Is there a known record for the most possible stalemates in one in a legal position, for with and without promoted pieces? I did some constructing myself for a start.

For without promoted pieces, I found 105, counting different promotions as different moves.

[Title "me, chess.stackexchange.com 2/24/2020, Stalemate In 1"]
[FEN "kN6/P1PPPPPP/2P5/4K3/6Q1/2NBB3/1R6/R7 w KQkq - 0 1"]

For promoted pieces, I remembered a position from elsewhere on this site. I switched two pieces, made come counting deductions, and I arrived at a number of 205 moves for with promoted pieces.

[Title "me, chess.stackexchange.com 2/24/2020, Stalemate In 1"]
[FEN "R6B/3Q4/1Q4Q1/4Q3/2Q4Q/Q4Q2/pp1Q4/kBNN1KR1 w - - 0 1"]
  • Your "with promoted pieces" currently has 9 queens and 3 bishops, so it seems illegal.
    – D M
    Jul 15, 2019 at 3:25
  • What about stalemate by repetition? If you were to give the black king two squares to move between, you would have a huge number of different combinations that produce a threefold repetition...
    – Aric
    Jul 17, 2019 at 10:32
  • 3
    @Aric that is called Draw by repitition
    – Marvin
    Jul 18, 2019 at 18:39
  • 2
    The first position isn't legal, there's no possible previous move for black. The third position is already a draw, since there's no possible sequence of moves that will lead to either player getting mated.
    – DanTilkin
    Aug 8, 2019 at 22:14

1 Answer 1


I managed to find the ultimate records in the Die Schwable Chess Problem Database.

[Title "Anthony Stewart & Mackay Dickens, Feenschach 9/1969, 113 Stalemating Moves"]
[FEN "6R1/PPPPPP2/7R/1K6/2N1BN2/2B4p/P4P1k/3Q4 w - - 0 1"]

Source Page

[Title "Anthony Stewart & Mackay Dickens, Feenschach 1/1969, 211 Stalemating Moves"]
[FEN "4Q2R/1K4Q1/3Q4/Q4Q2/2Q4Q/4Q3/1Q4Rp/1N1BB1Nk w - - 0 1"]

Source Page


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