Can you place a stone and cause both players to take a prisoner at once?

See this board I was playing yesterday (first time playing in 17 years!):

In the space I’ve circled, I’m confused on the rules. From what I understand, either player is “allowed” to play in any available space.

So, if black placed a stone there, it would immediately have no liberties. Does that mean that it would immediately be taken prisoner by white? The thing is, that move would also cause the unit of two white stones directly above the open space to have no liberties. So wouldn’t those two white stones also become black’s prisoners? I can’t tell if one, both, or neither things would happen.

What’s amazing is that you can see this exact pattern also occurred just above and to the left of this–in reverse! Really screwed up our game, since we didn’t know the rules governing play here. Please help!



Hello, Pauly, welcome to OGS.

If Black plays there, he will capture the two White’s stones and the black stone will remain on the board with one liberty (“whoever’s turn it is captures first”).

To further explore the position - as the new Black stone will have only one liberty, White can capture it back with the next move. This is NOT ko, as the situation does not repeat, it is different - 2 captures against 1.


Got it, thank you @AdamR for such a quick response. So helpful!

I guess someone better than me will come in soon, but because the black stone will remove white’s last liberty, it “overrules” its own “temporary” capture. The final result is not black being prisoner, since there is still one more action, capturing the two white stones.

Please see this link, it explains it nicely

(@AdamR, ah, few seconds :slight_smile: )