EdLee wrote:
Hi Bill,
Thanks for being patient with these 30k questions.
Quote:
- Click Here To Show Diagram Code
[go]$$ A
$$ +---------------------+
$$ | . O . O . . X . X . |
$$ | O O O O O O X X X X |
$$ +---------------------+[/go]
Black has 2 pts. White has 2½ pts.
The whole number 2 is clear (for both B and W).
How to derive the ½ for white?
The easy way is through the method of multiples (i.e., the mean value theorem).
- Click Here To Show Diagram Code
[go]$$B 2(A) Score = -1
$$ +---------------------+
$$ | . O . O . 1 X . X . |
$$ | O O O O O O X X X X |
$$ | O O O O O O X X X X |
$$ | . O . O . 2 X . X . |
$$ +---------------------+[/go]
This board contains 2 copies of A. The score is -1. That means that A is worth ½ pt. on average. The same is true for 2N copies, no matter how large N is.
and
are miai. It does not matter who plays first, or which copy they play in, the score will be the same.
Quote:
I can see one can argue there's a 50% chance W will get the extra 1 point.
But how to derive the (magical?) 50% odds?
There is more than one way to derive the probability. But IMO the clearest way to think about this is in terms of fuzzy logic, that is, in terms of
possibilities. Unless we know better, the possibility that Black will play first in A is equal to the possibility that White will play first.
In any event, the mean value justifies the probabilities, not the other way around.