Probability and Assassination Attempts: A Mathematical Case for Conspiracy

The idea that numerous factors in the attack occurred randomly is far less likely than that they were coordinated.

I was recently having some fun conversations with colleagues at the university about applying probability theory to rare events, and over the weekend I began thinking about how our discussions might apply to the recent Trump assassination attempt (and assassination attempts more broadly). This train of thought led me to some very unsettling conclusions, which I thought I would share on this substack. Perhaps you can convince me that I am wrong in the comments.

There is a little math involved in the argument, but I have tried to make it simple enough that just about anyone can understand it (I hope!). If you have any questions, please ask in the comments.

Three events must have occurred

I decided to begin with the basics of the Trump assassination attempt, and consider a minimum of 3 events that all had to occur in order for this assassination attempt to be possible. They are by no means the only ones that had to occur, and other events/combinations could be constructed, but I think these suffice for the present example.

The following three events had to occur:

• Event A: The secret service failed to secure the building/rooftop where the shooting occurred.

• Event B: The shooter chose that particular rooftop for the attack.

• Event C: The shooter was not discovered and stopped prior to the attack.

I'm sure that everyone can agree on these three points. As I said above, there are also other considerations, but for now let’s focus on these 3 fundamental events.

How probable were these three events?

Let's consider the probability of A, B, or C occurring in an isolated manner, as individual events that are independent of the others. In other words, what is the probability that each one would occur on its own? I will call the probability of event A occurring p(A), the probability of event B is p(B), and C is p(C).

Let's throw some numbers around, to use as an example. These are just ballpark wild guesses, they don’t have to be all that accurate to make the subsequent points.

• p(A)=0.1 ... basically there was a 1/10 chance that the secret service would fail to cover a particular potential viable attack position during this speech. 9 times out of 10 they would have secured it properly, but this time they did not.

• p(B)=0.5 ... it was a good shooting position, though quite visible and risky. If the shooter could have looked at the property beforehand, and considered all the options, it is maybe a 50-50 chance they would choose that rooftop.

• p(C)=0.1 ... ok, how does a 20-year-old kid carry off such a plan without discovery, bring a ladder and a rifle within shooting range of the target without being detected and/or stopped, and so on? I think most people would consider this to be a small chance. At least I would!

So I'm just throwing these numbers out there, without serious justification, to use as an example. Insert different numbers, if you like.

Hypothesis 1: The 3 events occurred by random chance

Let’s start with the simplest hypothesis, and the one that we will probably be expected to believe in the long-term. The probability of the attack occurring as it did is the joint probability that all three events occurred simultaneously, which is written as p(A⋂B⋂C). The symbol “⋂” means the “intersection” and simply denotes the co-occurrence of the events.

If A, B, and C occurred completely independently, by random chance, then we can calculate the joint probability as:

p1(A⋂B⋂C)=p(A)*p(B)*p(C)

where I write “p1” to denote that this is the probability for hypothesis 1 (random chance). Using our example numbers above, this would be:

p(A⋂B⋂C)=0.1*0.5*0.1=0.005

or a 1-in-200 chance.

Hypothesis 2: Events A and B are connected

Next we consider the case where A, B, and C are not necessarily independent of one another. For this example, suppose that the shooter could discover that this particular roof would be unprotected. Obviously the shooter would be far more inclined to choose this particular shooting position if he knew that it was unguarded. This changes the joint probability of A and B occurring together, p(A⋂B). In fact, we might assume that if the shooter knows the building will be unguarded, then the chance they will use it for the attack is certain. Therefore p(A⋂B)=p(A). In other words, the chance that A and B both occur is simple the chance that A occurs (that the building is unguarded). In this case we would calculate:

p2(A⋂B⋂C)=p(A⋂B)*p(C)=p(A)*p(C)

Using our example numbers:

p2(A⋂B⋂C)=0.1*0.1=0.01

or 1-in-100 chance. So using our numbers, this only makes the probability that all 3 events occurs twice as likely as if they were all to occur by random chance (as in hypothesis 1).

Hypothesis 3: All 3 events were coordinated

Now let’s consider the possibility that all 3 events were planned/coordinated. In other words, a conspiracy that implicates the protective services themselves. Of course even if they are skilled at arranging everything in advance, there is still a chance that the plot might still be foiled (for example, the shooter is randomly stopped by a patrolman and discovered). But let’s say this is all arranged quite skillfully, so that the probability of c, p(C), rises to p(C)=0.8. Meanwhile, the occurrence of A and B are guaranteed by the plotters, so they are practically 100% probability. So in this case

p3(A⋂B⋂C)=p(C)=0.8

or a 4-in-5 chance.

Relative probabilities

So far our probability considerations were restricted to just the 3 aforementioned events. There are of course many other factors that have to be included, in fact is is nearly impossible to account for everything. Therefore we cannot robustly arrive an an absolute probability that the assassination attempt on Trump occurred. However, we can compare the different probabilities against one another, and this relative probability is robust. To obtain the relative probability, all we have to do is divide one by the other. Thus we can compare hypotheses 1 and 2 using p2/p1=0.01/0.005=2. This means that the probability of scenario 2 working is twice as likely as scenario 1, all else being equal. And scenario 3 is p3/p1=180 times more likely to be carried out than the scenario in which everything occurs by random chance.

Consequences of improbable scenarios render themselves less probable

Ok, so we have considered the probabilities in each scenario by throwing out some numbers, and then comparing them with one another. However, there is actually more to the story than meets the eye. In particular, we can imagine that a manner of assassination attempt that is very unlikely to occur (such as p1 or p2) would have many more failures than successes if they were to be repeated often. In this case, on average the security services would be foiling very many attempts before every one that occurs, and we might imagine that it would cause them to become far more diligent in their protection measures. For example, p(A) would be expected to decrease substantially (they would be far less likely to leave attack positions uncovered).

Therefore in a regime in which all assassination attempts are carried off randomly, without any coordination between the requisite parts, the probability would go down even further owing to the consequences of having many failures for every success. While the quantitive magnitude of this feedback is difficult to estimate, we can at least understand the direction of the effect and take it into consideration in our analysis. In particular, the relative probability would increase, rather than decrease, by taking into account these kinds of feedbacks on probabilities.

Assassination attempts are far more likely to be carried out in collusion with protection services than without

What this implies is that it is significantly more likely that events are coordinated in carrying out an assassination attempt. Not just against Trump, but also in other cases. In other words, the relative probability that arrangements between the shooter and the protective services were made to coordinate the attack is dramatically higher than the chance that all of the elements came together randomly.

Any questions? Please add a comment…

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PS: Again, I will remind everyone that I simply plugged in numbers. They are made-up. If you don't like those numbers, then go ahead and plug in your own to see how they pan out. But I think we will see that the probability of coordination is always much much higher than all of these events occurring randomly.