Ok, so let’s explore these numbers a bit.

Suppose that I have a cup. I can either put a black stone in it, a white stone, or no stone at all. There’s three states that the cup can be in: containing a black stone, containing a white stone, or containing no stone.

Now, suppose I have two cups. Now there are 3^{2}=9 ways the two cups can be: black-black, black-white, black-empty, white-black, white-white, white-empty, empty-black, empty-white and empty-empty.

If I had three cups, there would be 3^{3} ways the cups can be filled (or not).

Now, suppose I have 361 cups. Now there are 3^{361} ways to either place black stones or white stones in the cups, or leave them empty. In other words, there are 3^{361} ways I can place stones randomly on a goban. Not all of these are legitimate board positions in Go, (there could be groups without liberties for example), but this is a mindbogglingly large number, around 10^{172}.

It turns out that about 1 in every 80 random board positions is actually a legal Go position. If you know a little bit about computer programming, you can easily check this yourself by generating boards filled with random stones, and seeing if there are any groups without liberties. I might make a simulation myself later to illustrate this.

So, to get some decent approximation of the number of legal Go positions, we have to divide 10^{172} by 80. To get, roughly, 10^{170}…

So, the number of legal Go positions is mostly huge because many Go positions are simply random stones thrown on a go board.

Now, let’s tackle the number of Go games.

If I have a Go position with 1 Black stone and 1 White stone, then there is only 1 possible way we can reach that position, by playing the black stone first, then the white stone.

If we have a position with 2 Black stones (a and b) and 2 White stones (c and d), then we reach that position in 4 ways: a-c-b-d, a-d-b-c, b-c-a-d and b-d-a-c. In particular, there are 2 possibilities for the first black stone, and in each of those possibilities only 1 possibility for the last black stone, and similar for white. So in total there are 2 x 1 x 2 x 1 ways to reach our board state.

If we have a position with 3 Black stones and 3 White stones, there are 3 x 2 x 1 x 3 x 2 x 1 ways to reach this position. Usually this is written as (3!) x (3!).

Now suppose I have a random Go position, then I can expect about 1/3 of the stones to be Black, 1/3 to be White, and 1/3 to be empty, since it’s random. So that’s 120 Black stones, 120 White stones. In total, that gives me (120!) x (120!) different ways to reach that end position.

This in itself is around 10^{397}, and that’s just one of the 10^{170} random board positions.

That means that the total number of possible games of Go, is at least somewhere around 10^{397} x 10^{170} = 10^{577}. Why at least? Because we’re currently only counting games where none of the stones are captured. There’s no long ko fights, no filling the entire board and then capturing everything and start over again, etc.

Anyways, I hope that explains a bit why these numbers are so large.