The sequences of two tickets from F are described by F # F, which has four tickets: (1,1), (1,2), (2,1), and (2,2).
Division relabels these tickets as 1/1 = 1, 1/2 = 0.5, 2/1 = 2, and 2/2 = 1, respectively. Therefore there are three outcomes: 0.5, 1, and 2.
Outcome 1 is written on two tickets, so its probability is 2/4 = 1/2. The probability of 0.5 is 1/4 and the probability of 2 is 1/4.
1. The process described in this problem divides one probability distribution by another. This can be done provided the divisor has zero probability of being a zero (for then the quotient is not defined).
2. Indeed, in this case we divided a distribution by itself. We did not get a constant value 1, though. This is a hint that probability distributions, although they can be added (as in the preceding problem), subtracted, multiplied, and divided (as in this problem), do not obey all the usual laws of arithmetic.