RobertJasiek wrote:
- Click Here To Show Diagram Code
[go]$$B Count?
$$ -------------------
$$ | O . . . . . . X .
$$ | O X 1 X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
How to calculate the count? Is it also 5? I think that yes but I am not used to microendgame calculations. Is the next follow-up move a sente or reverse sente?
I'm no endgame expert, but here's my take on the inquiry above.
From this position, it's gote both for white and black to play at 'a':
- Click Here To Show Diagram Code
[go]$$B
$$ -------------------
$$ | O a . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
Therefore, it's a 50% chance that black plays here and ends the position:
- Click Here To Show Diagram Code
[go]$$B Position A
$$ -------------------
$$ | O B . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
In this case, that's 6 points for black with 50% probability. What about the other 50% chance? Well, that comes down to the count of this position, where white played first:
- Click Here To Show Diagram Code
[go]$$B Position B
$$ -------------------
$$ | O W . . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
So what's this count? Recursively apply the procedure above... GIVEN this position, there's 50% chance that black blocks:
- Click Here To Show Diagram Code
[go]$$B Position C
$$ -------------------
$$ | O O B . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
So 50% chance of getting 5 points. But what about the other 50%? Well, it's the value of this position:
- Click Here To Show Diagram Code
[go]$$B Position D
$$ -------------------
$$ | O O W . . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
Repeat procedure, again. 50% chance of black playing first and ending, this time for 4 points:
- Click Here To Show Diagram Code
[go]$$B Position E
$$ -------------------
$$ | O O O B . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
And 50% chance that it's the value of this position:
- Click Here To Show Diagram Code
[go]$$B Position F
$$ -------------------
$$ | O O O W . . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
which is 50% chance of this:
- Click Here To Show Diagram Code
[go]$$B Position G
$$ -------------------
$$ | O O O O B . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
(3 points) vs. 50% chance of this (1 point):
- Click Here To Show Diagram Code
[go]$$B Position H
$$ -------------------
$$ | O O O O W . . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
At this point, it's questionable as to whether the white move is gote. Black gets a point by responding, so it's relatively large. So I'd say that it's likely going to elicit a response from black. In other words, the board position will end like this, given the scenario above:
- Click Here To Show Diagram Code
[go]$$B Position I
$$ -------------------
$$ | O O O O O B . X .
$$ | O X X X X . X X X
$$ | O X . O O X X X .
$$ | O O O O O O O X X
$$ | . O . O O O O O O[/go]
for a final position of 1 points. That's a terminal state.
Now roll the evaluation back up to the beginning.
0.5*Value(A)+0.5*(0.5*Value(C)+0.5*(0.5*Value(E)+0.5*(0.5*Value(G)+0.5*Value(H))))
Which comes to:
0.5*6+0.5(0.5*5+0.5*(0.5*4+0.5*(0.5*3+0.5*1))) = 5 points...
That's my interpretation, anyway, but in a real game, I'd say, something like, "meh, maybe a bit over 5 points?".
How far off am I in this calculation?