Math is hard, Barbie
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:
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 threeday event, so that 80,000 is reduced to 26,667 before being reduced another onethird 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 onethird 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.
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 subMensa level. And finally, I suggest that it is not "jumping to a conspiracy" to observe obvious and glaring statistical improbabilities.
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.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 sixsided die are 1 in 6, then the odds of rolling two sixes on two different sixsided dice are (1/6) * (1/6) = 1/36.
Which is flatout wrongunless I am making some boneheaded error, which is always possible, and why I showed my workand 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 hereI'm sure someone will quickly point it out, if sobut 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%
Proportion of the population NOT at LV is the inverse or 99.9959%
Likelihood of one person being at both events is then: 1  (.999959^52,800). Which is 88.8%. The number of times this apparently happened is 3, so it's 0.888^3, or 70%.
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 LVParkland connection: 39% 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 BorderlineLV coincidence has the lowest odds as I run the numbers, actually quite low, at 0.046% 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 LVGilroy and LVParkland coincidences both happening is 0.70*0.39 = 0.27 or 27%. 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".
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 threeday event, so that 80,000 is reduced to 26,667 before being reduced another onethird 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 onethird 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.
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 subMensa level. And finally, I suggest that it is not "jumping to a conspiracy" to observe obvious and glaring statistical improbabilities.
Labels: anklebiters, conspiracy, mailvox
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1 – 200 of 421 Newer› Newest»on the other hand there is the story of that poor bastard that survived the blast in Hiroshima... only to travel to Nagasaki just in time to do the whole thing all over again.
i'd love to see the probability on that.
@VD
I once read that astronomers treat the 1/10^20 events as "an unprobable to happen in the Solar System".
Wow.
Ow. My brain.
I had to grate my teeth while reading through the bad math. Why would complicating factors make things More likely. There are so many red flags for obvious errors in the calculations.
1776 to 1778. Let's go.
There is a high probability that VD did not do the calculations at first, because he was waiting for the first gamma to stick his cobweb filled cranium above the parapet.
on the other hand there is the story of that poor bastard that survived the blast in Hiroshima... only to travel to Nagasaki just in time to do the whole thing all over again.
It seems ironic, but it's not hard to believe from a statistical or practical perspective. With Hiroshima being destroyed, all the 200k survivors had to go somewhere. Nagasaki was one of the twelve largest cities left and Tokyo was being firebombed. Nagasaki was also closer than Nagoya, Yokohama, or Kyoto, so I would expect there were dozens of Hiroshima survivors in Nagasaki by the time it was bombed.
I don't know if linking is allowed but Jack Posobeic linked to an ex of Gilroy shooter talking about him and their relationship. I gave up reading and skimmed. Both with mental disorders. Her writing just screams give me attention.
Where was David Hogg this time?
If your initial conclusions fall under "absurd" Likelihood of one person being at both events is then: 1  (.999959^52,800). Which is 88.8% it's time to google how to do the math.
thanks Vox
I hate math, but Vox, that was easy to follow.
Despite the possibility of attention whoring, what is the probability of those three persons being publicly identified so rapidly? Were journos wandering around yelling "Hey! Was anybody here also in Vegas for the other shooting?"
I agree with the calculation of 1 in 516 million but not sure that it represents the odds of any American at both events but represents the odds of one specific American being at both events. 'Barbies' point is that there were 350 million trials which brings the odds over 50%.
In Holocaust class
The probability of these particular people being at multiple events is one to one. They exclusively attend mass shootings. Groupies, pfft.
As Vox noted, this is the best case probability. Other ANDs which reduce travel or event connections make the number a lot smaller. The O(26) value makes it easier to describe all of the Earth's physical properties from behavior of a pound of sand. It's more useful to examine the links among these outlers, to look for commonalities and causalities.
not sure that it represents the odds of any American at both events but represents the odds of one specific American being at both events.
They are the same thing in this regard.
'Barbies' point is that there were 350 million trials which brings the odds over 50%.
And that point is incorrect. There were not "350 million trials". After the first event, the subset was necessarily reduced to 17,787. No one who did not survive the first event was capable of surviving both events.
The first thing I saw wrong with the initial calculation: The assumption of independence is a risky thing in doing probability. What is the probability of overlap of anyone who participated in a Nathan's Famous Hot Dog eating contest and next week's attendance at any particular Hindu temple in the United States? It is not so directly calculated from the unconditioned probability of a random American being a participant in the Nathan's Famous Hot Dog eating contest and the unconditioned probability that the average American is a Hindu who attends services at a temple, because they aren't really independent.
There was a book about that. Last Train To Hiroshima. About 100 people survived both bombings, iirc.
Are crisis actors publicizing their appearances at major events to pad their resumes? All the probabilities involved make that idea more likely than random chance.
Man, have to give that dude respect for the hardcore commitment to the Gamma.
Only difference is nobody really gives a sh*t, especially when you can't even bother to get the math right.
Of course its possible to explain the presence of the same person at all three events in three little words... "paid crisis actor". Its a miracle if you happen to get a royal flush in a 5 card stud game three times in a row but not so impressive if you are allowed to secretly stack the deck before dealing each and every time.
Doktor Jeep gets the win on this thread
I agree with VDs math but there are some other missing factors. What percent of the us population go to music festivals? What percentage go to food festivals? What's the overlap there? The possibility of someone going to Gilroy in CA also going to a concert in Vegas is better than the statistic probability that any one person in the us would have. We would be giving equal weight to an elderly woman in a nursing home as to a foodie, country music fan that lives in SoCal.
Dear Uncephalized:
Given your colossal and obvious failure to "Correct the Math," I suggest you look up Bayes Theorem and check out the concept of conditional probabilities.
It could only help.
In probability theory, this is known as the birthday problem. The solution is not very intuitive.
No. That's not relevant here because there is no equivalent to the finite number of birthdays in a year. This is not the place to talk about things you have heard of, but don't actually understand.
I agree with VDs math but there are some other missing factors.
Obviously. However, the sum total of them will tend to lower the probability, not increase it. For example, how many of the people at both events were not even US citizens, given that four of the fatalities in Las Vegas were Canadians?
How mindnumbingly bad at math do you have to be to not find anything wrong about the numbers of 88% and 70%? This isn't even advanced statistics. Its literally assumed knowledge before you turn up to your first college class. There is no hope for dumbasses that confidently spout numbers that are not only wrong, but so completely wrong that it could very reasonably be made a crime in most civilised countries. The very basic model that Vox is using here is generous towards the probabilities. There is no coincidence.
Uncephalized's blog profile states he is a mechanical engineer. These are embarrassing mistakes to be making for an ME. As an electrical engineer I had to take Probability and Statistics for Engineers and Scientists. IF I was going to try and correct someone on probability, specifically someone more intelligent than me, I would at least break out my P&S textbook and college notes first.
Even treating the events as independent, the odds would be equal to the sum of the odds of being at either event, which is about 1:45k (.002%). Already very low odds that one person would be at two shootings, but the odds of it happening three times would be this cubed, if I remember right, or about 2*10^13.
Not exactly something I'd lay money on.
@31. I have independently calculated that the probability of that degree being fake is 88.8%.
@26: Valid point. We're talking about a target population under 330 million.
On the other hand, one may also assume that someone who has survived such an incident at one performance might avoid similar performances.
What troubles me more is the timing. Those of us who are firearm owners know that every single time it looks like we might get a break, one of these incidents happens. And there is a case before the Supreme Court dealing with the transport of firearms that could define a standard of scrutiny...which would put a spike in the willful blind eye that the lower courts have given the Heller decision.
I'm not given to conspiracy theories, but when it becomes this predictable, my suspicions are raised. I just haven't figured out how a notquitesane killer can be activated on command.
the odds would be equal to the sum of the odds of being at either event
Not the sum, the product.
If someone was numerate and didn't posture, would that make them... cephalized?
Very charitably, one could possibly assume that the sort to go to the GilroyGarlic Festival and the sort to go to country music concerts might have a significant overlap, but even if one did assume that, neither even has any overlap with Parkland whatsoever, and that would only change the math from 'utterly impossible during the life of the universe' to 'almost certainly not probable under ordinary cirsumstances.'
Uncephalized's blog profile states he is a mechanical engineer. These are embarrassing mistakes to be making for an ME.
That's not embarrassing, that's downright worrisome. I wouldn't want to drive over any bridge he designed.
To be fair, probabilities can be a bit tricky if you are not familiar with them.
Hammerli 280 wrote:To be fair, probabilities can be a bit tricky if you are not familiar with them.
They shouldn't be for a mechanical engineer.
And, perhaps one should not adopt a superior posture in "correcting" someone if something is "tricky." How hard would it have been in this case for him to get some local peer review? A buddy saying, "Nah, man, 88% is too high; check your math" would have gone a long way.
