Please. I'm pretty confident that ending has never arisen in any real chess game. (It actually is quite difficult to find real game instances containing even one bishop underpromotion. Know any?) If my rules failed to work in all chess endings that required >=2 underpromotions to get there, nobody would care. In fact, if you changed the rules of chess to forbid making more than 1 underpromotion per game, I doubt that would have ever altered any real game. Here is the old rule proposal back again with some minor typos fixed: --- WARREN'S PAWNLESS ENDGAME RULE: In a pawnless endgame with 6 men or fewer (this includes kings): Compute the total material points for both sides using the (very nontraditional!) N=4, B=3, R=7, Q=24, bonus if own two rooks=6, penalty if own "loner queen" faces 2 or more enemy officers=6 (if faces>=4 then penalty=15) bonus if own two bishops of opposite square-colors=5, penalty if own two knights=3, bonus if own both a rook and a non-rook=2, Let M>=0 be the difference between these two material points totals. If M>6.5 then WIN. If 0<=M<6.5, then DRAW. Allegedly the only exceptions are: the KQknn endgame, which is generally a draw although 18-5=13>>6.5; the KNNBkn and KNNBkb endgames, generally wins; and KBBNkb with the BBs having same square-color (not of great importance!) which is a draw. (Annoying man also pointed out KBBBk with all three B's same square color.) I am not sure what the KQQkqr ending usually is (win or draw?) but by redefining the word "non-rook" in the final bonus or adding a "have 2 queens" term if necessary, it should be able to handle this also. With our "if faces>=4" clause this also seems to handle all the common 7-man cases (specifically I think every case mentioned in http://en.wikipedia.org/wiki/Pawnless_chess_endgame is now handled? if I did not screw up) for example KBBNNkq is a win for the 4 pieces: 3+3+4+4-3+5=16 versus 24-15=9.