Counting A Three Stage Ko
If Black fills the ko the score is 3 points (for Black). If White takes the three legs of the ko with White 1, White 3, and White 5, and then fills the ko, the score is -4 points (i. e., 4 points for White). Each move in the ko gains 1 2/5 points. So the count in the original position is 1 3/5 points.
Very long iterated kos are possible, but you seldom see one with more than four legs.
Kiko: I don't understand the math here. Three points for Black vs. four points for White means playing the four moves (1, 3, 5, and connect) is worth 7 points. 7/4 = 1 3/4. How do you get 1 2/5?
Bill: If Black fills the ko (1 move), the result is +3. If White takes the ko three times (3 moves) and then fills the ko (1 move), the result is -4. 5 moves make a difference of 7 points. On average, each move is worth 7/5 points.
Karl Knechtel: Or you can do a full analysis of the move tree like I did at Endgame Problem 40 / Attempts, and come to the same conclusion. You can do the same thing for a simple ko with one leg to get the 1/3 point value; that case is a lot simpler. Of course, counting moves between the results is a lot nicer mathematically, but I wonder if it would be any easier to program ;)
The complicated calculation is analogous to summing an infinite series. It reminds me of the story of the "fly and bicycles" problem being posed to Von Neumann...
At any rate, it appears that in general the plays in an n stage iterated ko have a miai value of (3n-2)/(n+2).
Bill: It's more complicated than that, Karl, too complicated to go into here.