Ed,

Interesting calculation, but I think you can do better. Since this is only a count of board positions WITHOUT any captures, any given pawn can only occupy one of six squares, NOT sixty, right? Also, once a white pawn is advanced, this decreases the number of available squares for the black pawn in the same rank by the number of squares the white pawn is advanced and vice versa.

What you've calculated is that there are at MOST x board positions without caputres, but what we really need is to know that there are at LEAST x positions. ;-)

Still interesting, though.

Thad

Yes, as stated on the page, I wanted to calculate a reasonable upper bound that could not be exceeded.

The lower bound can be estimated easily by not placing the kings on the board at the end, and restricting the pawn count as you indicated. It gets messy though, when you consider a position that has a legal en passant capture, even though not executed, can also look the same as a position where the en passant capture opportunity has expired. These identical board setups are actually two different positions.

The same is true of castling rites. If I move my king to the left, then move it back to the original square, but it looks like I can castle, it is really a different position than an indentical setup wherein I have not moved the king.

If someone wants to work out the numbers to that level of detail, they are welcomed to it!

I also made a formula error for the Bishops. I listed them as a summation of a choose formula, rather than of the form (T/2)^2 where T is the total number of vacant squares that 2 opposite color bishops can occupy. The formula should not be T choose 2 as I indicated at the top of the page, since that implies the Bishops can be both on the same color squares.

I did explain this part of the calculation properly but I did not redraw the formula yet.

More on the pawn count...

Originally I had the non-captured pawn permutaion count at 774,468,423,574,600,034,220 which, as Thad pointed out, seems a little high. When using the concept of the "rammed pawns" constrained to a rank, there are actually 30 different configurations possible per file.

You can have an unmoved pawn in the a-file on a2, and the black pawn can be one of 5 different squares in the same file, a7,a6,a5,a4 or a3. If the white pawn were on a3, then there are 4 squares for the black pawn, etc.

There are always 5-r permutations for the black pawn for each rank of displacement, r, the white pawn has moved.

So, there are 5+4+3+2+1 = 30 possible configurations per file. The total number of permutations is therefore 30^10, which is 590,490,000,000,000. This is only 590 trillion, a lot less than 774 quintillion!