Although most players are familiar with the concept of half an eye, that concept can be extended to a few other fractional values, such as the ⅓ eye, ⅔ eye, ¾ eye, 1¼ eye and 1½ eye.
Generally, we can say that in situations where the fractions add up to a value of 2, the group is alive even if the opponent moves first, if it is 1 or less, the group is dead even if the player moves first. Between 1 and two it depends on sente and may be ko.
The most familiar fractional eye is the half-eye.
Each of the marked points is half an eye. Since Black also has a full eye, his total eye value adds up to two, and he is alive (White cannot kill)
If we remove one of the half eyes, Black’s eye value adds up to 1½ and life or death depends on who moves first. Black can move and get two eyes, White can move and reduce Black to one eye.
In this situation Black also has 1½ eye. Black can make it two, or White can make it one, by playing a.
Using ko, we can construct one third or two thirds of an eye.
Consider these positions: the intersections marked will be an eyes if Black can win the kos. This suggests they are worth some fraction of an eye, but how much?
Using the method of multiples (§ Ko), we now argue that the above eyespaces are worth ⅓ and ⅔ eyes respectively; this turns out to yield sensible results within a single group.
Let us take multiple copies of the first position and think about them in groups of three.
In any such triplet, if White starts we can expect her to connect one, Black to take another, White to connect a third with and Black to connect his. If White plays a ko threat for , Black can simply answer it with , then at , at , leaving the situation as before with one White threat less. (We ignore the more complicated situation when White has infinitely many threats from a double ko elsewhere.)
If Black starts, we have the same result, in the order ––-.
In either case, Black saves one of the three groups,
as though he had transferred 2 fractional eyes to that group
(even though complete eyes are clearly not transferable)!
If there are more than three, the same argument shows that Black will on average save one in three of them; at the very end there may be one or two kos left, but that does not affect the limit. Since it takes three of these fractional eyes to save one group, we try considering them worth a ⅓-eye, and the same argument suggests that such we have a a ⅔-eye when Black leads the ko.
We now show that the values ⅓-eye and ⅔-eye yield sensible results when they occur within one group.
Consider this position first: we have a whole eye, a ⅔-eye and a ⅓-eye, making a total of two, so Black should be alive.
This group has three ⅓-eyes and a whole eye, making 2 in total, so it should also be alive.
Indeed, if White moves first to connect one of the kos, Black can take another to create the previous position; again he lives with double ko.
Removing one of the three ⅓-eyes leaves Black with 1⅔ eyes. Since this value is between one and two, we expect the position to be unsettled.
Indeed, Black can live (still in double ko) if he plays first, while it is ko if White moves first.
Similarly, with a whole eye and a single ⅔-eye Black can live (outright) if he plays first, while White can start the ko.
Reducing it again to a single ⅓-eye leaves 1⅓, meaning it should still be unsettled.
Indeed, White can kill by moving first, while Black can get ko.
With 2 ⅔ = 1⅓ eyes, it should be unsettled.
Indeed, Black can connect to make one eye and fight a ko for the second, but White can reduce this to the following case of death in double ko.
This group has ⅔ + ⅓ = 1 eye, so it should be dead.
Indeed, Black is dead in double ko: he needs to win both kos for life, but cannot because whenever he takes a ko, White will retake the other.
This group has only three ⅓-eyes, making 1 in total, so it should be dead.
Indeed, if Black starts by taking a ko, White can connect another, reducing it to the previous position.
These are trivially dead, as expected
We can also construct eyes that are ¾ or 1¼ of an eye. Again, if the sum of such values reaches two, the group is alive.
By playing either point marked a, Black can create a 1½ eye. Since both sides are symmetrical and therefore miai, the fact that Black can make a 1½ eye combined with the half eye at b means he is alive even if White starts (and takes one of the stones). Each side is ¾ of an eye, so with the half eye, it is ¾ + ¾ + ½ = 2 eyes.
Again, Black is alive. If he plays b he obviously lives, but if White plays there he can play a and then either make two eyes on the right or, if White prevents that, capture the white stone at b.
Since a is ¾ of an eye, that means b is 1¼ of an eye. This is because Black can move to two eyes, while White can move the half an eye, so the average is 1¼.
- Eyespace Values in Go at http://www.msri.org/publications/books/Book29/files/landman.pdf in Games of no Chance, Volume 29 of the MSRI Book Series
 Since this is double ko life, it is not absolute: White may capture in the double ko as a threat in a ko fight elsewhere on the board, and Black may choose to ignore that threat. Note that although that situation would be a triple ko, it is not no result even under Japanese rules, as the double ko is not a double ko seki and cannot lead to a full-board repetition. (PJT: But could we combine positions with ⅓-eyes to create a cycle and, if so, does that say anything about the values?)