Probabilities With Renaming

Suppose the outcome of each trial can be one of the events E_i 
(i=1 to n) with the respective probabilities Pr{E_i}.  Prior to 
the first trial the events are renamed by applying the inverse 
of a particular permutation "M" to the indices.  Thus, if event 
E_i is renamed E_j, then M[j]=i.  Since there are n indices there 
are n! possible permutations, which we can denote by M_i (i = 1 to n!).  
Each of these has a known probability denoted by Pr{M_i}.

Let g_1, g_2,...,g_k denote the outcomes of the first k trials, where 
g_i represents the index of the ith permuted event label.  For example, 
if the outcome of the third trial is E_7 then g_3 = 7.

Given the values of all Pr{E_*}, Pr{M_*}, and g_i (i=1 to k), what are
the probabilities that g_(k+1) = s for any given index s?

The answer is the ratio of 

      the sum of the probabilities of all the possible
      sequences of events (including choice of permutation)
      with the given values of g_i, i=1 to k+1

divided by 

      the sum of the probabilities of all possible sequences 
      with the given values of g_i, i=1 to k

Symbolically this can be written in the form

Pr{g_(k+1) = s}  =


 SUM(i=1 to n!) Pr{M_i}*Pr{E_(M_i[g_s])}*PRODUCT(j=1 to k) Pr{E_(M_i[g_j])}
 --------------------------------------------------------------------------
        SUM(i=1 to n!) Pr{M_i}*PRODUCT(j=1 to k) Pr{E_(M_i[g_j])}


For example, consider the case of n=2 with Pr{E_1}=1/3 and 
Pr{E_2}=2/3.  Letting M_1 denote the identity permutation and M_2 
the reversal, we are given that Pr{M_1} = Pr{M_2} = 1/2.  Also we 
have k=1 with g_1 = 1.  The probability that g_2 equals 1 is

         0.5*(1/3)*(1/3) + 0.5*(2/3)*(2/3)
         ---------------------------------  =  5/9
              0.5*(1/3) + 0.5*(2/3)


Now suppose p(A)=1/3, p(B)=2/3, and we are told that the events are
being renamed, with a 50% chance that they will be kept that same,
i.e., that p(A')=1/3 and p(B')=2/3, and a 50% chance that they will
be switched, resulting in p(A')=2/3 and p(B')=1/3.  After the die is
rolled we are told that event A' occurred.  What is the probability
that on the next roll, A' occurs again?  More generally, what if
p(A)=p, p(B)=q=1-p, p(A'=A)=r, and p(A'=B)=s=1-r ?  The answer is
that A' occurs again with probability

                                    pq
              pr{A'} =   1   -  -----------
                                  qs + pr

Thus if we have p=1/3, q=2/3, r=s=1/2, the result is 5/9.