From: "Mauricio Fernández" <mfp@...>
Date: 2005-07-17T19:18:29+09:00
Subject: Re: [QUIZ] Sampling (#39)

On Sun, Jul 17, 2005 at 05:45:10PM +0900, ruby quiz wrote:
> > > 0.3688. The probability that this happens for 20 such 200-numbers
> > > intervals is 0.3688**20 = 2.1e-09.
> > 
> > I think you are wrong. A rough estimates of this
> > probability gives me 1e-2171438.
> > 
> > Paolo
> After we get exactly one number in the first interval,
> we then are looking for exactly one number in the
> second interval, which is only one interval out of 19.
> In other words, the probability should
> be(18.0/19)**18. and so forth. The overall probability
> therefore is
> (19.0/20)**19 * (18.0/19)**18 * ... * (2.0/1)**1
> which my RubyShell informs me is 0.377353602535307e-8,
> or thereabouts :-)

I believed it was 

PI(i*200, i=1..20) / 4000^20 = PI(i, i=1..20) / 20^20

>> 1.0 * (1..20).inject(1){|s,x| s * x * 200} / 4000**20
=> 2.32019615953125e-08

-- 
Mauricio Fernandez