"In fact, Game B is a Markov chain, and an analysis of its state transition matrix (again with M=3) shows that the steady state probability of using coin 2 is 0.3836, and that of using coin 3 is 0.6164.[3] As coin 2 is selected nearly 40% of the time, it has a disproportionate influence on the payoff from Game B, and results in it being a losing game." . . . .