My brother-in-law is a meteorologist and was happy to chat with me about the conditional probability of being struck by lightning (see my previous post here):

However, I can say right off the bat that there isn’t going to be one right answer. There is substantial variation in both the microphysics and dynamics of moist convective systems. As a result, there is wide variation in the lightning frequency from different storms. Take, for example, so-called MCS’s (Mesoscale Convective Systems) that originate on the Front Range of the Rockies and propagate eastward in the nighttime hours, eventually affecting places like MSP, MSN, or ORD in the morning hours. These are profligate lightning generators, often producing essentially continuous lightning. Meanwhile, a completely different animal from MCS’s are baroclinic frontal convective systems. Often these will pass through with only occasional lightning (occasional lightning is officially defined by the weather service as no more than once every 2 minutes). Then, perhaps most interesting of all, there are some storms that generate copious cloud-to-cloud or cloud-to-air lightning but very little cloud-to-ground lightning. The specific reasons for the wide variation in lightning frequency and type are still the subject of research (we have generally poor observations of microphysical processes, especially those processes in violent convective storms).

So, the type of storm will affect your conditional probability of being struck by lightning.

My brother-in-law also helped me to compute the probability of being struck by lightning conditioned on whether there is a storm.

P(L|S) = P(S|L) P(L) / P(S)whereL = event of getting struck by lightningS = event that there is a stormWe need to estimate values for P(S|L), P(L), and P(S).P(S|L) is easy; It’s 1 (you can’t have lightning without a storm).Estimates for P(L) and P(S) can be obtained from the public sphere.A little searching reveals that ~ 400 people per year are struck by lightning. If we assume the strike incidents are independent and uniformly distributed across the 365 days of the year, then the number of people struck on any given day is roughly 400/365. An estimate for P(L) for any given day is then the number of people struck on any given day divided by the total US population. Using 310 million as the US population we get P(L) for any given day is (400/365) / (310×10^6) = ~ 3.5 x 10^(-9).An estimate of P(S) can be obtained from the so-called “thunderstorm-day” statistic, which is the number of days in some prescribed time interval in which thunder is heard at a given location (and hence a storm must have been within striking distance of that location). A little searching reveals that the thunderstorm-days per year of Washington D.C. is 32.8, based upon the last 30 years of data. If we assume the stormy days are independent and uniformly distributed across the 365 days of the year, then P(S) for any given day is 32.8 / 365 = ~.09. However, this is not a reasonable assumption: the estimate of P(S) for Washington, D.C., for a given day in July is ~.25

Given the above:

P(L|S) = 1 * 3.5×10^(-9) / .09 = 3.9×10^(-8)

I tried to extend this analysis to compute the conditional probability of interest:

P(L | O, S)

where O = event that you are outside. When applying Bayes Rule and rearranging the probabilities, I eventually get to expressions that depend on P(L), such as this one:

P(L | O, S) = P(S | L, O) P(O | L) P(L) / (P(A | S) P(S) ) = P(L) / (P(A | S) P(S) )

(assuming that P(S | L, O) = P(O | L) = 1 for our purposes). However, this is problematic, since estimates for P(L) (like the one that my brother-in-law reports above) implicitly capture the likelihood that people are outside, P(L | O, S). This is circular reasoning.

If you can shed additional light on this problem, please leave a comment.

March 9th, 2012 at 4:46 pm

I’m not sure that lightning implies storm with probability 1. See http://en.m.wikipedia.org/wiki/Lightning.

March 10th, 2012 at 7:18 am

Good point, Paul. Lightning implies storm with probability ~1(?) These are back of the envelope calculations. Thanks for keeping me honest.

March 10th, 2012 at 8:29 am

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