I always find finishing off my opponent tricky in Nine Men's Morris.

If they have 3 pieces and can fly, how do I guarantee a mill to secure the win? What's the minimum number of pieces I can do this with?

3 Answers 3


You can't guarantee a win. In a 1996 paper called Solving Nine Men's Morris, Ralph Gasser describes how he used search techniques to determine that, given best play, the game of nine men's morris should end in a draw.

In figure 11 of that paper, he gives a graph of the percentage of won positions for the player to move, against the number of stones each has on the board. The highest percentage of won positions is 83% when the player to move has 3 stones against 3.

The data also reveals that a player with 3 stones left cannot force a win against an opponent with 6 or more stones left. Although many of those positions should still end in draws, rather than wins for the player with more stones.

A player is much more likely to be able to force a win when they have 7 or more stones left, than when they get down to 6. A player with 4 stones against 3 can only force a win if it is their go, and they can immediately make a mill.

All of this assumes two players who are playing completely optimally, which is not usually the case. In 1990 six games were played between the British champion Mike Sunley and a heuristics-based computer program. Two games were drawn and four were won by the computer.


There is a winning opening: Any corner of any quadron. If opponent does not play same quadron, you follow the corners of the quadron until you have a double mill opportunity. If opponent plays same quadron, proceed to next quadron with same strategy. If opponent plays same quadron all 3 times, there is almost always a way to connect existing pieces to a double mill opportunity, unless opponent knows that, in which case he can block.

You are now way above beginner level.

  • 1
    That's actually a really bad strategy that will have you losing most of the time against halfway decent players. Reasons are: 1) Don't place your stones in corners 2) Don't force early mills as they hinder you in the midgame where it's all about mobility and your stones will be clumped together. Just threatening a (double)-mill might be of advantage here.
    – Christoph
    May 5, 2021 at 7:55

Always keep in mind that one way to win a game is forcing your opponent into a position where they are unable to make a valid move!

So instead of reducing them to three stones and fight a tedious end game against flying stones you actually might want to box the opponent in, winning by them not being able to move.

The other strategy is to set up two independent mills before reducing them to 3 stones. That however is easier said than done and requires a material advantage that is reflected in the quoted paper with the drastic increase in win rates at the seven stone mark.

I personally prefer the first strategy and never reduce them two three stones in the first place.

  • What is the problem with this question? Why the downvotes?
    – Christoph
    May 6, 2021 at 9:59
  • This answer is saying to win by reducing your opponents options but not actually saying how to do that. That isn't much different than saying you should win by playing winning moves.
    – Joe W
    May 6, 2021 at 14:53
  • I disagree. I am just pointing out, that there is a lesser known second winning condition in this game where a win is easier to achieve than fight a flying stone endgame.
    – Christoph
    May 7, 2021 at 8:01
  • Disagree or not, I am guessing that is the reason for the downvotes because you don't say how to actually do what you suggest.
    – Joe W
    May 7, 2021 at 12:32

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