A square game

Probability Level 2

You and a friend play a game where you take turns painting the little squares on the sides of a $3 \times 3 \times 3$ Rubik's cube.

Each time you paint a square, you aren't allowed to paint a square that shares an edge, or vertex of a square already painted.

The last person to paint a square wins!

Who has a winning strategy?

Clarification : You can paint a square on an edge only if none of the five other squares along that edge are painted. Similarly, three squares share a vertex.

Image credit : thinkgeek.com

The player who goes second The player who goes first

2 solutions

Geoff Pilling
Mar 2, 2018

One winning strategy is where you always paint a square that is directly opposite to the one that was just painted. (Opposite meaning that if you draw a line from the center of this square through the center of the 3x3x3 cube you hit the "opposite" square on the other side) Since the one just painted shared no edge, face or vertex with a painted square, then, by symmetry neither does the one you just painted! :-)

That was the strategy that I came up with too.

At first I thought this was about painting with multiple colors (I was pre-conditioned by the 2x2x2 coloring problem I imagine). Once I read the problem carefully and decided that it was simply about "marking" with a single color of paint it was not too hard.

Steven Perkins - 3 years, 3 months ago

David Vreken
Mar 3, 2018

The game is at most 6 turns long, since there are 6 faces on a cube, and you can't paint a square that shares a face with a square that is already painted. The game is also at least 6 turns long, since an unpainted face will always have the center square available to paint, as it does not share a vertex or edge with a square from another face. So irregardless of anyone's strategy, the game is always 6 turns long, and since 6 is an even number, the player who goes second will always win.

Good point, David... Didn't realize that when I wrote up the problem... Guess that kinda trivializes things a bit! :-/

Geoff Pilling - 3 years, 3 months ago

I still like the problem even though it was not what you originally intended, although the "winning strategy" wording threw me off at first.

David Vreken - 3 years, 3 months ago

The problem doesn’t say they can’t share faces, only edges or vertices, so the game can be more than 6 turns long. @Geoff Pilling , you might want to add that. Or not, the playing opposite solution still works without the face requirements.

Jason Carrier - 2 years, 8 months ago

I believe the problem used to say that you can't paint a square that shares a face with a square that is already painted, but it has since been edited, so now my solution is no longer valid.

David Vreken - 2 years, 8 months ago

A square game

2 solutions

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