The Sean Bell shooting and probability
Yesterday, there were several protests in New York City. The participants were “outraged” over the recent acquittal of two black cops and one Lebanese cop who shot and killed Sean Bell, who was black.
Much was made about the fact that the three cops shot at Bell’s car 50 times. This number was touted repeatedly by some as evidence that the cops had used excessive force.
Let’s look at this from the probabilistic viewpoint. It turns out that when a cop fires his weapon at a person, he only hits his target about 30% of the time. Anybody who has ever fired a weapon before, especially in an altercation, will know that this is a pretty good rate, but of course not good enough to guarantee that just one shot will be enough to stop a target.
So about how many times must a cop fire so that he is at least 99.9% sure of hitting his target?
Well, if he fired just once, he has a 30% of hitting, or a 70% chance of missing. If he fired twice, what is the chance of hitting at least once? Hitting at least once can happen in three ways: hitting with the first bullet and missing with the second; missing with the first and hitting with the second; or hitting with both. The only other possibility is missing on both. The probability of all these scenarios is 1 (something has to happen). So the chance of hitting at least once is 1 minus the chance of missing both. Or 1 - (0.7)(0.7) = 1 - 0.49 = 0.51.
This means that only firing two shots gives the officer a 50/50 chance of hitting his target. Not very good odds. He must fire more times to increase them.
It turns out that the same formula can be used for any number of shots. The probability of hitting at least once in three shots is 1 - (0.7)^3 = 1 - 0.34 = 0.66. The probability of hitting at least once in n shots is then 1 - (0.7)^n.
We want 1 - (0.7)^n to be at least 0.999. Or, written mathematically, 1 - (0.7)^n > 0.999. Now we have to recall high school algebra and solve for n. Subtract 1 from both sides and cancel the negative signs, which gives (0.7)^n > 0.001.
Now the hard part. If you don’t remember, just take my word for it, but now we use logarithms. So that we get n log(0.7) > log(0.001), or n > log(0.001)/log(0.7) = 20 (rounding to the nearest shot).
That’s right. In order for the cop to be pretty sure of hitting his target (and therefore ensuring his target does not hit him), a copy has to shoot at least 20 times.
Thus, given that three cops were firing, 50 total shots does not seem that unusal.
Note: one cop shot 31 times, on 11, and the other 8. Of course, the above analysis ignores all external evidence, such as how the probability of hitting decreases when aiming at a moving target, awareness by one cop of shots fired by another, whether the cops were well motivated, etc.
16 comments May 8th, 2008
