Uncephalized demonstrates why it is tremendously foolish to attempt to “correct” one’s intellectual superiors in taking exception to my observation about it being astronomically unlikely that any individual present at one mass shooting in the United States will be present at another one:
I’m responding to Vox’s OP: The odds against one person in a country of 320 million being in the vicinity of two such events are astronomical.
Which is flat-out wrong–unless I am making some boneheaded error, which is always possible, and why I showed my work–and leads me to think Vox didn’t bother doing the math at all before jumping to a conspiracy as his explanation. I may be in error here–I’m sure someone will quickly point it out, if so–but by my math these coincidences are far from astronomically unlikely.
Las Vegas 2017 attendance: 20,000
Gilroy 2019 attendance: 80,000
I don’t know how many attendees were actually physically present at each event at the time of the shootings, but I’ll assume two thirds, so 14,520 and 52,800.
Proportion of US population present at LV shooting: 14,520 / 350,000,000 = .000041 or .0041{02465df28c20f4aad57bcc5b49594f9c07e66389abf8fcd2ca75e40861c18fe4}
Proportion of the population NOT at LV is the inverse or 99.9959{02465df28c20f4aad57bcc5b49594f9c07e66389abf8fcd2ca75e40861c18fe4}
Likelihood of one person being at both events is then: 1 – (.999959^52,800). Which is 88.8{02465df28c20f4aad57bcc5b49594f9c07e66389abf8fcd2ca75e40861c18fe4}. The number of times this apparently happened is 3, so it’s 0.888^3, or 70{02465df28c20f4aad57bcc5b49594f9c07e66389abf8fcd2ca75e40861c18fe4}.
In other words, through purely random chance it is more likely than not that 3 people who were at the LV 2017 shooting would also be present at the Gilroy shooting.
Making similar estimates about the LV-Parkland connection: 39{02465df28c20f4aad57bcc5b49594f9c07e66389abf8fcd2ca75e40861c18fe4} likely assuming an average family size of 4, 3000 attendees of Parkland and we are looking for a direct family member involved in both events.
The Borderline-LV coincidence has the lowest odds as I run the numbers, actually quite low, at 0.046{02465df28c20f4aad57bcc5b49594f9c07e66389abf8fcd2ca75e40861c18fe4} or about 1 chance in 2200, partly because the guy was actually killed, a much lower number than “was also there”.
I don’t know enough about the San Bernardino or Toronto events to start even making assumptions.
In this model the probability of the LV-Gilroy and LV-Parkland coincidences both happening is 0.70*0.39 = 0.27 or 27{02465df28c20f4aad57bcc5b49594f9c07e66389abf8fcd2ca75e40861c18fe4}. Just better than 1 in 4.
The very large numbers present at these festivals make for counterintuitive probabilities. The Borderline connection is the only one that even gives me pause, but even 1 in 2200 is not what I’d call “astronomically low”.
Uncephalized is correct about one thing. I didn’t bother actually doing the math beforehand because I didn’t need to do it. And I didn’t need to do it in order to have a general idea about the size of the probabilities involved because a) I am highly intelligent and b) I have an instinctual grasp of mathematical relationships. I knew the probabilities were astronomical, in much the same way you know a tall man’s height is over six feet without needing to actually measure him. The boneheaded error Uncephalized has made here is that he simply doesn’t know how to calculate the probability of independent events. But it’s really not a difficult concept. For example, if the odds of rolling a six on a six-sided die are 1 in 6, then the odds of rolling two sixes on two different six-sided dice are (1/6) * (1/6) = 1/36.
But before we calculate the probability of these two specific independent events, let’s get the base numbers right. The Gilroy Garlic Festival is a three-day event, so that 80,000 is reduced to 26,667 before being reduced another one-third as per Uncephalized’s assumption to account for the timing of the event. This brings us to an estimated 17,787 people present at the time of the shootings. Note that reducing the estimated 20,000 Las Vegas attendance by the same one-third gives us 13,340, not 14,520.
It’s never a good sign when they can’t even get the simple division right. Now for the relevant probabilities.
- Gilroy probability: Dividing 17,787 by 350,000,000 results in a probability of 0.00005082, or one in 19,677.
- Las Vegas probability: Dividing 13,340 by 350,000,000 results in a probability of 0.00003811428, or one in 26,237
- Gilroy AND Las Vegas probability: Multiplying 0.00005082 by 0.00003811428 results in a probability of 0.0000000019369677096, or one in 516,270,868.
You will observe that 88.8 percent, or 1/1.13, is very, very far away from 1/516,270,868, and 0.0000000019369677096 cubed – to account for all three dual survivors reported – is even further off from 70 percent. As I originally stated, the odds against anyone having been at both events are astronomical, even if we leave out relevant factors such as the fact that 11 percent of all US citizens have never left their birth state throughout the course of their entire lives and that Las Vegas averages 24,381 visitors from California every day.
I suggest that “astronomical” is a perfectly reasonable way to describe a probability of one in 516 million cubed, or if you prefer, one in 137,604,570,000,000,016,192,784,160. I also suggest that you refrain from attempting to correct me if your IQ is sub-Mensa level. And finally, I suggest that it is not “jumping to a conspiracy” to observe obvious and glaring statistical improbabilities.
