How many stalemate positions in numbers without duplicates are possible on chess board?

Is it a infinite or finite number in terms of stalemate positions count?

  • 5
    It is a finite number because the game iself is finite. But a very, very large number. Are you asking for an estimate ?
    – Evargalo
    Feb 7 at 15:37
  • Yes. Can a Chess engine give us this exact finite count? Feb 7 at 16:06
  • 4
    Chess engines aren't programmed to find and count stalemate positions, but you could write a special program to do so. I suspect that there isn't enough computing power on earth to answer the question in a reasonable amount of time. I could be wrong. Feb 7 at 17:29

1 Answer 1


It's a finite number since it's a finite chessboard. But the number is very large.

Consider the following stalemate position:

[fen "8/8/8/8/8/7K/5Q2/7k w - - 0 1"]

If you consider horizontal, vertical, and diagonal symmetry, there's 8 near-identical (but not really duplicate because the pieces are on different squares) positions like this.

For each of those, there are 53 places where you could put a white rook and have it still be stalemate. 53*8=424 of these stalemate positions with a rook.

And you could put a second white rook on one of the remaining 52 squares. Let's not stop there; with underpromotion we could have up to ten rooks on those 53 squares. Which gives us, in total, 199,978,590,456 ways to do this stalemate position just by using symmetry and adding rooks.

There are 60 squares a knight could go in this position and have it still be stalemate. With up to 10 knights possible, that's 745,424,376,384 ways you could make a stalemate position just by adding knights.

You could have knights and rooks. If you have 10 knights on the board that still leaves a minimum of 43 places a rook could go, or 903 places two rooks could go, for each of those 745,424,376,384 positions that had up to 10 knights. If you have 10 rooks on the board that leaves a minimum of 50 places a knight could go, or 1225 places two knights could go, for each of those 199,978,590,456 positions that had up to 10 rooks.

We're already approaching a quadrillion stalemate positions here, and we haven't even touched, say, positions with 3 rooks and 7 knights, let alone positions with pawns, bishops, more queens, or blocked opposing pawns. And these are only what we get by adding to this one position and its mirrors; there are countless other ways to stalemate.

  • 2
    Heck, even pure stalemates (in the sense: no piece is superfluous) will occur in a gazillion cases. Feb 8 at 8:59
  • 1
    Moreover, if we are looking for an exact number and not only for an estimate, we have to remove some of the positions you counted. For instance, you cannot add Ne2 and Nf3 in the diagram because the position becomes illegal (whichever move White just made, there is no ante-move for Black). This just adds to already daunting complexity...
    – Evargalo
    Feb 8 at 10:45
  • 2
    @Evargalo You can legally add Ne2 and Nf3. Black's last move was Qf2 and White has just played Qxf2. But yeah, if you're looking for all positions you will inevitably come across some which are illegal, and checking whether a position is legal is not easy even for a computer.
    – D M
    Feb 8 at 12:30

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