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The count of a local position is the net amount of territory (minus prisoners) or area (if area scoring is used) we calculate for it. When the position has been played out and the count is definite, we just call it the score.
Count is like the expected or average score, but it is a stronger concept. If there are no kos (either now or later), and the overall count is in your favor, and you have sente, you will win with correct play. You can't say that for averages or expected scores.
Conventionally, the count or score is just given as Black's score minus White's score, since it is only the difference that matters anyway.
See Method of Multiples for one way of determining count and further explanation.
Maybe the simplest way to understand the count is with the aid of game tree diagrams. (In these particular diagrams, all stones are alive.)
If instead White plays, the resulting position is C. We can represent the position with the following game tree.
A / \ B C
The game tree starts at A and if Black plays then the resulting position is shown on the left branch at B but if instead White plays, the resulting position is shown on the right branch at C. Note, we are only representing local positions, i.e. small regions of the board where there are still moves to be made.
The game tree might be more complicated, as follows.
A / \ B C / \ D E / \ F G
Here, position G is what would happen if you started with A and then White played and then Black played and then White played again. Similarly, E is what would happen if White played twice in a row. The positions B, E, F and G are the final positions where there are no more useful moves left. For these positions, the count is simply the score. Since we are dealing with local positions, we have to modify the usual score concept. It will suffice to count only territory inside an imaginary rectangle drawn around the position and any captures made locally.
The next type of positions to consider are simple gote positions. For these positions, the count is the average of the count for the position after Black plays and the count for the position after White plays. For example, the count for D is the average of the count for F and G. Then also C is the average of D and E, and A is the average of B and C. The reasoning behind this is that it is equally likely that either Black or White will play in the position first, so the count is simply the average.
Then there are simple sente positions. Suppose when Black plays in position C creating position D, it is sente. This means that White will immediately respond, creating position G. In this case, position F (which is very good for Black) will not be reached. Because a player usually gets to play their sente plays, the count for position which is sente for one player can be assumed to be the same as the count for the position after the sente is played.
So, if we suppose that C is sente for Black and that the score for positions B, E and G is +1, -3 and -2 respectively, then we know the counts for those positions and so the tree looks like this:
A / \ B(+1) C / \ D E(-3) \ G(-2)
The number in parentheses is of course the count for each position. Then we can work out that the count for C is also -2 by the sente position rule. Then we can work out that the count for A is the average of +1 and -2 by the gote position rule. The final game tree is:
A(-.5) / \ B(+1) C(-2) / \ D E(-3) \ G(-2)
It is not always obvious whether a play is sente or not, but it is easy to work out from the game tree. If we take the example from above and assume that the play from C is not sente for Black. We start with the diagram on the left and then work out the counts for positions C and D with the simple gote rule to get the digram on the right.
C C(-1.5) / \ / \ D E(-3) D(0) E(-3) / \ / \ F(+2) G(-2) F(+2) G(-2)
When Black makes a play from position C, if our game tree is correct then he gains 1.5 points on average. If Black did choose to make this play, then it was because a 1.5-point play is the highest-value play on the board. In that case, it is now White's turn and the local play for White (from D to G) is worth 2 points, higher than all other remaining plays. White would therefore play again straight after Black, making the sequence sente for Black by definition. It was wrong to assume that it is equally likely that Black and White will play in the position D, so we can simply remove F from the tree and assign C the same count as G.
Another exception is the reversal, in which there is a sequence of several plays.
A / \ B(-1) C / \ D E(-5) / \ F(0) G(-2)
Suppose that this is a game tree with these final scores. Then the intermediate values are as follows.
A(-1.5) / \ B(-1) C(-3) / \ D(-1) E(-5) / \ F(0) G(-2)
Note that the count of A is the average of that of B and G. It is not the average of that of B and C (simple gote), nor is it the count of D (simple sente). The plays from C and D are gote, but they are hotter than the play from A, and so play does not stop until G.
Again, we can work this out by first assuming that the plays are simple gote. C and D are correct. Our first attempt at A yields this tree:
A(-2) / \ B(-1) C(-3) / \ D(-1) E(-5) / \ F(0) G(-2)
-2 is the average of B and C. But that is not right, because -2 is less than D (-1). So we try A as sente (removing E for clarity):
A(-1) / \ B(-1) C / D(-1) / \ F(0) G(-2)
In that case A is a 0 point sente, because B - A = 0. But D is a 1 point gote (D - G = 1). So play will continue to G, and we get the final result for A.
A(-1.5) / \ B(-1) C / D \ G(-2)
- BQM 127 — Miai value of entering 3-deep 2-wide corridor:
Bill: IMO, this page in its current form has too much information. It is above beginner level. Instead of putting everything on this one page, how about making links? That is why we have hypertext. :)
 RobertJasiek: This is so because White plays at A with the gote move value 1 and Black replies at C with the gote move value 2. The increasing gote move value increases the count from A (-2) to D (-1) while violating the value behaviour of gote.