Hazard Return Period & Lifetime Risk

Explore how annual probabilities compound over decades and see how building codes solve return periods backwards.

Worked example — Hazard Return Period & Lifetime Risk

Expected Count n/T	0.30
Required Return Period	285

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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A 100-year flood is a 26% risk over a 30-year mortgage 🖖

A '100-year flood' does not mean an event that happens on a regular schedule once a century. It describes an annual exceedance probability of 1/100, or 1% in any given year. Over a 30-year mortgage, the cumulative probability of experiencing at least one such flood is 1 − (1 − 0.01)³⁰ = 26.03% — better than a 1-in-4 chance. Over a full 100 years, the chance of at least one event is 63.40%, while the probability of it never happening is 36.60%. The return period is simply the reciprocal of the annual probability, not a schedule.

The 63% constant: waiting one return period never gives certainty 🖖

If you set the exposure window n equal to the return period T, the cumulative risk 1 − (1 − 1/T)ᵀ converges rapidly to 1 − 1/e ≈ 63.21%. For T = 10, the risk is 65.13%; for T = 50, 63.58%; for T = 100, 63.40%; and for T = 500, 63.25%. Waiting one full return period yields a roughly 63% chance of experiencing the hazard, never 100%. The mathematical independence assumption — that each year is an independent draw from an unchanging distribution — is load-bearing here. In reality, climate trends and river catchment changes can alter the underlying probability distribution over time.

Building codes are this equation solved backwards 🖖

Inverting the relationship gives T = 1 / (1 − (1 − R)^(1/n)), where R is the acceptable risk and n is the design life. Structural and seismic codes have long been written around this inversion. Allowing a 10% risk over a 50-year design life requires designing for a 475-year return period event (the Design Basis Earthquake). Lowering the acceptable risk to 2% over 50 years pushes the required return period to 2,475 years (the Maximum Considered Earthquake). These famous return periods are not geological periodicities; they are a human risk tolerance passed through one line of algebra.

Problems solved in full

Two 100-year floods that hit the same town during one 30-year mortgage 6 steps
Two "100-year floods" hit the same town during one 30-year mortgage. How unlucky is that, and how many towns does it take before it happens to somebody more often than not?
1. A return period of T = 100 years is an annual exceedance probability of p = 1/T = 0.01. The Annual Chance card reads 1.00%.
  
  $p = \frac{1}{T} = \frac{1}{100} = 0.01$
2. Treat the 30 years as 30 independent draws. The chance of a clean run is 0.99³⁰ = 0.7397, so the chance of at least one flood is 26.03%. Both numbers are on the page, in the Chance of None and Lifetime Risk cards.
  
  $P (0) = (1 - p)^{30} = 0.9 9^{30} = 0.7397 ⟹ P (\geq 1) = 0.2603$
3. Exactly one flood means choosing which of the 30 years it lands in: 30 × 0.01 × 0.99²⁹ = 0.2242.
  
  $P (1) = (1 30) p (1 - p)^{29} = 30 \times 0.01 \times 0.9 9^{29} = 0.2242$
4. Everything left over is two or more: 1 − 0.7397 − 0.2242 = 0.03615. The Chance of ≥2 Events card reads 3.61%, roughly one mortgage in 28.
  
  $P (\geq 2) = 1 - 0.7397 - 0.2242 = 0.03615$
5. That is one town. Now take N towns, each on its own river with its own 1% threshold. The chance that not one of them doubles up is 0.96385 to the power N, and setting that to a half gives N = ln 0.5 / ln 0.96385 = 18.8.
  
  $(1 - 0.03615)^{N} = 0.5 ⟹ N = \frac{ln 0.5}{ln 0.96385} = 18.8$
6. Nineteen towns. At a hundred towns the chance that at least one of them takes two 100-year floods inside the same 30 years is 1 − 0.96385¹⁰⁰ = 97.5%, and the expected number of such towns is 100 × 0.03615 = 3.61.
  
  $1 - 0.9638 5^{100} = 0.9748$
Answer

Two or more is a 3.61% event where you live and a 97.5% event across a hundred catchments. Every card on this page conditions on one place with one threshold, and the quantity that carried 3.61% up to 97.5% was the number of rivers you were willing to look at. So "two 100-year floods in one lifetime somewhere" turns up in the news most years without being evidence that anything has changed; testing that needs the record of the one gauge, not the count of headlines. The arithmetic above also assumes the towns are independent, which catchments are not, since a single storm system floods neighbours together. Positive correlation pulls the true figure below 97.5%, and the page cannot tell you by how much: it models one site.

