After reviewing these posts , the Janes' chapter on the Monkey prior, and the rec.puzzles FAQ for both probability|apriori and decision|Monty Hall, I conclude that the solution requires an approach different from either r.e.s. or webster2000 quoted above. The Binomial Monkey Prior serves to remind us that there is a set of priors which are less restrictive than either solution. The original problem in fact now begins to resemble the Monty Hall problem!!!
To get down to the essentials, consider a variant of the original problem
in which there are only two chips in the urn.
After one chip has been observed by removing it from the urn, what is the
probability that the remaining chip is the same color?
(in this table, a_k is the hypothesis that k out of 2 chips are white.)
----------------------------------------------------------
a_k p(v|a_k) p(a_k) p(v)p(a_k|v) p(a_k)|v)
0 0 (1-g)^2 0 0
1 0.5 2g(1-g) (1-g)g 2*integral g - integral g^2|0,1 = 1/3
2 1 g^2 g^2 2*integral g^2 | 0,1 = 2/3
---------------
p(v)=integral column sum|0,1
=integral g dg|0,1
= 1/2
The last two steps were to sum the possible values p(v)(a_k|v) over both the
values of a_k and the range of g (0 to 1) to obtain the value of p(v)
Using this, p(a_12|v)= g^2/p(v) is integrated over the range of g (0 to 1)
to obtain the final estimate of:
p(a_2|v) = 0.666666
Now applying the same reasoning to the case of 12 chips
a_k p(v|a_k)
1 0
2 0
11 (11/12)^11
12 1
The second from last p(v|a_k) is the critical step for which webster2000
and r.e.s. claimed differing values. This assumes webster2000's value was
correct even though his symbolism was misleading.
Stated in words: if the hypothesis a_11 is that there are 11 white chips
and one non-white chip in the urn, the probability of the observation v,
where v is the observation that 11 white chips are seen, is the independent
AND combination of 11 events each with probability 11/12. That is the
precise meaning of p(v|a_11) and it can only be (11/12)^11.
Completing the table:
-------------------------------------------------------------------
a_k p(v|a_k) p(a_k) p(v)p(a_k|v) p(a_k)|v)
1 0 (1-g)^12 0
2 0 2g(1-g)^11 0
11 (11/12)^11 12(g^11)(1-g) 12(1-g)(11/12)g)^11
12 1 g^12 g^12 (1/0.10646)integral g^12|0,1
= 0.722546
-------------------
p(v)= integral column sum |0,1
= integral(1-k)g^12+kg^11|0,1
= (1-k)/13 + k/12= -0.27753 + .38399 = 0.10646
k =12*(11/12)^11=4.6079
As above, the last two steps were to sum the possible values p(v)(a_k|v) over both
the values of a_k and the range of g (0 to 1) to obtain the value of p(v)
Using this, p(a_12|v)= g^12/p(v) is integrated over the range of g (0 to 1)
to obtain the final estimate of:
p(a_12|v) = 0.722546
John Bailey
June 21, 1999
corrected June 24, 1999
revised June 26, 1999
Comments?