Three invariance arguments

Suppose we have two infinite collections of items L_n and R_n indexed by integers n, and suppose we have a total preorder ≤ on all the items. Suppose further the following conditions hold for all n, m and k:

1. L_n > L_n − 1

2. R_n > R_n + 1

3. If L_n ≤ R_m, then L_n + k ≤ R_m + k.

Theorem: It follows that either L_n > R_m for all n and m, or R_n > L_m for all n and m.

(I prove this in a special case here, but the proof works for the general case.)

Here are three interesting applications. First, suppose that an integer X is fairly chosen. Let L_n be the event that X ≤ n and let R_n be the event that X ≥ n. Let our preorder be comparison of the probabilities of events: A ≤ B means that A is no less likely than B. Intuitively, it is less likely that X is less than n − 1 than that it is less than n, so we have (1), and similar reasoning gives (2). Claim (3) says that the relationship between L_n and R_m is the same as that between L_n + k ≤ R_m + k and that seems right, too.

So all the conditions seem satisfied, but the conclusion of the Theorem seems wrong. It just doesn’t seem right to think that all the left-ward events (X being less than or equal to something) are more likely than all the right-ward events (X being bigger than or equal to something), nor that it be the other way around.

I am inclined to conclude that countable infinite fair lotteries are impossible.

Second application. Suppose that for each integer n, a coin is tossed. Let L_n be the event that all the coins ..., n − 2, n − 1, n are heads. Let R_n be the event that all the coins n, n + 1, n + 2, ... are heads. Let ≤ compare probabilities in reverse: bigger is less likely. Again, the conditions (1)–(3) all sound right: it is less likely that ..., n − 2, n − 1, n are heads than that ..., n − 2, n − 1 are heads, and similarly for the right-ward events. But the conclusion of the theorem is clearly wrong here. The rightward all-heads events aren’t all more likely, nor all less likely, than the leftward ones.

I am inclined to conclude that all the L_n and R_n have equal probability (namely zero).

Third application. Supppose that there is an infinite line of people, all morally on par, standing on numbered positions one meter apart, with their lives endangered in the same way. Let L_n be the action of saving the lives of the people at positions ...., n − 2, n − 1, n and let R_n be the action of saving the lives of the people at positions n, n + 1, n + 2, .... Let ≤ measure moral worseness: A ≤ B means that B is at least as bad as A. Then intuitively we have (1) and (2): it is worse to save fewer people. Moreover, (3) is a plausible symmetry condition: if saving one group of people beats saving another group of people, shifting both groups by the same amount doesn’t change that comparison. But again the conclusion of the theorem is clearly wrong.

I am less clear on what to say. I think I want to deny the totality of ≤, allowing for cases of incommensurability of actions. In particular, I suspect that L_n and R_m will always be incommensurable.