Tic-tac-toe game on the cube 3×3×3

Question

Consider the tic-tac-toe game on the $3\times3\times3$ cube. We know that in this case Player I has a winning strategy. But Player I may not play according to his winning strategy and the game probably ends with the winning of Player II or a draw. My question is: Is this a game with no draw? More precisely, is there a position in this game that is a draw? In other words is this a game in which either Player I or Player II must win and there is no draw?

Answer 1

tl;dr: draws are not possible.

Suppose there is a completed cube where no-one has won. First, look at the corners of the top face. These can't all be the same player (that forces the sides of the top face to be the other player, but then the centre can't be either). They also can't have two diagonally opposite x's and two diagonally opposite o's (again, the centre couldn't be either). If they have two of each and are not diagonally opposite, say the two left-hand corners are x's and the right-hand ones o's, look at the bottom corners. At most one of the left-hand corners can be x (by considering the left-hand face), and at most one can be o (by considering the diagonal plane from top right to bottom left). Similarly we have one of each player in the bottom right corners. But this means we have two diagonally opposite corners of the cube which are x, and the same for o, so the centre cube can't be either.

Thus the only possibility is that the top face has three corners belonging to one player and one to the other. These corners determine the rest of the face, so it must look like

x o x
o o x
x x o

possibly rotated and/or with x's and o's swapped. The same goes for every other face. But assuming the top face looks like that, the front face must be

x x o
o o x
x o x

and that forces the left face to be (imagine rotating the cube 90 degrees so it's at the front)

x o x
x o o
o x x

Now the bottom-left-back and top-right-front corners are o, and the top-left-back and bottom-right-front corners are x, so there's no way to fill in the centre cube.

Incidentally, the Hales-Jewett theorem means that if you choose any fixed $n$, and play the game of $n$-in-a-row tic-tac-toe on a $d$-dimensional cube, if $d$ is sufficiently large then draws are impossible.