.
Common mistake
Adding the yearly chances is the most common slip: 2% a year for 30 years seems like 60%. The chance of at least one is $1 - 0.98^{30} \approx 45\%$. Check with an extreme: 5% a year for 30 years would "add" to 150%, which is impossible.