A curious thing about infinite sequences of coin tosses

Suppose that at locations ...,−3,−2,−1,0,1,2,3,... (in time or one spatial dimension) a relevantly similar independent fair coin is tossed. Let L_n be the event that we have heads at all locations k≤n. Let R_n be the event that we have heads at all locations k≥n. Let A≤B mean that event B is at least as likely as A, and suppose all our events L_n and R_n are comparable (i.e., A≤B or B≤A whenever A and B are among the events L_n and R_n). Write A<B when A≤B but not B≤A. Assume ≤ is transitive and reflexive. Then, intuitively, we also have:

1. L_n+1<L_n
2. R_n<R_n+1.

After all, L_n+1 and R_n require one more heads result than L_n and R_n+1, respectively.

Now here is a surprising consequence of the above assumptions. Say that n is a switch-over point provided that L_n≥R_n but L_n+1<R_n+1. When there is a switch-over point, it's unique (since for m≤n, we will have L_m≥L_n≥R_n≥R_m, and for m>n, we will have L_m≤L_n+1<R_n+1≤R_m). Then:

3. Under the above assumptions, either (a) there is a switch-over point, or (b) L_n<R_m for all n and m, or (c) R_n<L_m for all n and m.

But each of these options is really rather absurd. Option (a) says that the probability distribution of our sequence of relevantly similar independent fair coins has a distinguished switch-over point. Option (b) implies that infinite sequences of heads stretching leftward (if we're talking about a spatial arrangement; backward, if temporal) are always less likely than infinite sequences of heads stretching rightward (forward, if temporal). And option (c) is just as absurd as (b). And of course in each of the three cases we have a violation of very plausible symmetry conditions on the story: in case (a), we have a violation of symmetry under shifts, and cases (b) and (c) we have a violation of symmetry under flips.

So something is wrong with the assumptions. Classical probability theory says that what's wrong are (1) and (2): in fact, all of the L_n and R_n events are equally likely, i.e., have probability 0. This seems a very plausible diagnosis to me.

Why does (3) hold? Well, suppose that (b) and (c) do not hold. Thus, L_n<R_m for some n and m. If m≥n, then we L_m≤L_n<R_m, and if m<n, then we have L_n<R_m<R_n. In either case, there is an b such that L_b<R_b. By the same reasoning, by the falsity of (c), there is an a such that L_a>R_a. As n moves from a to b, then, L_n decreases while R_n increases, and we start with L_n bigger than R_n at n=a and end with L_n smaller than R_n at n=b. This guarantees the existence of a switch-over point.

This result is basically a generalization of an observation about Popper functions for such infinite sequences in a forthcoming paper of mine.