# Endgame Problem 40 / Solution

Sub-page of EndgameProblem40

The atari at is slightly better, on average, than the sagari at a. The sagari leaves Black with 4 points. The position after , while still hot, has a count of 4.0375. In a real game, the sagari may be correct.

Of course, now we have the question of how we derived the value of 4.0375.

If White plays first we have a position with a count of 1.075.

If Black plays at instead, the local score is 7. As these are gote, the average count is 4.0375.

How do we get the count of 1.075?

After the count is -1.85. (Per convention White scores are negative.) If Black plays the score is 4. The average is 1.075.

How do we get the count of -1.85?

After the count is -0.2.

After the count is -3.5.

The average count is - 1.85. is tempting, but the count after is -1.6, for an average loss of 0.25 point.

After the local score is -5 (counting the Black stone already captured). If instead, the local score is -2. The average is -3.5, as advertized.

The count of the next position is -0.2 (counting the captured Black stone). elsewhere. fills the ko.

The local score is -3. White has made 2 net plays to get here. , elsewhere. fills the ko.

The local score is +4. Black has made 3 net plays to get here.

The difference in score between these positions is 7 points. The difference in plays is 5 plays. The miai value of each play is 7/5 = 1.4 points.

So the initial position in the ko has a count of 4 - 3 * 1.4 = -3 + 2 * 1.4 = -0.2.

### Discussion

Karl Knechtel: b was also proposed in the problem statement; what is the miai value here? Also, what situations would make the sagari correct? (I assume they have to do with the ko threat environment?)

Bill: The sagari may be better if this ko is unfavorable for Black.

The atari at leaves the same count as the sagari, but White can still make the ko.

Endgame Problem 40 / Solution last edited by BillSpight on May 4, 2004 - 16:51