$N_{\geq 2} = 100 \times 0.03615 = 3.61$
The 475-year earthquake, and the building that outlives its design life 6 steps
A code asks for no more than a 10% chance of exceedance in 50 years. Work out the return period that demands, then decide whether the promise still holds for a building still standing at 100.
1. This is the first problem run backwards. There the hazard was given and the risk came out; here the risk is fixed by a committee and the hazard is what has to be found. Fifty clean years, with probability 0.90.
  
  $(1 - p)^{50} = 1 - 0.10 = 0.90$
2. Take the fiftieth root and subtract from one. The annual exceedance probability is 0.0021, and the Annual Chance card reads 0.21%.
  
  $p = 1 - 0.9 0^{1/50} = 1 - e^{(l n 0.90) /50} = 0.0021050$
3. The return period is its reciprocal: 475 years, which is what the Required Return Period card prints. No geology chose that number. A committee chose 10% and 50 years, and one line of algebra returned 475.
  
  $T = \frac{1}{p} = \frac{1}{0.0021050} = 475.06$
4. The shortcut most people reach for is 50 divided by 0.10, or 500 years, and the direction of its error matters. A 500-year event delivers 9.53% over 50 years rather than 10%, so the shortcut asks for a slightly larger design event than the code does. It also improves as the risk falls: at 2% in 50 years the exact answer is 2,475 against a guess of 2,500.
  
  $\frac{n}{R} = \frac{50}{0.10} = 500 ⟹ 1 - (1 - \frac{1}{500})^{50} = 9.53%$
5. Now leave the building standing for 100 years on the same hazard. Survival multiplies, so a clean century is 0.90 squared, which is 0.81. The exceedance probability is 19.00%, not the 20% that doubling would suggest and nothing like the 10% the code promised.
  
  $(1 - p)^{100} = [(1 - p)^{50}]^{2} = 0.9 0^{2} = 0.81 ⟹ 19.00%$
6. To hold 10% over 100 years the annual probability has to come down to 0.105%, a return period of 949.6 years. Double the service life and the required return period doubles with it, near enough, because at probabilities this small the exponent does all the work.
  
  $p^{'} = 1 - 0.9 0^{1/100} = 0.0010531 ⟹ T^{'} = 949.6$
Answer

475 years for a 50-year life, and 19.0% exceedance if the same building is still there at 100. The 10% belongs to the fifty years somebody wrote into the brief rather than to the structure, and it grows roughly in step with the years the building is kept.

Hospitals, dams and school buildings routinely outlive the design life used to size them, and nothing about the structure changes when they do. What changes sits in the exponent. The tool will show it: put 475 into the forward mode with an exposure of 100 years and the Lifetime Risk card reads 19.00%. So the honest way to quote a design event is with its exposure attached. 475 years is shorthand for 10% in 50, and detached from those fifty years it is a large number that sounds like a guarantee.

$T = \frac{1}{1 - ( 1 - R ) ^{1/ n}}, T (0.10, 100) = 949.6 \approx 2 T (0.10, 50)$

References (2)

What a "100-year flood" means, and the misreading the tool exists to correct Holmes, R. R. and Dinicola, K. USGS Fact Sheet FS-229-96, "The 100-Year Flood". U.S. Geological Survey.
Where designing to an annual exceedance probability comes from Cornell, C. A. (1968). Engineering seismic risk analysis. Bulletin of the Seismological Society of America, 58(5), 1583–1606. doi:10.1785/bssa0580051583

By the 42Math team · Updated 2026-09-03

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

30-Year Mortgage (T=100) - A 1-in-100 annual hazard carries a 26.03% risk over a 30-year mortgage.
100-Year Window (T=100) - Over a 100-year lifetime, a 100-year event has a 63.40% risk of occurring, and a 36.60% chance of never happening.
Seismic Code DBE (10%/50y) - Designing for a 10% risk over 50 years yields a required 475-year return period (Design Basis Earthquake).
Max Considered MCE (2%/50y) - Designing for a 2% risk over 50 years requires a 2,475-year return period (Maximum Considered Earthquake).