One error you are both making is the the total population of the US is not the right number to start with. It would be a number less than the adult, festival/event going populations of CA and NV. Not sure what that number is but its <350,000,000
Likelihood of one person being at both events is then: 1  (.999959^52,800). Which is 88.8%
Didn't even slow down at this "revelation"? Wasn't given pause at a near 90% probability of a random someone being at two geographically and temporally distant events? DunningKruger on 'roids.
Could Uncephalized actually be Leonor Flores?
The math shows the chance that, prior to both events occurring, a preselected person at random from the US population would go to Vegas (Independent Event A) and then also go to Gilroy (Independent Event B), with the assumption that every person in the USA would have the equal chance at being at either event, and that each person at both events were from the USA.
But that's not the question. The question is: What is the chance that any of the people who were already at Vegas also were at Gilroy, assuming that the people who were at Vegas each had the same chance of anyone else in the country being at Gilroy.
All of these assumptions aren't true, but they will give us an estimate.
Any single person of the 13340 who attended Vegas could have been any single person of the 17787 people at Gilroy. Given P(US Person At Gilroy) = (1 / 19677), then P(US Person Not At Gilroy) = 1  (1/19677). We then multiply the probabilities by the # of people, since we make the assumption of independence for each person at Vegas (again, not a good assumption): (1  (1 / 19677)) ^ 13340 = ~51% chance of nobody at Gilroy being at Vegas, or ~49% of any one of them being there given the massive number of assumptions.
The assumptions might make this probability more or less likely than it probably should be. The math can be a lot better if we exclude a percentage of children and folks that never leave their state, if knew the home city of each person at Vegas, their love of Garlic, etc.
Uncephalized's approach is correct.
We are not trying to compute the chance that a single person, taken at random from the US, would then attend both events. This is what Vox has computed.
We are trying to compute how many people from the 17k who attended one event (E1) also attended another event (E2). Or alternatively (what Uncephalized computed), the chance that none of the 17k people at E1 went to E2. If I use Vox's attendence numbers, I get that the average number of E1'ers that went to E2 is 0.67 people, and that there is a 49.2 chance that at least one person from E1 when to E2.
There are two key points to arrive at this. First, the probability of going to E1 is irrelevant. Everyone there is already there. That is, P(E1) for people at E1 is 1.0. Secondly, you are not trying to compute the chance at a single person attended E2. You are trying to see how many of the 17k E1 attendees did. Since each person at E1 has a chance to attend E2, you must sum the probabilities.
It is different from the birthday paradox, but it has the similar characteristics, and also produces an answer that is surprising.
I am commenting only on the math, given simplified assumptions of attendance probability.
Hammerli 280 wrote:@26...I'm not given to conspiracy theories, but when it becomes this predictable, my suspicions are raised. I just haven't figured out how a notquitesane killer can be activated on command.
Isn't that what MKUltra was all about?
Wtf... How did the guy not intuitively know that the odds would be astronomical? And he's supposed to be a mechanical engineer. That's strange.
@43 It's not the probability of a random someone being at two different events. It's the probability of any one of 14,520 people at one event being at another event with 52,800 people.
Vox
I can't express my respect for you highly enough, and I have learned so much from your writings and those of the posters.
I have learned that in many key parts of history you knew more about the history of (((my own))) people than I did! And that most of what I had been taught and assumed, was close to the exact opposite of the truth.
(For example, I was genuinely brought up to think that the USSR was a tragic saga of Slavic nonJewish brutish gangs of Ivans persecuting harmless bespectacled Shlomo who wanted nothing more than to quietly live in humble poverty, studying the Torah and praying to God. It is almost funny  in a bitter way  to unpick just how much disinformation is in the previous sentence)
On this math question, though, I think this is the oneinathousand freak exception that proves the rule. I think the other poster is broadly correct in his math, and Vox has made a very rare error. It is a probabilistic anomaly  almost a paradox  which is counterintuitive at first glance, even to UHIQ's.
It is similar to the "birthday paradox". With 24 random people or more in a room, the probability is above 50% that at least 2 share a birthday. Even though this seems insane at first glance  there are 365 days, so how can 24 ppl yield a likely matching birthday!? but when you start to calculate it out, it holds true, because every "pair" have their own chance of matching, and before long the combined probability rises past 50%. Remember, like the "terrorist only has to succeed once!", the birthdays only have to match once. This is not an identical puzzle, but it is similar in how counterintuitive the solution is.
If I have to roll a 6, and i have 6 chances to roll the dice.
Obviously, the chances are high that i'll manage it. But also, obviously, it's not 100% that I'll manage either.
The prob of rolling a 6 in 1 roll is 1/6
But just adding 1/6 to 1/6 to 1/6 for each of the 6 rolls would yield 1 after 6 rolls  ie 100% probability  and that can't be right, because rolling a 6 is not certain, even with 6 rolls!
There are deeper ways to calculate it, but a handy shortcut is to aim to work out the probability of NOT rolling a 6 at all  and then just subtract the answer from 1 to find the probability of rolling at least one 6.
No 6 on the first roll  5/6. Same for every future roll. Now we've reduced a complex, multithreaded issue into a typical simple probability question, where I need ALL of the events to go the way I want (NOT rolling a 6) so i simply multiply the probs. 5/6 x 5/6 etc = 5/6 (power of 6) = 0.335, approx a third.
That's the chance of NOT rolling a 6 at all.
So chance of rolling a 6 is 1  0.335 = 0.665 or about two thirds.
Apply the same logic to mass shootings, using Vox's numbers
Gilroy shooting 17,787 present
Vegas shooting 13,340 present
US pop 350,000,000
Let's start with Gilroy. After that shooting, we have 17,787 people who could each then be at Vegas.
The chance of ANY US person being in Vegas is 13,340 / 350 Mn = 0.0000381143
So starting with the first of our 17,787 Gilroy ppl  call him Gilroy #1  he has that tiny chance of being at Vegas. "ie rolling a 6"
What are the chances of him NOT being at Vegas? 1 minus that tiny number = 0.99994918
Now we look at Gilroy #2  him NOT being at Vegas is another 0.99994918
So too Gilroy #3.
Remember, we need to "win" every "roll" for there to be NO Gilroyeans at Vegas
So it's 0.99994918 multiplied by itself 17,787 times, = 0.99994918 (power of 17,787) = 0.5077
That's the chances of there being NOBODY at both. It is over half, but only just. So almost 50% chance that at least 1 person was at both.
Over many mass shootings, you would expect to occasionally get some double attendees.
All the best
MC
PS: My math probably needs some finetuning  but I think it's broadly correct
PPS: Having MORE than 1 double attendee is much more unlikely. But I don't think it would get to tiny probabilities.
I think you're wrong, Vox.
"Gilroy probability: Dividing 17,787 by 350,000,000 results in a probability of 0.00005082, or one in 19,677."
This probability is the probability that, if you pick a single person out of the US population, that person will have survived Gilroy. It is, obviously, not the probability that someone in the US has survived Gilroy, as we know that happened 19,677 times. Similar situation with the Las Vegas numbers. Multiplying them, the resulting number is also not the probability that it happened in the US, but the probability that it happened to a randomly selected individual taken from the population. Randomly pick someone from the US, and there is a one in 516,270,868 chance that they survived both. Looking at the US as a whole, with its 350,000,000 trials, that gives us around a 50.7% chance of that happening for that particular set of shootings once. It's important to emphasize /that particular set/ because for any shooting at a venue with an equal amount of people, there will be the same probability that someone will have survived the other as well. For every pair of equallysized shootings, flip a coin, heads there is a survivor of both, tails there isn't. If there are just two shootings, you flip once. If there are three, you flip three times. If there are four, flip six. It only takes four shootings with similar crowds to come up with three shared survivors on average.
Wow. I had stats 30 years ago, and I could still do better than that moron.
This is Liberal education. Probability is a difficult subject.
Midwits, smh
Civil engineers are the usual bridge designers. I wouldn't go near a bridge built by an ME like myself!��
This person basically said, "if you were at LV for the shooting then you'll very likely also have been at Golroy too!" and didn't even bother to figure out why that happened. The question then would have been why did it only happen 3 times?
MC is more correct. I looked at the wrong number for Gilroy survivors (no biggie), and misstated what the 50.7% referred to (a biggie, but also not so much as we're right around 50% anyway regardless).
So here's what I'm wondering..
Assuming it is not a significant statistical coincidence that they were in the area of another mass shooting, assuming instead that they were there for some purpose and with preknowledge of the shooting, what would be the purpose of them being there that couldn't have been accomplished without them being there?
Likelihood of one person being at both events is then: 1  (.999959^52,800). Which is 88.8%. The number of times this apparently happened is 3, so it's 0.888^3, or 70%.
