Why doesn't greek gift work in Pozsonyi vs Pataki 2005? Or does it?

Question

original game. i'm asking about move 21.

[Event "Ch Hungary (team) 2005/06"]
[Site "HUN"]
[Date "2005.12.04"]
[Round "?"]
[White "Istvan Pozsonyi"]
[Black "Gyozo Pataki"]
[Result "0-1"]

[fen "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1"]


1. b3 d5 2. Bb2 c6 3. e3 Bf5 4. Nf3 Nf6 5. c4 e6 6. Be2 h6 7. O-O Nbd7 8. d3 Be7 9. Nbd2 O-O 10. Qc2 Bh7 11. Bc3 Bd6 12. Qb2 Qe7 13. b4 e5 14. a4 Rfe8 15. b5 d4 16. exd4 exd4 17. Nxd4 c5 18. Rfe1 cxd4 19. Bxd4 a6 20. Bc3 axb5 21. axb5 Rxa1 22. Qxa1 Bxd3 0-1

Below is my modified version starting move 21. See for yourself in engine. I think greek gift works. But I didn't let the engine run to 30/99 depth or anything. Of course the original is 4 points better (modified is only 2 points up. Original is 4 points up).

Question: Am I wrong in thinking that greek gift works i.e. black still has an advantage?

[Event "chess stackex 2021/10"]
[Site "HUN"]
[Date "2021.10"]
[Round "?"]
[White "Istvan Pozsonyi (mod)"]
[Black "Gyozo Pataki (mod)"]
[Result "0-1"]

[fen "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1"]


1. b3 d5 2. Bb2 c6 3. e3 Bf5 4. Nf3 Nf6 5. c4 e6 6. Be2 h6 7. O-O Nbd7 8. d3 Be7 9. Nbd2 O-O 10. Qc2 Bh7 11. Bc3 Bd6 12. Qb2 Qe7 13. b4 e5 14. a4 Rfe8 15. b5 d4 16. exd4 exd4 17. Nxd4 c5 18. Rfe1 cxd4 19. Bxd4 a6 20. Bc3 axb5 21. axb5 Bxh2+ 22. Kxh2 Rxa1 23. Qxa1 Bxd3 24. Bxd3 Qxe1 25. Qxe1 Rxe1

From chesstempo:

the only problem with the greek gift in this position is that it’s too much gift and not enough greek. that bishop isn’t that pinned.

Answer 1

As you've shown in your analysis, the line after ...Bxh2 does indeed give Black a decisive advantage. However, Black also has a completely winning position without going for ...Bxh2, so why bother calculating the complications when there's an easy way to have a good advantage?

Answer 2

What's the point of the Greek Gift? Do you gain anything by decoying the White king to h2? If not (and I don't see any gain; in fact you lose a little since the final ...Rxe1 does not come with check), there's no need to give up the bishop; you might as well play 21...Rxa1.

Answer 3

You ask why it doesn't work. To make things simpler, we move Pd3 to e3 such that all tactics in the e-file become a nonissue. When does Bxh2+ work anyway? The standard conditions (obviously there may be myriads of concrete alterations when it works a different way than standard) requires:

• A: You get your knight to g4.
• B: You get your queen to h4 (or maybe d6 or whatnot).
• C: Either those two mate the king on their own with Qh2 etc., or Black has reserves.

Here A is already violated - the wB can take on g4, thus any Bxh2+ tactics must rely on the pin in the e-line, and that, as we already saw, can be exploited much easier. No chance for Bxh2 with Pe3. Thus we can say:

Bxh2 works, but in a non-standard way. But White must try to play Nf3 (why throw the rook into the bin) with the intent to hand back the pin to Black, get the B back with Nd4 and cut the losses to a pawn. Now Bxh2+ nets a second pawn as the Q can unpin the bBe2 with check, and incidentally, White gets into dire mating straits. (The whole line, which I didn't computer-check: 21...Rxa1 22.Rxa1 Bxd3 23.Nf3!? Bxe2 24.Nd4 Bxh2+! 25.Kxh2 Ng4+ followed by Qh4 or Qg5, depending on where the wK goes. This line is more or less Greek Gift.) If you play Bxh2+ first and Bxd3 later, then a White Bxd3 cuts the losses in the sense that Black just needlessly gave a B for a P in comparison to the Bxd3 first variant, as already pointed out in the other answers.

I hope I made clear why this is a rather nonstandard Bxh2 issue. For the standard one (including elaborating on my simplified ABC) there exists tons of teaching material giving you a hint when it usually works and when not.