Bayes in Riddler mode

A very classical (textbook) question on the Riddler on inferring the contents of an urn from an Hypergeometric experiment:

You have an urn with N  red and white balls, but you have no information about what N might be. You draw n=19 balls at random, without replacement, and you get 8 red balls and 11 white balls. What is your best guess for the original number of balls (red and white) in the urn?

With therefore a likelihood given by

$\frac{R!}{(R-8)!}\frac{W!}{(W-11)!}\frac{(R+W-19)!}{(R+W)!}$

leading to a simple posterior derivation when choosing a 1/RW improper prior. That can be computed for a range of integer values of R and W:

L=function(R,W)lfactorial(R)+lfactorial(W)+
lfactorial(R+W-19)-lfactorial(R-8)-
lfactorial(W-11)-lfactorial(R+W)


and produces a posterior mean of 99.1 for R and of 131.2 for W, or a posterior median of 52 for R and 73 for W. And to the above surface for the log-likelihood. Which is unsurprisingly maximal at (8,11). The dependence on the prior is of course significant!

However silly me missed one word in the riddle, namely that R and W were equal… With a proper prior in 1/R², the posterior mean is 42.2 (unstable) and the posterior median 20. While an improper prior in 1/R leads to a posterior mean of 133.7 and a posterior median of 72. However, since the posterior mean increases with the number of values of R for which the posterior is computed, it may be that this mean does not exist!

One Response to “Bayes in Riddler mode”

1. But the riddle asks for $R=W$

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