With the power of mathmagic, you can explain anything!
@58 No  that's what makes the problem tricky.
The question is not "Choose one person at Vegas. What is the odds of that specific person being at Gilroy"?
It's "Choose every person at Vegas. What is the odds of any one of them being at Gilroy?"
Joe Smith wrote:This person basically said, "if you were at LV for the shooting then you'll very likely also have been at Golroy too!" and didn't even bother to figure out why that happened. The question then would have been why did it only happen 3 times?
This. 88% is a damned near certainty. You couldn't get that if you offered free avocado toast to the hipsters.
So almost 50% chance that at least 1 person was at both. Over many mass shootings, you would expect to occasionally get some double attendees.
Now apply this logic to four mass shootings and the probabilities that anyone from the first three shootings was at the fourth one.
In other words, why weren't there Parkland and Columbine and Virginia Tech survivors at Gilroy? And why weren't there Las Vegas and Gilroy survivors in El Paso, given the odds you give?
Uncephalized
"...Likelihood of one person being at both events is then: 1  (.999959^52,800). Which is 88.8%. The number of times this apparently happened is 3, so it's 0.888^3, or 70%..."
a lot of confusion would've been avoided had he written explicitly:
Likelihood of at least one person having been at both events ... is 88.8%
We've been looking at this as a math problem but precise use of language can make a big difference
:)
"Mathmagic" hurts my head.
We aren't just talking about someone from Las Vegas being in Gilroy, but what chance do they have of being in Miami or New York as well.
If the probabilities are as high for LA, Tuscon and what have as they are for Gilroy and for every other locale then there wouldn't be any of the Las Vegas crowd left in Las Vegas. They would all have to be somewhere else to account for such high probabilities of someone from the LV festival being somewhere else.
VD wrote:So almost 50% chance that at least 1 person was at both. Over many mass shootings, you would expect to occasionally get some double attendees.
Now apply this logic to four mass shootings and the probabilities that anyone from the first three shootings was at the fourth one.
In other words, why weren't there Parkland and Columbine and Virginia Tech survivors at Gilroy? And why weren't there Las Vegas and Gilroy survivors in El Paso, given the odds you give?
I don't know the answer to that.
Perhaps, people don't move around across states nearly as much as would be needed to make the initial assumptions valid at all.
Once we disregard the assumptions of randomness, factors like how many Garlicloving attendees at Gilroy would consider shopping in a Texan Walmart is a muchmore relevant issue than all the maths we have been looking at.
That's just a suggestion
PS:
I doubt anyone posting here wouldn't know that there is a Swamp that needs Draining, that those in charge would happily kill innocents in False Flags etc.
I and a few others were only commenting on the "Pure Math" angle  which is likely to not be that relevant anyway.
These calculations are based on the assumption that people are like gas molecules randomly bumping around. People's actions aren't random.
@67. There is no randomness at play with these crisis actors.
The odds of me adding anything to a math thread are approximately 1 in 350,000,000. But, instinctively, it doesn't add up. The mothman's appearances at disasters are more plausible than this pattern.
The guys math Vox dissects is clearly simplistic, but I' willing to accept this could legitimately be one of those counterintuitive probability scenarios like the one for two people in a modest sized group having the same birthday.
A weakness in Vox's otherwise more thorough take is that Las Vegas is a big destination, an attractor. Obviously to a much, much lesser extent the Garlic Festival is too. So it skews things from the implicit premise that 355M people are randomly distributed.
The odds of my being at a mass shooting in Las Vegas are lower, because I have no desire to ever go to Las Vegas. Maybe I could be talked into it, but the odds are higher for someone who likes the idea.
The odds of my being shot at a Jason Aldean concert are pretty much zero.
Math lessons aside (always welcome...), he seems to have missed the point entirely.
A weakness in Vox's otherwise more thorough take is that Las Vegas is a big destination, an attractor.
Hence my observation that about onequarter of Vegas's visitors are from California. But remember, we're not just talking about Vegas and Gilroy. We're also talking about Parkland and El Paso, Vegas and Thousand Oaks, Vegas and San Bernardino, Vegas and Parkland, and Toronto and Aurora.
Vox's initial observation is technically correct. But the question we want answered is the likelihood of ANYONE being present at both shootings. Uncephalized's method of calculation is correct.
Wow, Vox Popoli sure has a lot of secret Alpha Spergs
While there is some truth to all of these calculations, Vegas is a bad example to use just because so many people could be said to have 'escaped.' I myself know a family of five who were there and left immediately at the sound of gunshots. They even live in California. I'll ask if they went to Gilroy.
In addition, there are much fewer than 10,000 people at the festival at 5:30 pm. People come and go throughout the day.
What you should be calculating are the probabilities that these multiple witnesses are the ones who get on camera and get interviewed. That's where the probabilities become obviously skewed by enemy action.
Umcephalized's method is wrong. That's the incorrect way to calculate whether anyone from A was at B.
My dudes, the point here is that 88% is obviously the wrong answer. Unless it's a coded neo nazi message or something equally lame.
I'm not a math guy, but just from a common sense perspective i can tell you that these coincidences are, at the very least, very unlikely.
like Vox said, I don't need a ruler to know your tall.
Harambe wrote:My dudes, the point here is that 88% is obviously the wrong answer. Unless it's a coded neo nazi message or something equally lame.
Well, there were 14 words in your first sentence (if you count the %)
And in that sentence just there, too.
Just saying ...
What's the statistical probability that three cops shot and killed the suspect AND he committed suicide at the same time?
Did the suspect turn the assault rifle on himself, before or after the cops shot him?
https://www.mercurynews.com/2019/08/02/coronergilroygarlicfestivalshooterdiedfromsuicide/
Shouldn't you do this by region? Isn't there a lot of selection bias when talking about music festivals? One has to think that residents of California and Nevada are more likely to be at these festivals than a New York resident. The claim is not that a random person was at both at the festivals, but that a random person from the region who was at the first festival was also at the second festival.
One has to assume that not all demographics are equally likely to go to a music festival. I'm not sure what the band lineup was at Gilroy, but a classic rock band was playing when the gunshots started. There is probably a good deal of crossover between the two festivals. Modern country music is very similar to pop and classic rock.
this thread gave me cancer.
Now do Clinton's and suicides.
VD wrote:
In other words, why weren't there Parkland and Columbine and Virginia Tech survivors at Gilroy? And why weren't there Las Vegas and Gilroy survivors in El Paso, given the odds you give?
Pretty much because the assumption of this magical thing called randomness in probability is bollocks. Conservative/Dissident math blogger WM Briggs recently wrote a great book on the subject of overcertainty in both frequencybased and Bayesian probability called Uncertainty.
Each person at Event A is going to have some probability of going to Event B. Each of these probability is going to be different, depending on what is known about that person, what is guessed about Event B, etc. The math is then easy, do the combination and get the result. This means that Uncephalized's math is right, but the application of probability in both Vox's and Uncephalized's way is not wrong so much as not at all useful.
Probability really shouldn't ever be used to explain events that have already happened  you just look at Event A and Event B and see who was at both. And probability never can explain cause.
Instead, you have to come up with a hypothetical new event, and generalize the question to: "What are the odds of any one person at this specific mass shooting, which had X number of people, each of whom has these characteristics, is then at another mass shooting, in location X, with this number of people, etc." This is essentially Uncephalized's answer, but doing this after the fact, pretending that Gilroy hadn't happened yet.
Fundamentally, it comes back to the idea that "What are the odds of someone getting hit by lightning?" is not the same question as "What is the relative frequency of someone getting hit by lighting?" or any of the number of questions that a Bayesian could propose.
If you're standing on a golf course in a thunderstorm holding up your golf club, you're going to have better odds of being hit than if you're in the Bahamas on a cloudless day.
This is probably best viewed as a capture/recapture problem...but no calculator handy as I'm on the road right now, so no way for me to check to see who's more accurate (I'm reluctant to speculate, as intuition fails often with these types of questions).
10 years ago an old lady threw a pair of dice 154 times without rolling a 7.
I can wrap my head around one person being at 2 shootings as they happened. Weird things can and do happen and not in a linear manner. Multiple occurances though? That's suspect.
What are the odds survivors of mass shootings at concerts and festivals continue to attend similar events to blow off steam? I am going to assume they would rather order in pizza and binge watch Netflix.
10 years ago an old lady threw a pair of dice 154 times without rolling a 7.
Loaded dice or it didn't happen.
"0.99994918 (power of 17,787) = 0.5077"
No, .99994918 ^17,787 =. 000118375 or roughly a .012% or twelve thousandths ofa percent chance that there was a Gilroy attendee at Vegas concert.
