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My partner and I are newbies, and recently found ourselves in this position:

[FEN "8/8/8/6kp/p5p1/8/K7/6Q1 w - - 0 1"]

I expected white would easily win because of the queen (and point) imbalance, but white never made time to free the king for the fight on the other side of the board, and there doesn't seem to be any way for the queen to handle that pawn-king cluster by herself.

My question: how many pawns generally make up for a missing queen in the endgame? According to the point system, black would need nine pawns, but that seems like too many to handle. Or would a more skilled white player have no problem dispatching pawns at this stage?

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3 Answers 3

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The simple and obvious answer is that it all depends on the position of black's pawns and king. In general the further up the board the pawns the better for black provided the king is in contact with the pawns, preferably in front of them.

Worth pointing out that the position you give is winning for white because the pawns aren't far enough forward. From what I said earlier it will be clear that white's first task is to separate black from the pawns, then win them.

This is how it is done.

[fen "8/8/8/6kp/p5p/8/K/6Q w - - 0 1"]

1. Qh2 {Cunningly not allowing the king to come forward} h4 2. Qe5+ Kg6 {Step 1 quickly accomplished. Black king separated from the pawns} 3. Qf4 Kh5 4. Qf5+ Kh6 5. Qxg4

and it is game over.

Move the kingside 2 rows forward, however, and best play is a draw, although white can give black opportunities to go wrong. At the same time white has to make sure black can't promote a pawn by checking, pinning or threatening mate.

[fen "8/8/8/8/p7/6kp/K5p/6Q w - - 0 1"]

1. Qe3+ Kh2 {only move} (1...Kh4?? 2. Qf2+ Kg4 3. Ka3  Kg5 4. Qg3+ Kf6 {anything else and Qxh3 will be check} 5. Kxa4 {and the king is going to come over and take the pawns}) 2. Qf4+ Kh1 {another only move} (2...Kg1?? 3. Qg3 Kf1 (3...h2 4. Qe1#) 4. Qxh3 Kf2 5. Qh4+ Kf1 6. Qf4+ Ke1 7. Qg3+ Kf1 8. Qf3+ Kg1 9. Ka3 Kh2 10. Qf4+ Kh1 11. Qh4+ Kg1 12. Kxa4 Kf1 13. Qf4+ {and repeating the process is going to bring the white king over to help and deliver mate}) 3. Qf3 Kh2 4. Qf4+

The two keys to playing these kinds of position are calculation and endgame knowledge. For instance in the second position it is key for black to stop white from winning the h pawn. That is because with just the g pawn black is lost. However replace the g pawn with either an h pawn or an f pawn and the position changes back into a draw.

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First of all, White wins in the diagram you provide, although it's not obvious at a glance how. White plays Qf2 first to stop the king from advancing. Black can't just sit there forever because the White king will eventually eat that a-pawn and come back, so they have to play h4, and then another pawn move. If Black plays g3, White responds with Qf3, and now the pawns can't safely advance. And if Black plays h3, White plays Qg3.

Surprisingly, Stockfish still says White is winning even if we give Black pawns on the fourth or fifth ranks on every file:

[FEN "8/8/8/3pppkp/ppp3p1/8/1K6/6Q1 w - - 0 1"]

Of course, it depends on the position. Here, White's king is well positioned to stop the queenside pawns (for example, c3+ is met with Kc2, and then b3 allows Kxc3 while a3 allows Kb3, so none of those pawns can safely queen) while Black's king is on the other side. And the immediate Qe3+ helps a lot.

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The answers are pretty good, but what I'm not seeing in the other answers is the elementary endgame theory of queen vs pawn:

Theory:

Theoretically even 1 pawn can make up for a queen. Check out Queen versus pawn endgame (try watching some videos eg gotham chess or josh waitzkin):

For king and queen vs king and pawn, when the pawn is on its seventh rank, this can be theoretically drawn if the pawn is on the a or h files (rook pawn) or the c or f files (bishop pawn).

Practice:

Equipped now with this theoretical knowledge, we can guess that even just 2 pawns could beat (or draw with) a queen if advanced far enough or if king is close enough to support the pawns, especially if the pawns are connected.

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