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
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!
Another Riddle(r), considering a box with M myrtle balls and D dandelion balls. Drawing balls without replacement while they stay of the same color as the initial draw, else put back the last ball and repeat the process until all balls are drawn. The funny thing is that, unless M=0 or D=0, the probability to draw a myrtle ball at the end is always ½..! This can be easily checked by simulation (when M=2 and D=8)
![b^\star=\left[ 1+2a\pm\sqrt{1+8a}\right]/2 \in \mathbb N](https://s0.wp.com/latex.php?latex=b%5E%5Cstar%3D%5Cleft%5B+1%2B2a%5Cpm%5Csqrt%7B1%2B8a%7D%5Cright%5D%2F2+%5Cin+%5Cmathbb+N&bg=000000&fg=B0B0B0&s=0&c=20201002)