The simplified math is 17,787 * 13,340 / 350,000,000 = 67%. Multiply to the power of the number of people in both places.
Personally I am not surprised that out of the thousands of crisis actors at each event, a couple of them are the same.
I'm supposed to know how to do prob and stats, but I stink at it.
But it takes a special kind of idiot to look at 70% odds and think they did the math right.
Your point about the lack of out of state travel by a significant number of Americans is important. I don't know the numbers but I do know that for an overwhelming number of ppl, most of their out of state travel has been contained to one or two states that join their own.
Also, what is the likelihood someone would be at one type of event that seemed to have a mostly 35 & under audience and an event in another state that was more geared to extended families with small children & grandparents?
My hobbies/likes are somewhat finite. If I'm into Civil War Reinactment,knitting and blue Grass music,I'm going to only travel to do those 3 things on my limited time/money.
Oh my gosh! Let's nitpick this to death shall we? All this bloviation is like saying "well that's not quite the CORRECT shade of purple....there's some nuance lost in YOUR description...."
What skews the probability a bit from the straight math VD did is overlapping sets, but not everyone who likes country music was at the LV music festival and just how big is the set of people who like country music would go to garlic festivals?
That messes with the numbers a little bit, but not nearly enough to get such a high probability unless those groups have high overlap and small numbers.
I suck at maths like your typical women tend to do, and even I knew that was all wrong.
Wouldn't the factor of time distinguish the Gilroy/Vegas Event Problem from the Birthday Problem?
On a long enough timeline, the probability of any attendee a particular event and then attending a second event declines to zero. The Birthday Problem doesn't have that factor.
I can see what Vox means in @27 when he says that there is no equivalent to finite birthdays in a year. It doesn't seem possible to use the same methodology for both problems.
Always pleasant to start one's day with several dozen strangers calling you an idiot. Better than coffee for getting the heart started. I do appreciate the couple of people who bothered to check the math, instead of working from the assumption that Vox is always right.
I'll happily defend my work here. Vox, your approach is ridiculous. You are calculating the odds of any given random person in the US being at both events with no prior knowledge. I am taking the pool of people who have already attended the first event, using that as my base proportion of the US population, and then running however many thousand independent eventsassuming random sortingagainst that probability.
For those who are allowing the large numbers involved to blind them to the scenario, this is the same situation: You have a bag of 100 marbles. You pick out 10 and mark a red dot on them, then put them back in the bag, shake the bag, and draw out five more one at a time, putting them back in the bag after each one. What are the chances that at least one of the marbles you pull out had a red dot?
Vox's approach says it's 1/10 times 1/20, or 0.5%. This is obviously wrong because it's smaller than the odds of drawing a marked marble in one try10%. Clearly if you draw multiple marbles your odds of drawing a marked one go UP, not down. The correct way is to take the odds of drawing an unmarked marble on each try90%and raise that to the power of the number of attempts5 tries. 0.90^5 is 0.59, the inverse of which is 41%, which sensibly puts the odds of getting one success in 5 tries much higher than one success in one try.
Even using Vox's much less generous assumptions about the number of people presentwhich is fine; I don't have better informationthe chance of at least one person from LV being present at Gilroy is 19%. Lower than my 88%, sure, but nowhere near 'astronomically' low.
As for those pointing out that people don't randomly disperse like gas moleculesfair enough. I didn't model that way because 1) it would have made my job impossible and 2) it probably would have supported my position anyway, and I was trying to give the other side a generous shot. I have a suspicion that the type of people who go to a music festival are much more likely than background odds to go to other festivals as well, for instance, and California borders on Nevada and people travel frequently between those states.
@61 MC "Likelihood of at least one person having been at both events ... is 88.8%
We've been looking at this as a math problem but precise use of language can make a big difference
:)"
True. I probably could have worded it more clearly. I was focusing on the math. Thanks.
@Vox "Note that reducing the estimated 20,000 Las Vegas attendance by the same onethird gives us 13,340, not 14,520. It's never a good sign when they can't even get the simple division right."
Yep, I made an error there. It doesn't change the basic thrust of the argument and your approach is still silly.
When this Probability Drive hits 88% you're gonna see some serious shit.
HouellebecqGurl wrote:My hobbies/likes are somewhat finite. If I'm into Civil War Reinactment,knitting and blue Grass music,I'm going to only travel to do those 3 things on my limited time/money.
Yes. I don't think it right to draw from the entire country. Not only are there geographic constraints, but not all people in a given geographic location are equally likely to attend a music festival. It is not truly random the way a throwing dice or drawing numbers is.
As i have written, this looks like someone who won the lottery twice.
On a lottery that goes on once every three months. By someone who do not buy lottery tickets
"Now do Clinton's and suicides."
The Babylon Bee already calculated this.
8chan owner says manifesto not uploaded by El Paso shooter
Vox, your approach is ridiculous. You are calculating the odds of any given random person in the US being at both events with no prior knowledge. I am taking the pool of people who have already attended the first event, using that as my base proportion of the US population, and then running however many thousand independent eventsassuming random sortingagainst that probability.
In other words, you incorrectly interpreted what I said and tried to perform a different calculation, which you then screwed up the math doing.
Now, what did I say?
The odds against one person in a country of 320 million being in the vicinity of two such events are astronomical.
That is precisely true. And THAT is what you said was "flatout wrong". Forget the math. You can't even read correctly. It's obvious to the point of retardation that a person who has been involved in one event is more likely to be involved in two events than someone who has not been involved in any.
Does it mean that there's a chance all 52,800 people were repeat mass shooting attendees? My calculator broke...
MC wrote:Let's start with Gilroy. After that shooting, we have 17,787 people who could each then be at Vegas.
The chance of ANY US person being in Vegas is 13,340 / 350 Mn = 0.0000381143
So starting with the first of our 17,787 Gilroy ppl  call him Gilroy #1  he has that tiny chance of being at Vegas. "ie rolling a 6"
What are the chances of him NOT being at Vegas? 1 minus that tiny number = 0.99994918
Now we look at Gilroy #2  him NOT being at Vegas is another 0.99994918
So too Gilroy #3.
Remember, we need to "win" every "roll" for there to be NO Gilroyeans at Vegas
So it's 0.99994918 multiplied by itself 17,787 times, = 0.99994918 (power of 17,787) = 0.5077
That's the chances of there being NOBODY at both. It is over half, but only just. So almost 50% chance that at least 1 person was at both.
Over many mass shootings, you would expect to occasionally get some double attendees.
Nobody but a dedicated Nerd would care at this point, but there was actually a "copy and paste" typo in my post above
The four references to 0.99994918 should of course read 0.999961886 (as any fule kno)
The other numbers are unchanged, including the conclusion
all the best
MC
Does it mean that there's a chance all 52,800 people were repeat mass shooting attendees? My calculator broke...
To a certain extent, yes. There have been 248 mass shootings in the USA in 2019. If Uncephalized's approach was correct, a sizable minority of the survivors would be previous survivors of mass shootings.
"I'll happily defend my work here. Vox, your approach is ridiculous."
Good god! He doubled down, but that's not surprising.
The mathematics from this guy were horrifically bad. Even middling grade school students would sense something off here very quickly. He did say "I may be in error here but...." to be fair but to not recognise at any point that the numbers were pants on head retarded is worrying considering somebody mentioned earlier he claims to be a Mechanical Engineer. That seems highly unlikely or highly worrisome.
@Vox "The odds against one person in a country of 320 million being in the vicinity of two such events are astronomical.
That is precisely true. And THAT is what you said was "flatout wrong". Forget the math. You can't even read correctly. It's obvious to the point of retardation that a person who has been involved in one event is more likely to be involved in two events than someone who has not been involved in any."
I am pointing out that this is a ridiculous way to frame the argument in the first place. Your ability to correctly multiply two small numbers to get an even smaller number was not in question. The problem is that the number you arrive at has nothing to do, conceptually, with the point you are trying to drive home.
I hope your multiple irrelevant personal insults are making you feel better about yourself. They certainly aren't doing anything to bolster your argument. I'll point out that you don't know my IQ any more than I know yours.
Plugged some numbers into wolfram alpha during lunch. If we assume the following:
350,000,000 Americans
17,000 at Gilroy
13,000 at Las Vegas
Possible attendance groups at Gilroy is: 350,000,000 choose 17,000
Possible attendance groups with no Las Vegas attendance: (350,000,000  13,000) choose 17,000
((35000000013000) choose 17000)/(350000000 choose 17000) is about 0.5318171067206 according to Wolfram alpha
1  0.5318171067206 = 0.4681828932794 probability that at least one person was at both.
Of course, the assumption that everyone in the country has an equal likely hood of attending both events respectively is super flawed.
