Hendrik van Rooyen wrote: > "Nick Craig-Wood" <[EMAIL PROTECTED]> wrote: > > >>Paul Rubin <http> wrote: >> >>> The fencepost method still seems to be simplest: >>> >>> t = sorted(random.sample(xrange(1,50), 4)) >>> print [(j-i) for i,j in zip([0]+t, t+[50])] >> >>Mmm, nice. >> >>Here is another effort which is easier to reason about the >>distribution produced but not as efficient. >> >>def real(N, M): >> while 1: >> t = [ random.random() for i in range(N) ] >> factor = M / sum(t) >> t = [ int(round(x * factor)) for x in t] >> if sum(t) == M: >> break >> print "again" >> assert len(t) == N >> assert sum(t) == M >> return t >> >>It goes round the while loop on average 0.5 times. >> >>If 0 isn't required then just test for it and go around the loop again >>if found. That of course skews the distribution in difficult to >>calculate ways! >> > > > I have been wondering about the following as this thread unrolled: > > Is it possible to devise a test that can distinguish between sets > of: > > - five random numbers that add to 50, and > - four random numbers and a fudge number that add to 50? > > My stats are way too small and rusty to attempt to answer > the question, but it seems intuitively a very difficult thing. > > - Hendrik >
Yes, if the generating processes yield numbers from different probability mass functions. You could simply look at the likelihood ratio. Otherwise, the likelihood ratio will be 1. Duncan -- http://mail.python.org/mailman/listinfo/python-list
