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.

May 8, 2008 at 8:42 am

Most candidates entering the police service have never fired a handgun and many departments do not have the budget for the ammo required for proper training. In addition, most departments have changed from revolvers holding six rounds to semi-automatics than can hold up to three times as much. A revolver is relatively awkward to shoot rapidly and accurately except in the hands of an expert. A famous old-time western police officer in the revolver era advised to “take your time in a hurry” , because you can run out of ammo very quickly. Conversely, with a high capacity semi-automatic , there is more of a tendency for rapid target acquisition and to keep firing rapidly until the adversary is no longer a threat. The number of rounds expended in this encounter seems to be merely a sign of the times.

May 9, 2008 at 9:53 am

I get the same answer, but because you need to assume that the police would stop firing after the individual was shot, isn’t the expansion more like:

0.999 = 0.3 +(0.3)(0.7) + (0.3)(0.7)^2 + (0.3)(0.7)^(n-1)

or

0.999=0.3(sum(1 + 0.7 + 0.7^2+…+0.7^(n-1)))

or

0.999=0.3(1-0.7^n)/(1-0.7)=1-0.7^n

or

0.001=nlog(0.7)

Did I miss something?

May 9, 2008 at 9:59 am

Come to think about it, why would you assume that each policeman has to be certain that he/she has hit the target. Is not the correct model one where the target is hit once by any of the policeman and not by all of them?

May 9, 2008 at 11:01 am

This is, of course, assuming that all shots are fired at random, and are also independent statistical events. Now, I would believe a bullet hitting a little bit to the left of the target would make the cop aim the next shot slightly more to the right.

Wouldn’t this change the math, sligthly?

May 9, 2008 at 11:17 am

Thor,

By Mj?lnir, it surely would.

The only thing I would say differently, would be to eliminate the words “at random” and “independent statistical events.” Your second sentence is the meat.

The big assumption in the model is that the probability of hitting a target is the same for all shots fired. If not, then the model is out the window.

Bernie,

You might be right. But I’m guessing that each cop does not pause after each shot and say to himself, “Did I hit my target? Or did my partner?” The entire event is over in less than a minute.

However, you could use something like a (so called) negative binomial or geometric and get what you’re after.

Actually, the best kind of hierarchical model would be one which incorporates three shooters, each having the same average hitting probability, but with some differences in the distribution. I mean, you could go on and on, but I’m not sure we would gain much (politically, I mean).

May 9, 2008 at 7:21 pm

.999? Idunno…I feel like if I was a cop, I’d be happy with .95, which takes 9 shots. If I couldn’t hit someone in one clip, it seems to me I’d think about switching tactics. But I’d like to hear a professional opinion.

And if anyone were to offer one, maybe they could tell me why police don’t use tracer rounds. Seems to me, if I were to want to protect something with firepower, I’d have tracers every 2 or 3 rounds. The closer I can get to “lasers”, the better.

May 10, 2008 at 1:30 pm

Sounds like job hunting to be expert witness for the defense – particularly in the civil trial(s). I hope you have the time for it as trial(s) will “probably” be ongoing for some time. At a minimum, counsels for both sides should be aware of the stats.

May 10, 2008 at 5:36 pm

Thor,

the police in this case were shooting at men in a moving vehicle. It was also at night under street lights.

In excellent conditions where the shooter has some feedback as to where he is hitting there may be adjustment. In this case, not quite random shots in the general direction is probably closer to reality.

distractme,

tracer rounds cause fires. As to tactics, they have already gone past verbal orders, physical grabbing, clubs, tasers, another police unit trying to block the moving vehicle… Calling in the SWAT team or an air strike takes time!!

My personal opinion (no law enforcement background) is that the real argument will be over whether there was sufficient reason to start firing the guns in the first place. The allegations that the victims rammed a police car will undoubtedly be investigated in depth.

May 10, 2008 at 7:14 pm

If they were baysian what would their prior be and when would they stop firing?

May 11, 2008 at 4:34 am

Steven,

If the cops were good Bayesians, and the probability of hitting the target was the same for each shot, then they would come to the exact same conclusion.

The stats above are about

observableswith the probability fixed and known. We would only need priors and whatnot if we were trying to estimate the probability of hitting. And even after we’ve done that, we still have to make statements about the observables (shots) again.May 12, 2008 at 12:36 pm

This is brilliant. And I’ve known many good Bayesian cops in my life.

May 14, 2008 at 3:39 am

There are times -many times in fact – when using stats to prove your point is plain ridiculous. You cannot get around the fact that they were trigger happy goons and I’m pretty sure you’d change your tune if you were the one being shot at. Couldn’t happen? There was a very similar case in London where the cops riddled a suspect’s car with bullets but it was a case of mistaken identity. Why did they use so many bullets? Because they were after a cop-killer. It’s nothing to do with statistical hit-ratios it’s just about being trigger happy. An American without a gun is a eunuch.

May 15, 2008 at 6:07 pm

I’m an old vietnam vet, and if you look at the statistics on how many rounds of small arms ammunition were fired in vietnam to kill one VC you would not be surprised at all by this.

JamesG. Good on topic comment. We eunuchs commend you for your intelligent comment.

Regards,

Ray

May 15, 2008 at 10:11 pm

But what if the cops were white and from the suburbs? Wouldn’t that alter the results dramatically?

P.S. When does the conspiracy theory contest start?

May 11, 2016 at 7:40 am

update

https://theconservativetreehouse.com/2014/03/13/trigger-happy-much-south-carolina-70-year-old-vietnam-vet-shot-during-traffic-stop-walking-cane-identified-by-officer-as-weapon/

because the internet is forever…