@107 The numbers were basically right, *given* the assumptions (aka 'evidence' in probability speak). It's the assumptions that are wrong.
If:
Evidence E = "All Martians Wear Hats and Joe is a Martian"
Then the statement:
Pr("Joe Wears a Hat"  E) = 1.
That's the correct probability, since probability is just a number, and all probability is conditioned on (and ONLY conditioned on) the 'evidence'.
There is never, EVER, a Pr(Event). There is only Pr(Event  Evidence). Is the "evidence" part that is the sticker  and nobody's two "evidences" are the same.
Math can be hard but computers are good at math. I wrote a simple simulation based on the assumption that attendees are randomly chosen and ran it 1000 times. These are the results showing on the left how many attendees were at both events and on the right how many times that was seen in 1000 runs:
0: 104
1: 221
2: 263
3: 211
4: 118
5: 46
6: 25
7: 10
8: 2
Happy to share the source code (its only 38 lines of simple C++) if people want to see the assumptions explicitly.
Where's William M. Briggs when you need him.
I am pointing out that this is a ridiculous way to frame the argument in the first place. Your ability to correctly multiply two small numbers to get an even smaller number was not in question. The problem is that the number you arrive at has nothing to do, conceptually, with the point you are trying to drive home.
And you are completely wrong. I was, am, and always will be extremely suspicious of the fact that there have repeatedly been multiple people on the scene of multiple shootings.
You can utilize your incompetent math and irrelevant logic however you like. The only point that matters is that the correct math that you consider to be "a ridiculous way to frame the argument" points the logically inclined in the right direction. And you are not intelligent enough to even understand what I said in the first place.
The error is assuming the probability of Gilroy attendance is one when combing with probability of not attending V.
Calculation should be probability of G* (prob of not V ^ nG). Or .00005082 * (.999911688 ^ 17787) = .0000105636
So a 99.999 percent chance that no Gilroy attendee was at Vegas.
MC's example calculated the odds one of a random group of 17787 Americans was at Vegas (20% chance of that). The proper group is 17787 Gilroy attendees and you have to account for the .00005082 chance of an American being in that group to begin with.
These are the results showing on the left how many attendees were at both events and on the right how many times that was seen in 1000 runs:
Very good. Now run it to calculate how many attendees were at two of 750 events, to limit it to the last three years, on the basis of this logic.
Why the assumption attendance is random? I doubt that it's ignorance.
None of these calculations mean anything because none of the assumptions about prior probabilities are correct. Any estimation of conditional probability depends exquisitely on using the right cohort so you get your priors right. For instance, consider the probability that someone who went to a Grateful Dead concert in the 1980s was at a second Grateful Dead concert in the 1980s. The calculated probability is different if you calculate based on the cohort of Deadheads versus the population of the US. Similarly, what is the probability of someone attending a Billy Graham crusade in Asheville, NC in 1980 *and* attending a lecture by Anton LaVey in San Francisco in 1990? Much smaller than that of seeing a Deadhead twice at two Grateful Dead concerts, I suspect.
Because nobody knows these prior probabilities among the cohort of people who live in the West who have an interest in both country music and California arts and crafts festivals, VDs calculation and Uncephalized calculations are both useless.
In underdetermined cases such as this, common Pascalian probabilities are pretty much useless, and any inference of likelihood has to rely on categorical socalled "Baconian probability", which is basically a series of summations of positive versus negative evidence. It's a little like making a medical diagnosis, where one writer noted that if two diagnoses within a differential diagnosis are close enough in likelihood that an explicit calculation of probabilities is necessary, then they are too close together to determine which is most likely in any realworld situation.
"Good god! He doubled down, but that's not surprising."
Wall of text generator... activated.
> Why the assumption attendance is random? I doubt that it's ignorance.
I don't think anyone is pretending it's a good assumption, it's just that doing the math gets much more complicated if you don't make that assumption. You could attempt to do a simulation that tried to account for the probability of someone being out of state to attend an event but it makes the problem much more complicated.
It is unintuitive and surprising that the chances (assuming randomness) of at least one person being at both Gilroy and Vegas are actually not that low. Trying to make the math more realistic could change the probabilities in either direction but it gets complicated. For example certain demographics are more likely to attend these type of events (the very old and very young perhaps less likely than 2040 demographic) which would increase the chances of someone being at both but if they are geographically far apart or appeal to very different audiences then that would reduce the probability.
> Very good. Now run it to calculate how many attendees were at two of 750 events, to limit it to the last three years, on the basis of this logic.
I'd be happy to run a more elaborate simulation but need some numbers to plug into it. In this case the attendance of Gilroy and Vegas was assumed to be quite large (52,800 and 14,520) but if we were to include something like the Walmart shooting the numbers would be much lower and I'd expect the probabilities of someone being at both to be much lower too. I've done enough probability and statistics to know to be wary of trusting my intuition without crunching the numbers though.
@117 We don't need priors, we can just *look*, since both events have already occurred. Priors (our "Evidence") can be only be useful if we take a current event, imagine a future event, and then make a *decision* on that probability, since that is all the number is useful for. When that future event occurs, exactly as we specify, we either win or lose the bet (we also lose if the event doesn't take place).
Compute the odds on:
A student who was at the school during the Parkland shooting will be also be at a shooting where 10 or more people are killed by a purported single gunman in the state of Texas during the next 100 years.
*Everyone* that makes that bet is going to have a different set of priors.
Which is the point. There is not a single number that can be calculated of two people being at the two events, there are many, all depending on the assumptions made.
Stilicho wrote:The error is assuming the probability of Gilroy attendance is one when combing with probability of not attending V.
Calculation should be probability of G* (prob of not V ^ nG). Or .00005082 * (.999911688 ^ 17787) = .0000105636
So a 99.999 percent chance that no Gilroy attendee was at Vegas.
MC's example calculated the odds one of a random group of 17787 Americans was at Vegas (20% chance of that). The proper group is 17787 Gilroy attendees and you have to account for the .00005082 chance of an American being in that group to begin with.
I hear your point, but I stand by my numbers and calculations
*Assuming all behaviour was random*
The odds of at least one of ANY group of of 17,787 Americans you cared to select in advance, being in that Vegas shooting the next week are the same.
About 50%
It doesn't matter which group of 17,787 you select. It could be at random. It could be based on a characteristic. It could be based on "the 17,787 who were at Gilroy". It makes no difference. The odds of them having been at Gilroy in the first place makes no difference. in all cases, it's about 50% that at least one of the 17,787 will have been at Vegas.
I know it's counterintuitive in a big way.
If we can't agree, never mind.
All the best
MC
*PS It goes without saying that this "pure math" calculation is close to useless for present purposes. Obviously, there are massive "group" and "personal" reasons why people will or won't have been likely to have been at one event, or at more than one event.
On balance, people are probably much LESS likely to have been in different shooting events  hence Vox's point that most events DON'T have previous event attendees.
Whether this "tendency" is so strong to render the occasional doubleattendee very improbable  this is Vox's main contention  I can't say.
On the very narrow question of the "pure math" version of this issue  based on random human behaviour, and hence clearly fairly irrelevant to our purposes  Uncephalized is broadly correct, albeit not as clear with his language as he could have been.
In the reallife version of this question  too many variables for me to calculate
@1
"on the other hand there is the story of that poor bastard that survived the blast in Hiroshima... only to travel to Nagasaki just in time to do the whole thing all over again.
i'd love to see the probability on that."
Considering the number of people who travelled back and forth between the two cities, for business purposes (i.e. it was part of their regular routine), the fact that there were dozens of such people is not surprising at all.
Don't compare apples (small, specialinterest events) with oranges (entire cities)
@9
"Where was David Hogg this time?"
Out on his bicycle, going to get some "Broll", of course.
Taking the pool of 350 million people, choosing one at random, and calculating the probability that that specific person goes to the Vegas event, and then to the Gilroy event, does not give us useful information.
Calculating how likely a survivor of the Vegas event is to then go to the Gilroy event does.
The cases of people showing up at multiple shootings with far fewer bystanders are more interesting.
@16
"The probability of these particular people being at multiple events is one to one. They exclusively attend mass shootings. Groupies, pfft."
That's the point.
Taking the pool of 350 million people, choosing one at random, and calculating the probability that that specific person goes to the Vegas event, and then to the Gilroy event, does not give us useful information.
You're ludicrously incorrect. It's rather amusing how people marvel over my predictive track record, then when I happen to provide a glimpse concerning how I do it, all the midwits leap to explain how I'm wrong.
@33
"I'm not given to conspiracy theories, but when it becomes this predictable, my suspicions are raised. I just haven't figured out how a notquitesane killer can be activated on command."
Look up posthypnotic suggestion.
