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I have recently bought an exercise copy and there, in the cover page I got an amazing fact about chess board, "There are total $10^{120}$ board positions in a chess board."

But I was just thinking how to prove this.

My work:

I think to first find the no of board positions including all pawns(yet don't know how to do that) and then get a motivation excluding some no of pawns. Can anyone give some idea?

Blue
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AK001
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    seems like this would be highly nontrivial if we're actually interested in the number of reachable positions. if we just care about how many ways we could uniquely arrange the pieces on the chessboard, the calculation is fairly straightforward – user3716267 Sep 22 '21 at 18:30
  • Are you interested in how many ways to arrange the chess pieces or possible positions in a real game? – Brian Lai Sep 22 '21 at 18:32
  • My approach would be to do the most stupid estimation. First calculate how long a chess game can maximal be. This is given by the 50-Move-Rule. After 50 moves where no pawn is captured or is moved, the game ends in a draw, so try to prolong this as long as possible. Then in every move take the 32 chess pieces and arrange them at the 64 fields. Maybe this already leads to $10^{120}$, but this "method" is very stupid and would give the maximal sensable estimation one could give. – Cornman Sep 22 '21 at 18:38
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    The estimate seems high. Even if we assume that each piece is unique (obviously, pawns, rooks, knights are not), the number of arrangements of 32 pieces on 64 squares will be the order of $10^{53}$ – Vasili Sep 22 '21 at 18:43
  • I heard of an estimate of about $10^{43}$ legal positions. – Peter Oct 23 '22 at 14:16
  • $10^{120}$ is definitely far too high , if we only consider legal positions. – Peter Oct 23 '22 at 14:18

2 Answers2

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$10^{120}$ is the Shannon number, a lower bound estimate of the number of possible chess games.

For number of positions, this is on the Shannon number wiki:

"John Tromp estimated the number of legal chess positions with a 95% confidence level at $ 4.5\times 10^{44} \pm 0.37\times 10^{44}$ based on an efficiently computable bijection between integers and chess positions."

https://en.wikipedia.org/wiki/Shannon_number#cite_note-5 https://github.com/tromp/ChessPositionRanking

Brian Lai
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I think that's a rough estimate of how many games you can create using legal moves. That is a very strange estimate given by $20^{80}$ catering to the fact that in each position, there are an average of $20$ moves you can play, and an average chess game lasts for $40$ moves. I found this information in a Numberphile video. I guess, it's called Shanon number.

But, if you want to have an elementary estimate of the number of (legal) board positions, you can go like-

The two kings has to be there. So, place them on any two squares in $\binom{64}2$ ways.

Now, the rest $30$ pieces may or may not be included in $2^{30}$ ways. Arrange them in the other $62$ squares in $\binom {62}{30}$ ways.

So, the total number of ways is $2^{30}\times \binom{64}2\times \binom {62}{30}\approx 9.7\times10^{29}$

Note that this estimate doesn't eliminate the chance of two kings being on adjacent squares which is illegal. Also, this estimate may put two bishops in the same coloured square (which is half illegal since considering pawn promotions make it legal). That reminds us, we also haven't taken pawn promotions into account.

Sayan Dutta
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