OK, since Paulson has pulled a hit and run (the "hit" occuring when he implied he had done a rigorous calculation of the odds and the "run" when I asked him to show his work), I'll show the work of an actual rigorous calculation:
First of all, the correct treatment is as a Poisson process<http://www.math.ucla.edu/~hbe/resource/general/3c.2.05f/sec12-4-6.pdf> : P(k)=e^(-Λ)*Λ^k/k! Where P is the probability k = the number of times the rare event occurs Λ=λt λ= the rate per unit time t= the time interval over which the k rare events occur Assuming: The Chelyabinsk meteor and the 2012 DA events are statistically similar events. These events occur roughly every 100 years. Our unit of time is 1 hour. A human lifetime is 80 years. λ=1/(100year/1hour) 1/(100year/1hour) 1 / ([100 * year] / [1 * hour]) = 0.0000011415525 t=16 Λ=λt 0.0000011415525*16 = 0.00001826484 P(X=2)=e^(-Λ)*Λ^2/2! e^(-0.00001826484)*0.00001826484^2/2 ([e^-0.00001826484] * [0.00001826484^2]) / 2 = 1.6679914E-10 So, the odds of any particular 16 hour interval experiencing 2 of these rare events is about: 1/1.6679914E-10 1 / 1.6679914E-10 = 5.9952347E9 1 in 6 billion So in an 80 year lifespan the odds of experiencing such a coincidence is: 1-(1-1.6679914E-10)^(80years/16hours) 1 - ([1 - 1.6679914E-10]^[{80 * year} / {16 * hour}]) = 0.0000073057752 1/0.0000073057752 1 / 0.0000073057752 = 136878.01 about 1 in a hundred thousand. On Thu, Feb 28, 2013 at 9:30 AM, James Bowery <[email protected]> wrote: > If my counting units had been years then you'd be right to imply my degree > of error was wildly off the mark, but they weren't. If the two events had > occurred within the same hour instead of within the same day, my > calculation would have been an even greater "far cry" from the time base of > years but it is still reasonable to base the calculation on counting units > derived from the distance in time between the events. What if they had > occurred within the same minute? The same second? > > In fact, the two events occurred within 16 hours of each other, not 24 > hours. > > Otherwise, thanks for pursuing a less naive calculation but you failed to > show your work. "Taylor expansion" doesn't cut it. > > Please update it for 16 hours rather than 24 hours and show your work. By > work I mean something more specific than "taylor expasion" which is about > as vague as you can get. > > > > On Thu, Feb 28, 2013 at 2:36 AM, George Paulson < > [email protected]> wrote: > >> 271.8*16,000 comes out to 4,348,800 days. 4,348,800/365 comes out to >> 11,915 years. >> >> So like I said we can expect an event like this roughly every 10,000 >> years or so. >> >> That's a far cry from the one in one billion odds or the one in one >> million odds after discounting by a factor of a thousand, isn't it? >> >> >> ------------------------------ >> Date: Thu, 28 Feb 2013 01:04:34 -0600 >> Subject: Re: [Vo]:Russian meteor coincidence odds >> From: [email protected] >> To: [email protected] >> >> >> You quote me incorrectly. My actual words were "less than one in a >> million". I stated so because mine was a "naive calculation" that came up >> with 1/1332250000 to which I then applied a "discount by a factor of a >> thousand" precisely to address such arguments as yours. >> >> To normalize your calculation properly you have to multiply 271.8*16,000. >> >> Now, can you do that arithmetic for us to complete your "critique"? >> >> >> On Wed, Feb 27, 2013 at 11:13 PM, George Paulson < >> [email protected]> wrote: >> >> In an earlier message, James Bowery claimed that the odds of the Russian >> meteor and asteroid DA14 passing Earth on the same day were "one in a >> billion": >> >> http://www.mail-archive.com/[email protected]/msg76844.html >> >> "The odds of this coincidence are literally far less than one in a >> million. The naive calculation is based on two like celestial events that >> independently occur once in a hundred years occurring on the same day: >> >> 1/(365*100)^2 >> = 1/1332250000 >> >> Note: that is one in a billion. Discount by a factor of a thousand for >> whatever your argument is and you are still one in a million. >> >> This is not a coincidence." >> >> This is incorrect. It is more like the birthday problem, where we're looking >> for the number of "years" that pass until two wandering asteroids have the >> same "birthday". A birthday here is when they fly by the Earth. >> >> >> >> We can expect the fly by of a DA14 type object every 40 years. If we >> also assume that something like the Russian meteor passes by every 40 years, >> this gives us a 16,000 day "year", and with a Taylor expansion you get a >> >> >> >> 99% probability of there being a coincident "birthday" after 271.8 "years", >> or roughly 10,000 of our years. >> >> So we can expect an event like this once every 10,000 years. >> >> >> >