Realize that a posthypnotic suggestion does NOT need to be triggered only by the controller giving a command. The suggestion can, for example, be phrased so that when the subject recognizes a certain condition, he is to respond in a particular way (the military uses this form of posthypnotic suggestion in everything from marksmanship training (improving aiming accuracy) to tactics, to higherlevel things, like firstaid.)
It's simpler than you think. MUCH simpler.
If anyone wants to check the assumptions of my simulation or adjust the parameters there's a live version here: http://ideone.com/fQa4Xg (I had to reduce the number of trials to 100 to avoid it timing out on the live version).
The fundamental point is that calculating a 'postevent probability' is useless thing to do. Some of the people who were at Gilroy were at Vegas. Why?
If the causes can be clearly documented, e.g. "Bill lived near the festival and goes to Vegas every other weekend", we can chalk it up to bad luck. If the causes are nebulous, e.g. "Bill lives in Silver Spring, MD, was only ever in Vegas once for the shooting, flew out to California on the spur of the moment to see the Garlic festival even though he gets hives when he eats Garlic", then a good investigative journalist should take a look at Bill.
Given that investigative journalists have no interest in these people (or have been told to keep quiet), it's unlikely we'll ever see good answers. I wonder if the journalists that put together the 'gaming the lottery' research would be interested.
https://gamingthelottery.org/
@MC the probability of not V raised to 17787th power is about 50 percent (not 20 as I wrote above) .99996188572^ 17787 =.5076558. Multiplying by the odds of an American being a gilroy attendee to begin with yields. 000025799. Therefore the odds of are about 99.74%that there was no Gilroy attendee at Vegas. I made an error in the calculation. The logic is the same and remains valid. As I said, your model just accounts for the odds of one of a random group of 17787 Americans being at Vegas it doesn't account for membership in the group we are discussing (Gilroy attendees).
BadThinker wrote:@117 We don't need priors, we can just *look*, since both events have already occurred. Priors (our "Evidence") can be only be useful if we take a current event, imagine a future event, and then make a *decision* on that probability, since that is all the number is useful for. When that future event occurs, exactly as we specify, we either win or lose the bet (we also lose if the event doesn't take place).
No. The term "Prior" doesn't have to do with time, it has to do with whether or not the evidence is taken into consideration, i.e. "before" taking the evidence into consideration, not "before" the evidence occured. Thus, for instance, if I see a dead person who has an enlarged and hypertrophied heart, I can ask "What is the probability that this person died of complications of hypertension (in the absence of intervening obvious cause)?" The probability of this person having died of a complication of hypertension depends on the prior probability that a person in his cohort have an enlarged heart regardless of whether they died of complications of hypertension, and the prior probability of people in his cohort dying because of hypertension regardless of whether or not they have an enlarged heart (e.g. of a ruptured berry aneurysm). None of this looks to the future. The body's on the autopsy table in front of me. Similarly, when one uses this in DNA identification, one is making statements about probabilities regarding past events, e.g. what is the probability that this is the man who assaulted this woman, not who might be the man who might assault this woman in the future.
That's not embarrassing, that's downright worrisome. I wouldn't want to drive over any bridge he designed.
@37 VD
Hopefully he doesn't have the same emotional investment in designing bridges as he does when attempting to prove that the crisis actors interviewed by the fake news aren't full of it.
@56
"what would be the purpose of them being there that couldn't have been accomplished without them being there?"
Onsite coordination. Preparation of the battlespace (Both intelligence preparation, and such little things like, say, "accidentally" removing or deceptively placing signs that point the way out. Intrusion on the security/emergency radio channels (Intrusion is when a station which is NOT part of your organization gets on your net, and starts sending counterfeit messages(*), so as to confuse the organization that's using the radio network for as part of it's work, in this case, to improve the efficiency of Operations and Intel). A team of network intruders would be best, so that you don't have all of the false information and instructions of misdirection coming from the same voice. Once you've got more than, say, 5 people on a a radio voice net who don't work together frequently, it's rather easy for outsiders to intrude on that net and cause preplanned chaos if there are no authentication procedures in place. Most networks in which the stations are handheld radios have NO authentication procedures, other than signal encryption. And in such a case, if the network DOES have signal encryption, then most users on that net will automatically assume that any voice they can hear is a legit source, because they obviously have the same communications encryption key in their radio, so they "must" be part of the security team, right?
This is why losing a COMSEC cryptographic device is considered to be a FAR more serious issue than losing a weapon.
(*) such as: false observations, false sitreps, phony locations, and counterproductive instructions to both the security team, and then, if one also obtains access to the local responder's network, flooding them with similarly counterproductive messages.
All of these things are ambushes. And as shown in Vietnam and Iraq, among other places, not all of the people on the ambushing side need to have weapons. Others can assist those who do have weapons by channelizing those meant to be in the kill zone. Clearing fields of fire (lines of sight) ahead of time so that the shooter(s) have maximum effectiveness against those in the kill zone. Staying in a predetermined "no kill zone" during the gunfire, and then appearing in the midst of it all right after the shooting is done to further the chaos through
MC's math combined with wall of text is an absolutely hilarious read. There are 17k over 350 million different possible groups of 17k Americans... an astronomical number.
And he is saying that it's 50% that any one of these groups was the group in the event?
The gamma math... For the rest of the people probabilities still add up to one.
Nate wrote:this thread gave me cancer.
No, Nate, it was an overdose of lewd selfies.
@86
"What are the odds survivors of mass shootings at concerts and festivals continue to attend similar events to blow off steam? I am going to assume they would rather order in pizza and binge watch Netflix."
Bingo.
Q has been warning for months to avoid mass events such as these, for precisely the reasons evident 3(!) times in one week have shown.
Meanwhile, deep in the woods in Virginia, we lost an officer (and a good one at that) when a storm came through and lightning hit a tree within the battalion headquarters AO, landing on him, and the battalion S6 section's "office" tent.
99.9974% chance that there was no gilroy attendee at Vegas. Looked at the other way a .000025799 probability that there was a gilroy attendee at Vegas. About 1 in 50,000. Now I'm going to stop and get back to my workout.
After working through this guy's argument, I'm going to walk back my claim of mathmagic. It was counterintuitive precisely for the reason he explained, but I understand his reasoning now.
@95
"I can see what Vox means in @27 when he says that there is no equivalent to finite birthdays in a year. It doesn't seem possible to use the same methodology for both problems."
The equivalent to the birthday in the birthday problem is, in this case, location. And for all practical purposes, the number of locations on earth are "countably infinite" (Aleph 0), but can be grouped into, say, 10m x 10m squares, (of varying probability... for example the north and south poles have extremely low probabilities for anyone being at either of them, as well as various random points in the middle of the various oceans).
But the number of 10x10 meter squares is even within the places where North Americans congregate is so much higher than 365 that the parallel is similar in form only, but most certainly NOT in magnitude.
@stickwick, sure it's counterintuitive, but it calculates only the odds of one of a random group of 17787 Americans being at Vegas. It does not account for membership in the relevant group (Gilroy attendees). How would you account for membership in the Gilroy group?
A lot of people seem to have difficulty computing the probability that any one of the people in the first event was in the second event.
The probability of any one of the people who was in the first event being in the second is (if independent): P(Was in the second event). Hence the asked probability is (1P(Was in the second event))^#people in first event. I get 36% using the approximate numbers.
When I saw the original post, the only probability I calculated was the astronomical likelihood of there being a math battle over the numbers, looks like I was correct.
@Dole "And he is saying that it's 50% that any one of these groups was the group in the event?"
No, that's not what either of us is saying.
@Stilicho "How would you account for membership in the Gilroy group?"
That would be the exponent that you use to multiply the base probability. The way you're suggesting is doubledipping skewing the numbers way down.
@Stickwick "After working through this guy's argument, I'm going to walk back my claim of mathmagic. It was counterintuitive precisely for the reason he explained, but I understand his reasoning now."
If I'm 'this guy', then thank you for taking the time to think it through.
@Stilicho I think a simulation like this http://ideone.com/fQa4Xg makes it clearer what's going on. If you don't read C++ I can try and explain in plain English.
Let's give every individual in the US an identifying number from 1 to ~330 million (population of the US).
Then let's pick a random set of 14,520 Vegas attendees from that set of id numbers.
Then let's pick another random set of 52,800 Gilroy attendees from the set of id numbers.
Now we want to see if anyone was at both events so we find all the id numbers that are in both sets and count how many that is. One way to do this is to go through each Gilroy attendee and see if they are in the Vegas set. In Uncephalized's math which I think is roughly correct that is where the raising to the power 52,800 comes from because to find the probability of *nobody* being at both events, every person in the Gilroy set has to not be in the Vegas set.
Running the simulation 1000 times seems to give me results that roughly agree with the ~89% probability that at least one individual was at both events.
