from one of wierd earl’s link - the number of possible positions in Go on a 19x19 board is greater than the number of subatomic particles that it would take to fill all the space in the universe??
I wonder what the corresponding calculations for chess are?
PS: That is one heckuva geeked out numbers page.
Nice that the Straightdope gets a link right at the top of the page.
Zenster: Not quite as much. Chess is too mathematical/logical by far to have as many positions (and it’s just a 8x8 board). I’d give you a solution, if I hadn’t forgotten my calculator at home. 
According to this page, about 1.17 * 10**49 positions are possible on a chess board. Since this is an attempt to encode the board rather than a calculation of legal positions, the actual number will be considerably smaller.
I couldn’t get to the OP link but found another source that claims the number of possible Go positions to be 10**750. “Not quite as much” might be in the running for the understatement of the week. 
[hijack]Zenster, I travel a lot. In what part of what country do you reside? I’ll bring a guitar.[/hijack]
Great, the page is unavailable, now I’ve got to punch some numbers…
Radius of observable universe (R[sub]U[/sub]): 15x10[sup]9[/sup] light-years = 4.5x10[sup]18[/sup] meters. cite, using the larger value.
Volume of observable universe (V[sub]U[/sub]): 4* pi * R[sub]U[/sub][sup]3[/sup] / 3 = 3.8x10[sup]56[/sup] meters[sup]3[/sup]
Volume of proton (V[sub]P[/sub]): 5x10[sup]-46[/sup] meters[sup]3[/sup]. cite
Number of protons to fill observable universe: V[sub]U[/sub] / V[sub]P[/sub] = 7.6x10[sup]101[/sup]
Number of Go positons (19x19 board, every spot has a white stone, a black stone, or is empty): 3[sup]361[/sup] = 1.7x10[sup]172[/sup]
Well, that beats number of protons it would take to fill the universe, although “number of subatomic particles” gets a little fuzzy once you start trying to get the volumes of things like quarks. And neither of those come close to NoCoolUserName’s 10[sup]750[/sup].
Hmm, the number of “games” (unique set of steps going from empty board to full board) might do it. My first guess is that it would be 361!, right? Empty board, first player has 361 options, second player 360, first 359… My windows calculator gives that value as:
1.4x10[sup]768[/sup].
Although, since a game of Go can stop after any number of moves, the number of unique games would be the sum of ( 361! / (361 - x) ! ) where x goes from 1 to 361…
[brain implodes]
Just to help your brain get a little more imploded:
In Go, it’s possible to take a large number of your opponents pieces and then play in those spots again, but it’s not possible (legally speaking) to fill in every single intersection with a stone. I suspect the 10[sup]750[/sup] number (which I found on a Go site–it’s not my number :)) was approximate because of these considerations.
sciguy, I’m impressed not only by the fact that you actually did the calculations, but that you coded all the super-and sub-scripts correctly! Great post!!
**<- **[POINTS AT USER LOCATION]
Cool. Please email me. Electric and acoustic guitars on tap.