It is clear that by playing Game A, we will almost surely lose in the long run. Harmer and Abbott show via simulation that if formula_10 and formula_11 Game B is an almost surely losing game as well. 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. 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.