Half-point ko / Discussion

minimal ko  

... if white plays, she has invested one move more than black (i.e. 1 white move, 0 black moves). If black plays, he has to capture and connect (=2 moves) while white does not play there (=0 moves). Therefore, black played two more moves than white. A possible ko-threat play by White is not counted, because it is outside this position.

Thus the difference of the possible outcomes is 3 moves and 1 point value, thus a miai value of 1/3.

Did I get it right? :o)

Bill: Well put, Arno! :-)

Alistair Turnbull: Furthermore, it doesn't matter how long the ko fight continues, because in each capture-threat-answer-capture-threat-answer cycle the two players play the same number of moves.

typical half-point play  

In this diagram, white has half a point, under the assumption that it's even probability that each player will be the next to play locally (in turn based on the assumption that there are roughly equal-value plays elsewhere on the board). Therefore, whoever plays at 'a' gains half a point for that move. That's about the limit of my confident understanding of miai counting ^^; But assuming I have the basic idea right, I apply it to the minimal ko as follows:

minimal ko  

Now there is an even chance that white will play next. Define the result to be net 0 points difference. The alternative is that black eventually wins the ko, and gets +1 point as a result.

If, at each move, it is an even chance that either white or black will play again locally, the chance that black wins the ko immediately is 1/4; that white wins immediately is 1/2; the remaining 1/4 of the time, black plays and then white plays, so that the original situation is restored. Thus, when we sum the infinite series, we get a total chance of 2/3 that white wins, and 1/3 that black wins.

Thus the current result gives white a disadvantage of 1/3 point from the 'reference' position where white has won the ko. Thus, playing locally is worth 1/3 point for white. Symmetrically, it is worth 1/3 point for black, since the black play would shift the value from 1/3 to 2/3 in black's favour (the value would become 2/3 since it's 1/3 less than the +1 value for black winning the ko - 1/3 less by the same argument as above).

This way I don't have to think about counting moves or amortizing the value of a play over the moves in the sequence (which I find a bit counterintuitive). But I'm not sure it's any easier to understand ;) (Actually, I just make it sound complicated, I think.)

MK In fact this reasoning nearly convinced me. Could you please make the 'infinite series' part clearer?

victim 1/2 - 1/4 + 1/8 - 1/16 + ... = 1/3.

Another way to come to the same result: Let's call the position where Black can capture or White can connect, B. If White connects, we get zero. If Black captures, we get a position we call W, where Black can connect or White can capture. So B = 0 * 1/2 + W * 1/2, if the probability is 1/2 for each outcome.

Similarly, for the next move, W = B * 1/2 + 1 * 1/2. Now we have two equations with two variables.

B = W/2

W = B/2 + 1/2 = W/4 + 1/2

W*3/4 = 1/2

W = 2/3

B = W/2 = 1/3

By the way, this is not as useless as it seems to be at first sight. The computation works the same way for a bigger ko. So, as a first estimate of how big capturing a ko is, compute the difference between "White wins" and "Black wins" and multiply by 2/3. If you don't have much time, play a gote move bigger than that rather than playing the ko.

Two-stage ko  

Of course the number changes if it's a two-stage ko.

Capturing or connecting here is worth the same as two points in gote, as an approximation. Use the same methods as above to calculate that.

Score is 1  

Black plays first. White connects, since it's no use to fight the ko. Black 3 also connects (below 1), since again it's useless to try to win both kos. Finally White connects. Black got a stone, that is, 1 point.

Score is 1  

White plays first. Again, after Black captures with 2, White simply connects the other ko. Black 4 below 2. Again Black got 1 point.

So the value of the position above is 1. Therefore, the value of each of the 3 kos is 1/3.

Is this reasoning correct? It looks so simple that I feel something is missing.

victim It's fine. Of course one has to remember that it's still an approximation.

1/4 point reverse sente  

- = threat and reply

takes ko

elsewhere

takes ko

1/4 point reverse sente (cont.)  

- = threat and reply

elsewhere

takes ko

- = threat and reply

takes ko

elsewhere

fills ko ()

elsewhere

Black has gained 1 point with 4 net local plays, each worth, on average 1/4 point.

Bass: Doesn't this immediately generalize to a 1/n point ko:

1/n point "ultimate gote" endgame?  

If black is a komonster, and is able to win the ko n (or so) times so that he can connect at all the n (or so) places along the side, then gains a point with n (or so) moves. Maybe I used some premises not needed for the 1/4 point situation? Or maybe I am counting the "white first" position wrong?

tapir: Same thought here, but no. The ko is about more than one point because if Black doesn't win the ko n times he will lose much more. Bill explained very patiently at tapir/MinimalKo. (It is encouraging to see the same misunderstandings by other people though :))

RobertJasiek: The position makes a nice external ko though.

Bill: When I discovered the 1/4 pt. ko position, it was my hope that it generalize as Bass suggests. But it does not, alas! I have not completely analyzed the diagram, but I am pretty sure that gains 1/3 pt. Note the careful placement of the stone, else an initial play there would be large. :)

tapir: You are joking aren't you? If White is komaster she will capture those four stones and some more. How will you ever arrive at an average 1/3 pt. gain?

Bill: I was assuming that Black is komaster. Looking again, I see that Bass is assuming that Black is komonster. :)

tapir: I know, I posted what I believe is a very similar example on this subpage of mine. Still you use the difference between Black winning and White winning the ko to arrive at the net count and divide that through the local tally. At least that is what you have done elsewhere. Even only considering it as a two stage ko and resolving it by capturing 4 stones results in >10 points difference, disregarding all the rest, even this is much more than 1/3 pt. per move, and being able to capture more after resolving the ko as White certainly does not cool it down to 1/n or 1/3.

Ko under area scoring  

Herman: Suppose that a is the last half point ko, and b is the last dame in the 5x5 game. There is no komi. Black to play, a or b?
Analysis:

Defending the ko  

Black has 12 points, white has thirteen, white wins by one point.

Taking dame. @ , pass, @  

Black has thirteen points, white has 12, black wins by one point.

So, the correct way to play is to take the dame instead of filling the ko.

Note that this depends on the fact that black has a ko threat, and white does not! If there was no ko threat, suppose there was already a white stone at in the last diagram), then this line of play fails (see next diagram)

pass, @  

With already there, doesn't work. Here, black has 11, white has 14, so white wins by 3.

RobertJasiek: Also see Ko and Dame Endgames.