Noob question: If Pn(An==”left”) --> 1⁄2, doesn’t that mean that as n gets bigger, Pn(An==”left”) gets closer and closer to 1/2? And doesn’t that leave open the question of how it approaches 1/2? It could approach from above, or below, or it could oscillate above and below. Given that you say it converges to randomly choosing, I’m guessing it’s the oscillation thing. So is there some additional lemma you glossed over, about the way it converges?
It does not approach it from above or below. As N goes to infinity, the proportion of n<N for which An==”Left” need not converge to 1⁄2, but it must have 1⁄2 as a limit point, so the proportion of n<N for which An==”Left” is arbitrarily close to 1⁄2 infinitely often. Further, the same is true for any easy to compute subsequence of rounds.
So, unfortunately it might be that An goes left many many times in a row e.g. for all n between 1010 and 10100, but it will still be unpredictable, just not locally independent.
Noob question: If Pn(An==”left”) --> 1⁄2, doesn’t that mean that as n gets bigger, Pn(An==”left”) gets closer and closer to 1/2? And doesn’t that leave open the question of how it approaches 1/2? It could approach from above, or below, or it could oscillate above and below. Given that you say it converges to randomly choosing, I’m guessing it’s the oscillation thing. So is there some additional lemma you glossed over, about the way it converges?
It does not approach it from above or below. As N goes to infinity, the proportion of n<N for which An==”Left” need not converge to 1⁄2, but it must have 1⁄2 as a limit point, so the proportion of n<N for which An==”Left” is arbitrarily close to 1⁄2 infinitely often. Further, the same is true for any easy to compute subsequence of rounds.
So, unfortunately it might be that An goes left many many times in a row e.g. for all n between 1010 and 10100, but it will still be unpredictable, just not locally independent.