Re: On Game Space Size (Monte Carlo)

[John Clarke <johnc@xxxxxxxxx>]

Hi


> Now, if we discard the map and just create a random board position we will most often create a board with the same number of black, white and empty points only at rare occasions a board position is created with only one stone. THIS IS ALLRIGHT ! because the even distribution of black white and empty stones should weigh heavier than the empty board situation !

Absolutely.  I have only half been paying attention to this thread
but to estimate the number of legal positions just take

n_legal = P( random position is legal ) * 3^361 

> It is however of no importance to the Monte Carlo estimate.
> 
> Is there anyone who can verify the above for me ???

That's right.  Please post you final number - I'm interested to know
what it is.


Re: counting arguments:

If you consider just the positions marked X

...............
.X.X.X.X.X.X.X.
..X.X.X.X.X.X..
.X.X.X.X.X.X.X.
..X.X.X

etc...

then the probability of a black or white stone at one of the X's being
dead is (2 / 3) * (1 / 3)^4 = 0.00823... as someone already noted.

So the probability of at least one dead singleton i.e. at least one of
the X's is a single isolated dead stone is

 1 - (1 - 0.00823....)^145 = 0.698...

Combining this with a similar analysis for the edges I get an upper
bound
of 0.122... for the proportion of legal board positions.

This is only an upper bound, the true value will be lower.  I'd guess
it's a few percent, but it will be nice to know for sure.

So it reduces 3^361 a bit, but not enough to get excited about.

Regards
John