[Computer-go] Combinatorics of Go

[Kahn Jonas]

> Intriguing!
>
> A position is obviously illegal if any point is occupied by a stone 
> surrounded by opposite-colour stones.
> At the 4 corners, 25 out of 27 combinations will be legal. The proportion 
> (25/27)^4 will survive.
> At the 68 edges, 79 out of 81: (79/81)^68 will survive.
> At the 289 interior points, 241 out of 243: (241/243)^289.
>
> Multiply those, I get 0.012321913.
>
> So presumably the number is on the high side, because this calculation only 
> takes account of
> single stone blocks illegally on the board.

I think it's rather (much) on the low side. The probabilities are not
independent. And the correlations are positive: in particular, if a
single point is legal, it may be that it is next to a free intersection,
which immediately makes legal four (at the center of the board) other
points.

Jonas