# What is the number of legal positions in a chess game?

There have been several questions regarding possible number of chess games. for example - Database of every possible move in chess

However Can there be an estimate on the number of "allowable" chess positions.

For example White Pawns cannot occupy first row and Black Pawns cannot occupy eighth row. If a given chess piece is occupying a square, then another chess piece cannot occupy that. Using simple elimination rules coded, can we have an estimate on the number of unique positions that chess pieces can have. Since they would be reducing in number. An upper bound on positions before applying any elimination rules would be sum of (64 C 32 + 64 C 32 ... + 64 C 2). This is less than 10^20. With elimination rules, they should be significantly lower.

Any ideas on how elimination rules be written for chess pieces?

• Your estimate is too low. You shouldn't just compute combinations, it matters which pieces are on which squares. – Dag Oskar Madsen Jun 16 '14 at 10:36
• The maths goes like you choose the 32 squares possible, and find number of pieces you can place on each square. Then multiply 64 C 32 – QuIcKmAtHs Dec 27 '17 at 12:25

Upper bounds on the number of allowable positions are discussed on the Shannon number Wikipedia page.

There is an accepted upper bound of 5×10^52 by Vicor Allis, and a less verified upper bound of 2^155 (approx 10^46.7) by John Tromp.

By comparison, the number of atoms on earth is about 10^50.

• The article fairly estimates the positions to be close to 10^50. Link is nwchess.com/articles/misc/Chess_Board_Positions_article.pdf – shoonya Jun 16 '14 at 9:34
• However this can be computed to a far better estimate, especially the cases when Kings are already in checks, Kings are simultaneously in check. The key thing being estimating this can allow us to typically find the optimal moves for a given position. – shoonya Jun 16 '14 at 10:11

Old question, but for people who just want a quick estimate they can do with basic math this is an easy way to go about it

If we consider all 32 pieces, we can rearrange them P(64,32) or 64 permute 32 ways. If we then consider 31 pieces, we do the same as above with P(64,31) but this time we must factor in the choice of 31 out of the 32 pieces so we multiply P(64,31) by C(32,31) or 32 choose 31. If we continue along this manner we have the sum from n =2 to 32 of P(64,n)C(32,n) which gives 1.24e54 which is only a few orders of magnitude larger than the number given by Vicor Allis.

Note that in the assumptions, we allowed illegal positions, and double or triple counted certain positions by treating each piece as distinct. But the number is reasonable compared to that above and more importantly is easily attainable with pretty basic math.

• @DagOskarMadsen I did notice you have a PhD in math, so does this upper bound seem reasonable or is there a way to improve it? – Colin Hicks Jul 21 at 4:58
• It's a start. You also have to take promotions into account which will make this bound much larger. – Dag Oskar Madsen Jul 21 at 12:32
• @DagOskarMadsen ah yes. I changed C(32,n) to C(96,n) to allow for all 16 pawns to assume the identity of the 4 other piece types other than king but now the value is 1.454e79 which is fairly off the mark haha – Colin Hicks Jul 21 at 17:24
• Wikipedia mentions 2x10^40 as the best known upper bound without promotions. – Dag Oskar Madsen Jul 21 at 18:18