Probabilities With Renaming Suppose the outcome of each of a series of trials can be one of the events Ei (i=1 to n) with the respective probabilities Pr{Ei}. Prior to the first trial the events are renamed by applying the inverse of a particular permutation "M" to the indices. Thus, if event Ei is renamed Ej, then M[j] = i. Since there are n indices there are n! possible permutations, which we can denote by Mi (i = 1 to n!). Each of these has a known probability, denoted by Pr{Mi}. Now suppose the first k trials yield the events with the permuted indices g1, g2,..., gk. For example, if the outcome of the third trial is E7 then g3 = 7. (Here the “7” is the renamed index, so this is not necessarily the original E7 whose probability is given.) Given the values of all Pr{Ei} for i = 1 to n (representing the unpermuted indices), and the probabilities Pr{Mi}, i = 1 to n!  of the possible permutations, and given the results gi (i=1 to k) of the first k trials, what are the probabilities that gk+1 = s for any given index s? If we knew the indices were not permuted at all, we would immediately know the probabilities for each of the events, but since the indices are permuted, we must infer something about the probabilities of the permuted events from the k outcomes already observed. 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 gi, i = 1 to k+1} divided by {the sum of the probabilities of all possible sequences with the given values of gi, i = 1 to k}. Symbolically this can be written in the form For example, consider the case of n = 2 with Pr{E1} = 1/3 and Pr{E2} = 2/3. Let M1 denote the identity permutation and M2 the reversal, and suppose we are given that Pr{M1} = Pr{M2} = 1/2. Also we have k = 1 with g1 = 1. The probability that g2 equals 1 is Now suppose Pr{A} = 1/3, Pr{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 Pr{A} = p, Pr{B} = q = 1-p, Pr{A'=A} = r, and Pr{A'=B} = s = 1-r ?  The answer is that A' occurs again with probability Thus if we have p = 1/3, q = 2/3, r = s = 1/2, the result is 5/9. Return to MathPages Main Menu