The game of Cambridge Whispers starts with the first participant Albert flipping an un-biased coin and whispering to his neighbour Bertha whether it fell `heads' or `tails'. Bertha then whispers this information to her neighbour, and so on. The game ends when the final player Zebedee whispers to Albert and the game is won, by all players, if what Albert hears is correct. The acoustics are such that the listeners have, independently at each stage, only a probability of 2/3 of hearing correctly what is said. Find the probability that the game is won when there are just three players. By considering the binomial expansion of \((a+b)^n+(a-b)^n\), or otherwise, find a concise expression for the probability \(P\) that the game is won when is it played by \(n\) players each having a probability \(p\) of hearing correctly. % Show in particular that, if \(n\) is even, %\(P(n,1/10) = P(n,9/10)\).% How do you explain this apparent anomaly? To avoid the trauma of a lost game, the rules are now modified to require Albert to whisper to Bertha what he hears from Zebedee, and so keep the game going, if what he hears from Zebedee is not correct. Find the expected total number of times that Albert whispers to Bertha before the modified game ends. \noindent [You may use without proof the fact that \(\sum_1^\infty kx^{k-1}=(1-x)^{-2}\) for \(\vert x\vert<1\).]