IMC 2008 Q14

Let \(A\) and \(B\) denote the ends of the first length of rope and \(C\) and \(D\) denote the ends of the second length of rope. Then Pat chooses one of 6 different possible combinations: (\(A\), \(B\)), (\(A\), \(C\)), (\(A\), \(D\)), (\(B\), \(C\)), (\(B\), \(D\)), (\(C\), \(D\)). Sam now holds one untied length of rope and one tied loop of rope if, and only if, Pat has chosen (\(A\), \(B\)) or (\(C\), \(D\)) so the required probability is \(\dfrac{2}{6} = \dfrac{1}{3}\).
(Alternatively: Irrespective of whichever end Pat chooses first, Sam will hold one untied length of rope and one tied loop of rope if, and only if, Pat now chooses a particular one of the three remaining ends, namely the other end of the same rope. As each of the three ends is equally likely to be chosen, the required probability is \(\tfrac{1}{3}\).)