There are (64!)/(32!) positions just with all the prices on the board...
Er, you sure about that? I think I prefer the gibberish.
So long as we're talking about basic rules positions at any rate (@tygxc obviously isn't).
No I ment positions in total not legal positions, that is the number if you just put every piece in a random position.
Only if you label the pieces, e.g. calling your white horses Prancer and Charger.
one of the things they check when they check if the position is legal or not is stuff like bishops not on the same colour, pawns not on first and last row, kings not near each other ….
Last is normally king of side not to move not in check, but you haven't mentioned side to move. Some tablebases have extra constraints like no triple checks (but unless you can get hold of the generating code you'll probably need to find out by trial and error). The upshot is to significantly reduce the number.
The best available estimate of legal positions under basic rules to date is the paper you didn't like, as far as I know. And it's not 10^44 as @tygxc keeps trying to tell everyone, but (4.82 +- 0.03) * 10^44.
If you want to talk about legal positions under competition rules and equate them to game states (as @tygxc does) then you have to increase that number by a ridiculously huge unknown factor (as @tygxc doesn't - he instead divides it by something rather less huge but more ridiculous).
(! Is factorial, it is the number times every integer smaller, so 5! = 5•4•3•2•1=120, 6! = 6•5•4•3•2•1=720, 10! = 10•9•8•7•…•2•1=3,628,800)
I assumed so.
Yes if you label the horses that was a very basic number to start with, but as I said if you make the calculations with the horses and the pawn unlabelled(and the king and the rooks have a “can castle” option) and you include the positions with less than 32 pieces is comes to about the same number, a bit less, my point with 12/2,000,000 still stands.
No it doesn't.
You haven't quantified what "about the same number, a bit less" is, so you don't have a point. The number you originally came up with was (8! x 2! x 2! x 2!)^2 = 104044953600 times too high.
You then proceed to assume Tromp's measured ratio of legal to total positions will also apply to positions corresponding to the diagrams in your estimate, whereas you include all diagrams with pawns on the first and eighth ranks, 100% of which are illegal, and Tromp doesn't include any, so that would be invalid even if you were to quote the correct ratio.
@6919
1 e4 e5 2 Ba6? is irrelevant to solving chess.
We know that loses by force for white.
It is not necessary to burn computer engine time on something we already know.
Such trivial obstacles are needed if you want to not solve chess.
Relevant questions:
How to draw against 1 e4, 1 d4, 1 c4, 1 Nf3
How to draw against 1 e4 e5 2 Nf3, 2 Nc3, 2 Bc4, 2 d4
How to draw against 1 e4 e5 2 Nf3 Nc6 3 Bb5, 3 Bc4, 3 Nc3, 3 d4
Etc, etc.