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*:

*L*_{n}>*L*_{n − 1}*R*_{n}>*R*_{n + 1}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.