Problem: The Y2K Game is played on a $1 \times 2000$ grid as follows.
Two players in turn write either an S or an O in an empty square.
The first player who produces three consecutive boxes that spell SOS wins.
If all boxes are filled without producing SOS then the game is a draw.
Prove that the second player has a winning strategy.

Solution: The main insight is that a construct of the form
\[ S \; \square \; \square \; S \]
(here the $\square$ is blank)
will kill any player which plays inside it.
We call this a \emph{trap} accordingly.

\begin{claim*}
  The second player can force a trap to exist;
  in this case the game will never end in a draw.
\end{claim*}
\begin{proof}
  Actually the second player can construct a trap
  on her second turn by playing an $S$ far enough away
  from the edges of the board and the first player's initial move.
\end{proof}

\begin{claim*}
  The second player always has a move which prevents
  her from losing.
\end{claim*}
\begin{proof}
  Since there are an odd number of empty squares at the start of
  the second player's turn,
  there must be a square which is bordered by either
  two filled or two empty squares.
  The second player can then play $O$ in this square,
  which is always safe.
\end{proof}

Together these two claims finish the problem.

\begin{remark*}
  Actually, one can show that the ``only'' way to lose
  is to be forced to play inside a trap.
  Indeed, suppose playing in a certain cell $c$ loses.
  If we wrote $O$, that means $c$ is bordered by exactly one $S$,
  with a blank cell on the neighbor.
  But we could also write $S$; checking cases we find $c$
  is part of a trap.

  Thus a player can lose only if all blank cells
  are in traps; ergo, the number of blank cells is even.
  This never happens for the second player.
  Thus this gives an alternative solution,
  and moreover a reason to believe that all
  correct solutions must involve traps.

  A similar proof shows that for large $n$, with a $1 \times n$ board,
  the first player has a winning strategy for odd $n$,
  and the second player has a winning strategy for even $n$.
\end{remark*}