The views expressed here are my own, not those of my employers.

I have the impression some believe the conditional probability P_cond = P(“catastrophe in 2025”|“N near miss years before 2025, and 2025 is a high risk year”) increases with N, i.e. with the number of non-catastrophic high risk years before 2025. However, I think P_cond decreases with N because the evidence for P_cond being low increases with N:

• Consider the probability of a high risk year being catastrophic (instead of a near miss year) could be A) 50 %, or B) 5 %.
• Observing just near miss years over 20 years would be 375 k (= 0.358/(9.54*10^-7)) times as likely under B) as under A):
  • P(N = 20|A) = (1 - 0.5)^20 = 9.54*10^-7.
  • P(N = 20|B) = (1 - 0.05)^20 = 35.8 %.

In addition, one should update towards thinking the near misses were actually not near misses as P_cond decreases. Having lots of near misses without a catastrophe is paradoxical. If they really were near misses, one would observe a catastrophe before many of them. Without a catastrophe, the nearness of misses decreases as the number of misses increases.

Nevertheless, the unconditional probability P(“catastrophe in 2025”) = P(“2025 is a high risk year”)*P_cond tendentially increases with N. Using a more general rule of succession:

• P_cond = 1/(1/P_cond_0 + N) assuming no catastrophe, where P_cond_0 is the prior probability (equal to 1/2 for the standard rule of succession).
• P(“2025 is a high risk year”) = (1 + N)/(1/P_risk_0 + T), where P_risk_0 is the prior probability, and T is the number of years for which the catastrophe could have been observed.

As a result, P(“catastrophe in 2025”) = (1 + N)/(1/P_risk_0 + T)*1/(1/P_cond_0 + N) = 1/(1/P_risk_0 + T)*(1 - (1/P_cond_0 - 1)/(1/P_cond_0 + N)), which increases with N.

My sense is that some people believe P(“catastrophe in 2025”) increases with N because P_cond increases with N. In contrast, the above illustrates that P(“catastrophe in 2025”) will tend to increase with N because P(“2025 is a high risk year”) increases with N, despite P_cond decreasing with N.

In addition, the above implies P(“catastrophe in 2025”) is P_risk_0/(1/P_cond_0 + T) for no high risk years before 2025, and 1/(1/P_cond_0 + T) for infinitely many of them. Consequently, P(“catastrophe in 2025”):

• Becomes 1/P_risk_0 (= 1/(1/P_cond_0 + T)/(P_risk_0/(1/P_cond_0 + T))) times as likely as N goes from 0 to infinity, which would be 2 times for the standard prior P_risk_0 = 1/2.
  • In reality, N cannot be larger than T, so near miss years will not increase the probability of catastrophe as much as suggested above.
• Barely increases with N if this is large. It tends to 1/(1/P_cond_0 + T), which does not depend on N.