@burkeaw,
A probably very oversimplified approach that avoids MTBF calculations and statistics might be:
6 failures on 13 machines per year is 6/13/4 = 0.115 failures per machine per quarter
7/8 of the cycle is non-critical
the probability that a given machine fails in a quarter is 0.115
the probability that a given machine doesn't fail or fails outside the critical period in a quarter is (1 - 0.115) + (7/8 * 0.115) = 0.9856
the probability that 13 machines don't fail during the critical period in a quarter is 0.9856^13 = 0.828
the probability that at least one machine fails during the critical period in a quarter is 1 - 0.828 = 0.172