Theo Borm wrote: > > Think of a sort of "power roulette", played with 58 balls > simultaneously, with a wheel containing 36 red/black slots of unequal > size, and 1 green slot. I need to calculate the probability that each of > the 36 red/black slots contains at least one ball. > Ah, now we come to a more precise problem :-)
> I have a hunch that it won't work as my p vector typically contains > values like: > > p<-c(0.99, 0.005, 0.003, 0.001, 0.0005, 0.0003, 0.0001, 0.00005, > 0.00003, 0.00002). > So you should expect that only 1/50000 will be in the low-prob hole in the average? > keeping in mind that the first "green" slot (0.99) represents my > "remainder" class (which may be empty), and calculating the probability > that with 9 trials I get exactly one in each of the other slots: > > p=0.005*0.003*0.001*0.0005*0.0003*0.0001*0.00005*0.00003*0.00002*9! > > =~ 2.45*10^-27 > > so, I'm looking for very rare events. > Indeed you are! > to calculate this figure even marginally /accurately/ with a monte carlo > simulation would require in well excess of 4*10^26 iterations..... > Hmmm.... That's right. No way to do a Monte Carlo here! Alberto Monteiro ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
