The Problem
The positive integers between $1$ and $10$ are holding an election. They are sitting around a circular table – $1$, then $2$, then $3$, and so on in clockwise order. Starting with $1$ and going clockwise, each integer publically votes for a president (between $1$ and $10$). After all $10$ integers have voted, the player with the most votes wins the election. The higher integer wins in case of tie.
Every integer prefers itself to win; but if it can’t win, it prefers the other integers in counter-clockwise order from itself. For example, 3 prefers itself, then 2, then 1, then 10, then 9, then 8, and so on. Every integer is perfectly rational and knows that every other integer will behave perfectly rationally as well.
Who wins the election?
Thoughts
This is a curious puzzle, and it seems much tricker than it initially appeared. I wrote a program which computes the answer (assuming no bugs), but I have not been able to prove it.
The challenge is that this is a game with more than two players, and standard arguments I am used to with two players do not seem to apply. Also, a lot of my intuition ends up being wrong. For example:
Even if some majority set of players prefers outcome $A$ to outcome $B$, that doesn't imply that $A$ happens and not $B$. This is true even if the set of players are consecutive.
I had the intuition that voting for player $P$ on your turn should always help player $P$ win; in other words, if you have at least one move that makes player $P$ win, then voting for $P$ is one of those moves. But this is not true. Consider the same game with four players $1, 2, 3, 4$, and suppose $1$ votes for $2$. Then if $2$ votes for $2$, $3$ will vote for $3$ and $3$ will win. But if $2$ votes for $1$, then $3$ will vote for $2$ and $2$ will win. So in this case, $2$ can win by voting for someone other than themselves!
Also, arguments based on "pooling votes" don't seem to work. For example, joriki's answer argues that $n = 1$ can't win, because even if $1$ through $5$ all vote for $1$, $6$ through $10$ can all pool their votes and vote for $6$ to thwart them. However, this intuition doesn't work. For example, consider $n = 3$ people instead of $10$: the argument would say that $1$ can't win, because even if $1$ votes for $1$, $2$ and $3$ can pool their votes and vote for $2$ to thwart $1$. However, $1$ actually wins in this case. The problem is that even if $2$ votes for $2$, $3$ will vote for $3$ instead of $2$ -- $3$ does not have an incentive to pool with $2$.
Note on problem origin
This problem is a minor variant of a problem that just appeared on the 2020 Utah Math Olympiad (problem 6), for which I was one of the problem writers. The contest asked for who wins in case the preferences are in clockwise order, and here I ask about counter-clockwise preferences instead.
We have multiple proofs for the clockwise version. For example, I can argue that (very roughly speaking), WLOG each player who does not win can simply "pass on" their vote to the next player, as assuming that they don't win, the next player's preference is as good as theirs. This leads to player $6$ winning. However, I haven't found an analagous argument here. The adjacent players' preferences don't seem to "line up" in the same way.
$10$ wins.
Here’s an argument for $6$ winning in the original problem that can be readily adapted to the present version: None of $1-5$ can win, because $6-10$ would all prefer $10$ to any of those, so even if $1-5$ pooled all their votes, $6-10$ could still all vote for $10$ to thwart them. $1-5$ know this, and their next-best preference is $6$, so they all vote for $6$, and $6$ joins them for a majority.
In the present version, again none of $1-5$ can win, because $6-10$ would all prefer $6$ to any of those, so even if $1-5$ pooled all their votes, $6-10$ could still all vote for $6$ to thwart them. $1-5$ know this, and their next-best preference is $10$, so they all vote for $10$, and $10$ joins them for a majority.