Bear with me here friends cause this is where I come when i have questions that im simply not smart enough to answer.
We heard a lot in the wake of the Aurora shootings about how if there was only someone in the audience who was armed the whole thing might never have happened. To be absolutely clear, I’m not here to argue whether or not that’s actually true. In fact, for arguments sake lets say it is true. To me, the problem with this as a public safety strategy still seems to be one of odds. What are the chances that an armed hero will just happen to be in that theater at that particular time? And to take this into the realm of the hypothetical, which is really the realm im interested in, if you wanted to beat the odds and reasonably ensure that there would always be a hero in any given movie theater how much of the population would have to be armed ?
The over 18 population of Aurora CO is roughly 250,000. Lets say there were 250 seats in that theater. What percentage of the population of Aurora would have to be armed in order to ensure the presence of an armed hero? (If this is basic math that im just too thick to get, sorry, I write novels for a living and barely passed algebra in HS)
Also, to help keep this all nice and hypothetical lets assume that every single gun owning resident of CO is very well trained, has great eyesight and is super cool under fire. All were concerned with is how likely is that he or she will even be present
In order to guarantee with absolute certainty that at least one person in the theater was a hero, there must be only 249 non-heroes in the entire city. If there were 250 total non-heroes, then there is an extremely (extremely!) small chance that all of them happen to be in the theater at the same time.
To be reasonably certain that at least one person in the audience is a hero, and assuming that everyone in the town has an equal chance to be part of the audience, is a bit harder. I’ll leave that up to someone with more math ability than I have.
If we assumed the probably of a random person is armed is p, then the probably of someone not being armed is (!-p). If we assume that the probability of one person being armed is independent of the probability of anyone else in the audience being armed, then the probability of no-one in the audience being armed is (1-p)^250. If we want a 50% of someone being armed, we solve:
(1-p)^250 = 0.5
250 log(1-p) = log(0.5)
log(1-p) = log(0.5)/250 = -0.0027726
1 - p = 0.99723125135206948615102477791874
That would mean there would be a 0.2769% chance of a given citizen being armed, or about 692 cititzens would need to be armed, out of 250000. In fact, the probability of 250 non-heroes is better than (1-p)^250, as if the first person isn’t armed, the second person has a probability (non-heroes - 1)/heroes of being a non-hero, and the next one (non-heroes - 2)/heroes, and so on. I’ll leave someone else to do those calculations.
The correct way to solve this is to use Poisson distribution. Assuming that there are 250 adults in the cinema, all of whom could be armed:
if 1 in 250 adults in Aurora walk around with a gun, you odds are 50-50
if 1 in 200 adults have a gun, odds are 99.97%