Let's say we have a few situations. White has all their pieces (hypothetically) and black has only:
If white were to run out of time, would each of these situations be a draw due to insufficient material? I'm wondering because let's say black were to have a forced checkmate with one of the above situations, could white deliberately lose on time and claim a draw?
lichess gets it wrong.
As it happens, White would lose all three games. With all white pieces still on the board, black has still "enough mating material" - it would actually help if White had no pieces at all!
I guess we all agree in the following position black is checkmate:
5k1K/5n1P/8/8/8/8/8/8 w - - 0 1
But this position can be reached from the examples shown. Even without pawns, it is still possible to checkmate for black:
2RKR3/5n2/3k4/8/8/8/8/8 w - - 0 1
Obviously, these cases are rather theoretical. But there are real-world situations (you need something like KRR vs KN, note that KQ vs KN is in fact drawn under that circumstances!) where this may apply. That is why the 10.2 rule (claiming draw with less than two minutes, now annex G) was established and why any higher-rated modern tournament will play with time increments.
There aren't many examples of asymmetric insufficient material. Most of the time, either both players or neither can win.
Edit: The relevant section of the rules is Article 6.9:
Except where one of Articles 5.1.a, 5.1.b, 5.2.a, 5.2.b, 5.2.c applies, if a player does not complete the prescribed number of moves in the allotted time, the game is lost by that player. However, the game is drawn if the position is such that the opponent cannot checkmate the player’s king by any possible series of legal moves.
The common simplification "has not enough winning material" is not always a good one.
