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There have been several questions regarding the 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 another example, White pawns cannot occupy the first row and Black pawns cannot occupy the 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 the 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 are written for chess pieces?

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  • Your estimate is too low. You shouldn't just compute combinations, it matters which pieces are on which squares. 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
  • The number of legal chess positions is roughly 4x10^44: github.com/tromp/ChessPositionRanking
    – John Tromp
    Aug 8 at 15:39
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Upper bounds on the number of allowable positions are discussed on the Shannon number Wikipedia page.

There is an accepted upper bound of 1.8x10^46 by Shirish Chinchalkar, and a somewhat accurate estimate of 4x10^44 by John Tromp.

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

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  • 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
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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.

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  • @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? Jul 21 '20 at 4:58
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    It's a start. You also have to take promotions into account which will make this bound much larger. Jul 21 '20 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 Jul 21 '20 at 17:24
  • Wikipedia mentions 2x10^40 as the best known upper bound without promotions. Jul 21 '20 at 18:18
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It is really difficult to find out legal positions but total positions possible is 13^64 , assuming each piece can be positioned in every available square

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