“…Setting p = π 01 π 10 +π 01 , the conditional probability of occurrence of [Y 1 < Y 2 ] when [Y 1 ̸ = Y 2 ], the testing problem is equivalent to that of testing H 0 : p = 1 2 against H 1 : p ̸ = 1 2 , which is same as the earlier one with p 0 = 1 2 . Bandyopadhyay et al (2014) work on this problem using a two-sided symmetric stopping rule. However, the present solution, based on a two-sided asymmetric stopping rule, provides the following alternative approach.…”