There's lots of ways this is not a good model as it makes a lot of simplifying assumptions that are not realistic but on the pure math version originally being discussed it is true that it is much more likely than not that at least one person was at both Vegas and Gilroy events, given the numbers originally used.
@stickwick, sure it's counterintuitive, but it calculates only the odds of one of a random group of 17787 Americans being at Vegas. It does not account for membership in the relevant group (Gilroy attendees). How would you account for membership in the Gilroy group?
I don't know how I can explain it better than Uncephalized already has. Having worked through the argument, my conscience demanded I walk back my inference that Uncephalized made a mistake, but I have no interest in stepping into this probabilistic quagmire any further than I already have.
"That would be the exponent that you use to multiply the base probability"
Nope. I've already shown that that calculation is the probability that a member of a random group of 17,787 Americans was at Vegas. The exponent only accounts for group size. Your model does not account for membership in the group.
If I'm 'this guy', then thank you for taking the time to think it through.
Yes, sorry. This kind of thing can be wildly counterintuitive, and I made a snap assessment.
The correct answer is 88.8%. I'd bet my life on it.
More on the fakenews narrative.
Five HUGE questions blow apart the official fake news narrative about the El Paso WalMart shooting
@Stilicho I'm not clear on the distinction you're making. Would you mind explaining what you mean by 'membership in the group'?
Show your work Markku
Is this not a hypergeometric problem? There are N people in the entire population (350,000,000). There are K people who survived Las Vegas (13,340). There were n people in attendance at one day in Gilroy (17,787). There were reportedly k people who survived both shootings (3). Run it through the equation and I get 2.6% chance of there being three people who survived Las Vegas being in attendance at Gilroy. This is just through utterly random selection.
Ignore all the counterpoints and extra variables that people have brought up before this that throws this all into the weeds. All of the extra variables drive the chances of people having survived multiple shootings squarely into the “skeptical” realm. No argument at all there. The “odds against anyone having been at both events” is astronomical on an individual level. No argument at all there either.
@Stickwick no worries. There is a lot of that going around and it seems to be contagious. ;^)
@HoosierHillbilly I'd call that a nice succinct summation of my argument, yes.
And would add that the percentages that come out are so extremely sensitive to the sample sizes used that small changes in the assumptions make very big changes in the outcome, but nowhere near the low likelihoods that Vox is claiming are the relevant ones.
We have a HUGE thread on this at the other place. If I posted all of it, it would probably make three screens of text. In short, I arrived at similar math as Uncephalized, except I had the probability and the exponential in reverse order. I don't quite understand his logic of putting them in the order that he did, but since the probabilities are independent, it doesn't matter. First, we take the population of the first event. Due to how we select the population, the probability of being at the first event is 100%. Then, we take the second probability, for "there exists" or "there is more than one", and invert it. We apply that to every one at the first shooting, and finally invert again. Which gives 88.8% .
Come on, there was a mistake. To someone who has some statistics, the answer are easy. This is why too many papers are goofed up not enough stats. Look at #4450
Mistake: We take the probability of "there doesn't exist" and invert to get "there exists".
Another mistake: "there is one or more" and not "there is more than one"
@Hoosierhillibilly Agreed, the walls of text over basic probability are ridiculous. Is the probability you reported of exactly three people, or three or more people? I think it would be charitable to consider the cumulative probability.
@109 is the best and clearest explanation.
As it was described to me, and why I came here, was that the probability of 88.8% for ONE or more was the controversial one. I didn't want to touch three or more, because although I could handwave it with the additive rule, that's an approximation. What I'd really need is the binomial probability, and and I'm too lazy for that. If I still had Matlab I'd do it easily, but I don't anymore. Yes, I could make an environment for SciPy too, but I'm also too lazy for that.
@132
I don't think I was clear about what I meant related to 'future' events. In your example, you have a simple statement of probability: Pr(PQ), where P = "Man Died of Complications related Hypertension", and Q = "all of the priors (this man has a enlarged heart, lots of men similar to him don't have enlarged hearts, other reasons men similar to him have died, etc), plus the fact you have a dead man, and every other assumption."
Sure, you can use a Bayesian model to come up with a number that estimates your uncertainty in a parameter of a model, then come up with some way of saying that the parameter estimates some observable. All of this is wrapped up into a probability model used in the calculation of a number.
William M. Briggs talks about how to use probability models in detail in chapter 10 of Uncertainty:
"Probability models, just like their casual and deterministic brethren, can and must be verified. That means only one thing: making predictions of observables never before seen. I mean never as in never... If the model makes useful predictions  where usefulness is related to the decisions one makes with a model  the model is good, else it is bad."
So if you are using your model to figure out how to screen to prevent deaths from hypertension, you might take some actions based on the model, and then you might see fewer deaths that could possibly be attributed to hypertension. So maybe that could be a good use of the model, if you're using it to make a prediction about men *like* that Man, but even then you have to modify your P some...
But your probability model is different than the model being described here.
This model starts with a a P of "A person at one mass shooting is also at another mass shooting" and a Q of "???"
When you put past events into P, and you make up some Qs, you can build a probability model. But that model is useless. Because you can just look, and see that Bill was at Event A and also at Event B.
@161 The chance of there being exactly 3 is 2.6%. I don't want to play in the mud any more than the initial report of three people. Just muddy the waters.
Well, you have to calculate it as "three or more", not "exactly three". Because the question is, what number is surprising? If 3 is a surprising number, then so is 4, 5 and so forth. So, very traditional binomial probability calculation.
@165 my simulation makes it look like ~26%, are you sure you're not missing a factor of 10 somewhere?
No argument there mr. Marrku, but a combox is a terrible place to show math. Just leave the one and stop there. If you want to banter more, you know where to find me.
Vox, your math is wrong. You're calculating the probability that an American picked at random was at both events instead of the probability that an American from the first event was also at the second event.
Uncephalized: your calculation shows the probability that one of 17787 randomly selected Americans was at vegas.
It assumes all 17787 were at Gilroy, but we know that the odds of being at gilroy are not 1. You are not accounting for the probability of membership in the gilroy group, you are assuming it.
I accounted for the probability that someone would be a member of the gilroy group to begin with .00005082 and multiplied it by the probability that one of a group of 17787 would be at Vegas.
Another methodology would be to apply the same calculation to the gilroy group that you applied to Vegas the multiply the products. That yields a 25% chance that the groups intersect at one person. A 6.6% chance that they intersect at points. A 1.7% chance that they have three members in common. But I doubt the efficacy of this latter method because it accounts for the odds of randomly selected subsets of Americans overlapping/not overlapping.
There are so many variables to take into account that haven't been considered that this is just a pointless exercise.
For example would you not have to also know how many gatherings of large groups of people take place on a given day there are that do not get shot up before you could even attempt to think about individual people being at the events where terror events happen?
I never had cause to learn stats, but I feel I should have.
What I'm saying is that the "or more" part goes all the way to the total attendance of the lower attendance event. That's a lot of numbers. So, "exactly three" is much, much smaller than "three or more". You could of course bruteforce the exact answer, and calculate the probability of:
Zero people
Exactly one people
Exactly two people
The inverse is your answer.
I'd also bet Markku's life that I am right.
Well, answer me this, Stilicho:
There are 100 people who toss a coin twice. Of those who have tossed heads, what is the probability that such a person tosses heads?
If you ignored a probability of one of the events, why did you ignore it?
@172 i'm with you there sir. Just taking the initial claim of there being three people from the old post. As to your addendums, hear hear, and i'm with your methodology for those who want to keep going.
How about the odds of surviving a first shooting and then getting killed at a second shooting? That is different from merely being present at both events. And infinitely more improbable. Can the quants advise?
I get around 6% for three or more people from the first event attending the second event, about 60% for one or more.
Also, are these three probabilities same?
1) A reporter picks a random person on the street, and it turns out that he's a survivor of both events
2) A reporter is at the scene of the second event with the survivors of the second event, picks a person at random from them, and it turns out that he also was at the first event
3) A reporter puts out a call at the second event, for anyone who also was at the first event, to come and talk with him
@56 "what would be the purpose of them being there that couldn't have been accomplished without them being there?"
THAT one is easy, no math required! HEADLINES!!
All the primates look at the headline: "survivor survives a second shooting!" Hoot screech hoot hoot! Try to imagine what the primate are led to think! If the guy from LV was in CA... "Random person survives"  no news value.... Someone the primate thinks s/he "already 'knows' and has sympathy for"? BIG "news" value!
Manipulation!
Also, are these three probabilities same?
Yes. The reporter will always lie to you.
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@92 "Let's nitpick this to death shall we?"
So, unlike the usual blog readers here, who are reading carefully and working hard to increase their understanding of a difficult area of ... math? engineering? science?... you're just here for the nachos?
