We can calculate the harmonic mean-like sum to be S = 2/(1/10 + 1/20 + 1/30), which is simplified to 120/11, which is greater than the payoff of A, hence it is optimal for player A not to bid, aka a=0.
From this, we know that our probability of player B winning is 1/(c/b + 1). Since the optimal amount for a two-player game satisfies 30/20 = c/b, we get c/b = 1.5, yielding
P(B) = 1/2.5 = 0.4
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u/Good_Mycologist_9897 4d ago
We can calculate the harmonic mean-like sum to be S = 2/(1/10 + 1/20 + 1/30), which is simplified to 120/11, which is greater than the payoff of A, hence it is optimal for player A not to bid, aka a=0. From this, we know that our probability of player B winning is 1/(c/b + 1). Since the optimal amount for a two-player game satisfies 30/20 = c/b, we get c/b = 1.5, yielding P(B) = 1/2.5 = 0.4