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Message Subject Man inside bar when a gunman opened fire was with "probably 50 or 60 others" who were ALSO at the Las Vegas massacre! WTF!?
Poster Handle Tainted Meat
Post Content
Here's some hard data, working out the exact odds of being caught up in one of these 'mass shootings'.

According to shootingtracker.com, there have been at least 462 people killed and 1312 injured in 353 mass shootings this year, which is already more than the 383 people killed last year. They define a mass shooting as one in which four or more people are killed or injured, which is a broader definition than the government has used. (Compare that to the 30K or so people killed in auto accidents each year.) According to the Census Bureau, the U.S. currently has a resident population of 322,367,564, giving me a 0.00000143% chance of getting killed in a mass shooting this year by the broadest definition.

[link to blogs.princeton.edu (secure)]


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Thus, being generous, we can presume that the odds of being in TWO mass shootings would be approximately 0.0000000000071%.


The odds of this many people, all being involved in another mass shooting... hmmm, so X50 or so?

0.00000000000000000000000000000000000000000000000000000000000000000000000000​000000000000000000000000000000000000000000000000000000000000000000000000000000000​000000000000000000000000000000000000000000000000000000000000000000000000000000000​000000000000000000000000000000000000000000000000000000000000000000000000000000000​000000000000000000000000000000000000000000000000000000000000000000007%
 Quoting​: McShillin


That's the same page I found earlier.

I wasn't sure how to do the other math to estimate the percentage of it happen though.

That's absurd that people believe this shit.
 Quoting: CrsCrpr

The odds of getting killed in a mass shooting is different from being in a mass shooting.

In general, if there are no hidden correlations (e.g. same people traveling frequently to the same events), you get probabilities by multiplying them. For instance,

P1 = probability of one person being in one mass shooting

P2 = the probability for one person being in two mass shootings = P1*P1 = (P1)^2

Pn = the probability for one person being in n mass shootings = (P1)^n, where n is the number of shootings.

The probability P for the same 50 people ending up in two shootings is:

P = ( (P1)^2 * (P1)^2 * (P1)^2 * ...) ; 50 times
= ( (P1)^2 )^50 ; and by applying power the rule (a^n)^m = a^(n*m)
= (P1)^100

This, of course, neglects any correlations between people and events and represents a case when everything's unconnected and random.
 
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