Watching smart  and some maybenotassmart folks  work through a complicated math problem is EDUCATIONAL. I'm not quite keepin' up but I'm running hard, and the mathfolks are not yet out of sight over the horizon!
If you're bored? Go watch TV; we'll stay here and try to understand.
@Stilicho you are not correct. The probability we want to know is "any one of the people were in the second event GIVEN that they were in the first". Of course the membership is assumed. You can sacrifice the life to the dark lord.
Yep, it's called conditional probability. And there's the dreaded Bayes's formula too. http://www.stat.yale.edu/Courses/199798/101/condprob.htm
How many of those attending the LV event were locals who attended due to convenience and would not be inclined to spend time and money going elsewhere?
How many were on vacation and how often would that vacation time coincide with the second Gilroy event?
How many at Las Vegas who experienced the Joys of being in a sudden combat zone decided to avoid such events in future?
How many of the event participants were in contact with the media?
These are just a few of the variables (there are others) and as has been mentioned they significantly alter the final probability.
Vox's answer is the probability that a person taken at random was at both events. Subtract it from one and you get the probability that a person taken at random was not at both events. Raise that value to the power of 350,000,000 and you get the probability that no person was at both events (50.8%). Subtract that from one and you get the probability that at least one person was at both events (49.2%).
Another way to look at it is to ask how many of those Las Vegas festival attendees live in the greater Santa Clara Valley area. Maybe between SF/Oakland and Monterey, Santa Cruz and Stockton, basically within 90 minutes drive from Gilroy. Then remember that the demographic who attends a country western music festival isn't that far from the demographic who attends the Gilroy Garlic Festival.
I could even extend that to the Thousand Oaks shooting. I've been to that bar; they had different nights for different music styles, but that particular night was one of the weekly country/western nights they were best known for. So very similar demographic to Vegas.
However the other shootings don't have similar demographics to Vegas, Gilroy or Thousand Oaks, then that's where the narrative starts to fail and the probabilities get a bit ridiculously low. That is worth citing.
> Wow, Vox Popoli sure has a lot of secret Alpha Spergs
Incorrect. I don't care that people are wrong in this thread. People make mistakes; even pretty big ones, from time to time. But the fact that people are gleefully calling Uncephalized a moron, when in fact he is correct here on the math, is worthy of being called out, if you regard truth as a goal.
"In other words, you incorrectly interpreted what I said and tried to perform a different calculation, which you then screwed up the math doing.
Now, what did I say?
The odds against one person in a country of 320 million being in the vicinity of two such events are astronomical.
That is precisely true. And THAT is what you said was "flatout wrong". Forget the math. You can't even read correctly."
Uncephalized correctly interpreted the coincidence problem, and you performed a calculation that was irrelevant to the coincidence problem.
These are the two problems you guys are solving: "How likely is it that AT LEAST one person who attended one event, also attended the other?", and "How likely is it that a random person in the country attended one event, and also the other?"
The first problem is the suspicious coincidence highlighted in your original post. The second problem is the one you solved, which is not the same as the first problem. You're acting like Uncephalized is retarded for solving the problem you highlighted. He was correct in performing a different calculation, because your calculation was irrelevant to the problem.
I'm still doubtful that it's just coincidence. The probabilities calculated do not (and cannot easily) take into account a multitude of other factors which should lower the probability further. But it's not one in 500 million to find at least one person who was at both LV and Gilroy. Uncephalized was correct in choosing that problem to solve as it can help explain the coincidence highlighted and his methodology was correct.
"First, we take the population of the first event. Due to how we select the population, the probability of being at the first event is 100%. Then, we take the second probability, for "there exists" or "there is more than one", and invert it. We apply that to every one at the first shooting, and finally invert again. Which gives 88.8% ."
No. There is a very specific prohibition you are overlooking here.
The samples fail to meet the sufficient baselines, even if the distributions were normal.
In other words, the rules of statistics say you can do this to make the math work, but NOT find the actual probability. Vox's approach does not suffer this disqualifier.
@dcn The odds of an American being in the vicinity of both events is not the same as the odds of any American being in the vicinity.
And the odds of any one American being in the vicinity is the same as the odds of any person attending the second event given that he was in the first.
Another obvious error is the rough fudging of a day's attendees, when a far more relevant number is the number of people within a 10 acre radius of the shooter  a reasonable fatal hit rate over time. This will give you a mucg more accurate "survivor" pool, rather than including the thousands of people in attendance that day wholly unaffected by the shooting. After all, if you are stopped in traffic because of an accident in front of you, you would never be described as a survivor or even a witness of that accident.
"The samples fail to meet the sufficient baselines, even if the distributions were normal." You are talking out of your ass. I specifically restricted myself to one or more so that I wouldn't have to deal with any distributions. And even if I did, you are apparently only aware of Gaussian distribution. The Binomial distribution exists precisely for the problem you mention. But again, doesn't matter, due to the phrasing of the question.
I happen to remember this stuff because the book Machine Learning by Tom Mitchell reminded to always use Binomial distribution and not Gaussian, when below a certain sample size. Binomial always gives the right answer, no matter how ridiculously tiny the size.
Dole wrote:@dcn The odds of an American being in the vicinity of both events is not the same as the odds of any American being in the vicinity.
And the odds of any one American being in the vicinity is the same as the odds of any person attending the second event given that he was in the first.
I know. We have ~13,000 cases of people being in the vicinity of one event at the time of the shooting. We only care if one of those people also attended the other event, because that's literally what this coincidence problem is. So we calculate the probability that NONE of those ~13,000 individuals were one of the ~17,000 out of ~350,000,000 Americans who attended the other event (all 13,000 individuals have a chance to have been at the other event), and the invert it to get the probability that AT LEAST ONE of those people were are the other event. The result is around 50%, but as I said there are a multitude of other factors which would reduce it further, as well as the fact that this happened quite a few times. The point is it's not one in 500 Mill, that's the result of a calculation that has nothing to do with this problem. Vox is shitting on someone for trying to solve the correct problem.
https://infogalactic.com/wiki/Binomial_coefficient
n! when n=0, is 0. Therefore noverx, the binomial coefficient, is always zero.
https://www.varsitytutors.com/hotmath/hotmath_help/topics/binomialprobability
Note how the coefficient multiplies the whole thing. When zero multiplies something, the result is always zero. Hence, we don't need the distribution to calculate any question that is of the form "something happens exactly zero times", the inverse of which is "it happens one time or more". Which is why both of us independently decided to resort to the easy way of calculating the zero times, and then inverting that. No distributions involved.
dcn wrote:
I'm still doubtful that it's just coincidence. The probabilities calculated do not (and cannot easily) take into account a multitude of other factors which should lower the probability further.
...
Uncephalized was correct in choosing that problem to solve as it can help explain the coincidence highlighted and his methodology was correct.
How do you square these two statements?
You say he was correct for formulating the "correct" problem, then go on to say that he ignored a bunch of stuff that lowers the probability even if they could be known.
Then, you say "but he was correct, really."
Vox's initial statement was to the realworld problem, which Uncephalized and other mental masturbators have not solved by all their own admission.
Sperging over some theoretical "ideal textbook problem" is fun and all, but not really useful.
Bottom line, Uncephalized 88% probability is completely ridiculous, either as an analytical of a past event or as a predictor for the future. Vox addressed this point when he said:
How come we don't see such similar "coincidences" when we pair up literally any other two similar events? If the math works out this simply and the probs are that high, we should be seeing multiple double attendance at all kinds of real world "events," and we simply do NOT see that.
Found a conceptual error in Uncephalized's steps, ignoring some imprecise language:
Likelihood of one person being at both events is then: 1  (.999959^52,800). Which is 88.8%. The number of times this apparently happened is 3, so it's 0.888^3, or 70%.
The math to get at least 1 person at both events follows the birthday problem logic.
Pr(1+) = 1  Pr(0) = 0.888
The math used to get "3" people at both events takes the above equation and cubes it. Working the symbols back into the equation, that results in the following:
0.888^3
= (1  Pr(0))^3
= 1  3*Pr(0) + 3*Pr(0)^2  Pr(0)^3
Why would Pr(3) depend on the square and cube of Pr(0)? That's clearly wrong.
The correct equation, which Markku describes in @172, should have the form of:
Pr(3+) = 1  Pr(0)  Pr(1)  Pr(2)
I want to do some of that math, but I'm going to need to double check all the steps before embarrassing myself.
A lot of commenters trying to figure out the odds of reporters finding out about a particular person being at two shootings. This is not hard: tv stations have a news tip number. Just call it and say, " My friend X was at two shootings! Call me at $&@€£>!! Then when the reporter calls, produce crisis actor. Easy peasy.
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