Now, under chinese rules, this is a huge win for white, under Japanese rules, I suppose the case exists that if black passed 354 times and white just kept filling his own liberties, then this would be a close game.

So, under chinese rules (which I tend to prefer with confusing game states), I'm going to go ahead and hypothesize that if I run out of stones, and there aren't any prisoners to swap, then the person who plays their last stone (again, without there being a swap available on the board) is winning by so much that you should be able to declare victory.

Would be a nifty alternate win condition.

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Tactics yes, Tact no...

and the players were taking turns??

no self-respecting go player would ever play like this. we are talking about a game of Go, correct?

We're talking about theory and the example just server its purpose of showing it is possible to run out of stones.

(and, even ignoring the "no true Scotsman" fallacy, not all go players are self-respecting.)

But the point is to get this board b needs to pass 175 times, whereas in a normal game of go black passes once. You can also run out of stones if you start feeding them to your goat, or using them to build a rock garden. I don't think we need to to deal with degenerate cases.

1. If you're playing with a full set, you can never run out of stones so long as black and white have gotten to move equal numbers of times. If you run out of stones in the bowl, you can swap prisoners. It cannot be true that all your stones (181/0) are on the board and your opponents stones have liberties. (this, interestingly, is the premise of no pass go.)

2. If you split up full sets among multiple games to save money, then of course you can run out. Below 180, the fewer the stones you have, the more often you'll run out. Figuring how often j games with k sets of stones amongst them will completely run out of stones might be a fun project (for someone else, I mean, not for me).

I'm not convinced that it's impossible (though obviously it is not likely). Suppose that White has 151 stones on the board, 30 white stones are in Black's bowl. Black has 181 stones on the board, 31+ of which are captured but not removed (there will have been prisoner exchanges, but no more are possible). There are then 29 empty intersections that can be liberties for living stones. Such a situation can even be a near perfect tie (I'm too lazy to do math).

Is there some reason the situation I describe is impossible, or must be the result of bad play?

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Occupy Babel!

I think you mean b has 29 caps. Where are the 29 dame that resulted from b's capture?

Perhaps there was a large squeeze that was played out, resulting in the captures and killing the black group, but not removing it from the board. This might happen as part of a ko fight, if the circumstances are right.

There need not be that many dame on the board--if there was a capturing race featuring a big eye, or squeezes, or whatever else, you can create captures without creating a lot of liberties.

And I should have said this but those numbers are schematic.

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Occupy Babel!

no we're not.

you opened discussion with "what happens in a real game, if..."

I submit that in a real game, no one would ever need more than 180 stones.

The following scenario occurred to me:

Black and White play a "regular" game of go, each playing a stone at each turn, with no intention of passing until the end. At some point, White kills and captures a very large Black group, over 50 stones, say, and removes them from the board as prisoners. But so far, Black has no White prisoners.

Black could resign, but they decide to continue the game. But with so many Black stones removed, there is enough space for many more moves. And with a "head start" of his earlier stones, eventually White has over 180 stones on the board, while Black has far fewer.

Now, is White clearly winning? Not necessarily. It could be that it is White who blunders this time, and a large White group is clearly dead. But those dead stones are still on the board - during the game, they cannot be removed until Black has filled in all their liberties. In other words, Black still has no White prisoners to offer for exchange. So at this point, White may run out of stones to play, and no prisoner exchange is possible to alleviate the situation. And it could be that it is Black who is headed for the win.

Now, is this scenario possible in a "real" game, between "self-respecting" players? Well, if it is between two professionals, or between two strong amateurs, probably not. Most likely, Black would have resigned after his large group was captured.

But between weaker players, and especially beginners, it's not so clear. These kind of see-saw battles, where one player pulls ahead, then the other, are not unheard of.

So yes, I think that in a "real game", someone might need more than 180 stones. (And the game outcome is not necessarily a forgone conclusion at that point.) Unless, of course, you want to claim that a game between weaker players is not a real game

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And the go-fever which is more real than many doctors’ diseases, waked and raged...
- Rudyard Kipling, "The Light That Failed" (1891)

thanks, you made me smile today

OT, I love how the "no true Scotsman" fallacy has "For the practice of wearing a kilt without undergarments, see True Scotsman." in a tagline under the title.

I feel like this is a question for KGS statistics.

Has there ever been a game in which one side played more than 181 moves?

If so, has there been a game in this set where the number of shared prisoners was less than the number that one side played over 181 moves?

I feel like it's possible and probably happened. (Especially in Computer: New Zealand matches)

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Tactics yes, Tact no...