I came across the book Panicology, where “Two Statisticians Explain What’s Worth Worrying About (and What’s Not) in the 21st Century”. The back cover chastens the reader:
Terrorism?
More Americans have been killed by lightning or by peanut allergies than by terrorist attacks.
I’ve read this comparison in different forms many times; it is true, but misleading. The implication is that, because you don’t spend much time defending against lightning strikes, and that is reasonable, you are foolish to spend much time defending against terrorist attacks. Now let’s apply Drake/Backus equation reasoning to lightning and terrorism separately to see why these are not the same.
A model of getting hit by lightning during a given storm might include the following components
Pr(lightning hitting me given lightning generated and I’m outside)
Pr(lightning hitting me given lightning generated and I’m in the basement)
Pr(lightning generated given I’m outside)
Pr(lightning generated given I’m in the basement)
Pr(I’m outside rather than in the basement)
The probability of me getting hit by lightning is then 
. Now, I doubt my presence outside or in the basement affects the voltage potential between Earth and sky so much as to influence the frequency of lightning strikes, no matter how much iron is in my diet. This means that B and C are independent, so 
. Let 
, and factor out B. The probability of me getting hit by lightning is ![B \cdot [A_1 \cdot C + A_2 \cdot (1-C)]](http://s.wordpress.com/latex.php?latex=B%20%5Ccdot%20%5BA_1%20%5Ccdot%20C%20%2B%20A_2%20%5Ccdot%20%281-C%29%5D&bg=ffffff&fg=000000&s=0 "B \cdot [A_1 \cdot C + A_2 \cdot (1-C)]")
. Assuming I’m safer in the basement (
) I can minimize the chance of getting hit by minimizing C, the probability that I am outside. However, as long as 
is very small, C doesn’t matter, so I might as well go outside if I like.
How does this model fare when applied to terrorism? Not as well. The analogous model of being a terrorist victim on a given flight would include
Pr(terrorist successfully affecting me given terrorist attempt and I vote for less airport security)
Pr(terrorist successfully affecting me given terrorist attempt and I vote for more airport security)
Pr(terrorist attempt given I vote for less airport security)
= Pr(terrorist attempt given I vote for more airport security)
Pr(I vote for less airport security)
Again, the probability of the bad event is . However now 
; in fact, we expect 
because the terrorist has a greater chance of success given an attempt: . If we have no airport security then becomes large, so my choice C really matters.
Terrorism is not lightning. Terrorists respond to our choices about security; lightning does not respond to my choices about where I spend the rainstorm. Deciding whether peanut butter responds to one’s choices is left as an exercise for the reader.
Peanut butter has never approved of my choices.
I agree in spirit (terrorists are strategic and all that). But … as long as A1 or B1 are very small, the argument stands. The fallacy is in estimating the terrorism risk by looking solely at actual deaths. Which of course is nonsense. (Perhaps a good analogy would be nuclear facilities safety protocols.)
Eduardo, the argument stands if $latex A_1 \cdot B_1$ is small. My point hinges on the fact that most critics of airport security focus on $latex A_1$ being perceived as small but ignore that likelihood that $latex B_1 > B_2$. We have seen successful air-based terrorist attacks, not just on 9/11, so $latex A_1$ is not minuscule. If we set airport security lower one can reasonably expect /both/ $latex A_1$ and $latex B_1$ to increase, assuming a constant level of hostility on the part of the would-be terrorists.