How many different chess positions are there?

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DeepFlight12

I'm lost here. Define unique

Victorinox2

https://www.youtube.com/watch?v=Km024eldY1A

DirtyGriffinIsBrown

I did a combination calculation and got ~3.00 x 10^46. This number is not 100% accurate as it includes positions where multiple pieces exist on the same square and illegal positions (eg. pawns on 1st and 8th rank ect.).

 

Edit: the number may be ~ 1.64 x 10^ 28 depending on whether the final number is multiplied by 1.83 x 10^18. Sorry for the ambiguity; I hope this was helpful.

folksgl

Since the question does not specify legal vs illegal positions, the answer can be found exactly with some relatively simple math. Forsyth–Edwards Notation dictates that a position consists of 6 unique parts. Two of these parts involve the move number so we can discard those, as they don't distinguish one position from another. Another section of the fen indicates the passant target square that can be captured from the position, which we can discard since it doesn't distinguish this position from other positions, only the possible permutations that can be reached from that position. That gives us a 3 part modified fen string to give us all unique chess positions possible.

In order to find all possible combinations of the 3-part fen string, we can make the following calculations:

Number of possible pieces per square = 6 black pieces + 6 white pieces + 1 empty = 13 possible states per square.
Number of possible board states = states per square (13) ^ number of squares (64) = 1.96053476E71

So the first part of our fen string has 1.96053476E71 possible states.

The second part of the fen string is the color to play, which can be in 2 possible states: white or black.

The final part of our 3-part fen is the castling rights of both players. We can calculate the possible states of this section similar to the first section:

Number of possible states per slot = 2 (can/can't castle)
Number of possible slots = 4 (white king/queenside and black king/queenside)
Number of possible castling rights = possible states (2) ^ 4 (number of slots)
So the third part of our fen string has 16 possible states.

So the total number of possible chess positions is:
1.96053476E71 * 2 * 16 = 6273711245784354346581112333554112493580904960131705202866910842951966752

IMKeto

4...

Ziryab
folksgl wrote:

Since the question does not specify legal vs illegal positions, the answer can be found exactly with some relatively simple math. Forsyth–Edwards Notation dictates that a position consists of 6 unique parts. Two of these parts involve the move number so we can discard those, as they don't distinguish one position from another. Another section of the fen indicates the passant target square that can be captured from the position, which we can discard since it doesn't distinguish this position from other positions, only the possible permutations that can be reached from that position. 

 

By discounting whether an  en passant capture is possible, you have artificially (and in violation of the rules of chess and the definition of a position) lowered your estimate.

This position was reached after 28...f7-f5, but it would be a different position if Black had just played some other move.



SciFiChess

A solution to the original question as well as some related questions are posted on my blog. There is a Quiz and a page with Solutions.

SciFiChess

I have noticed that authors of math text books make the odd numbered problems easy and the even numbered problems hard! evil.png

Lumenoak

I've looked at this from the perspective of possible board states for each piece. For example, each bishop can only occupy one of 32 spaces. Each pawn can reach between 21 and 32 spaces before it is promoted.

Promoted pawns have a total of 256 (64 × 4 unit types) on the board states.

Allowing captures except for the kings gives a total number of piece states of 1.1857500663886417E67 as an upperbound. This is still not the true upper bound as the calculation allows pieces to share a state, ie rest on the same board square. It also includes situations where the only difference is switching the kingside/queenside/promoted pieces in the same positions.

Chef-KOdAwAri
There are more possible games of chess than atom in the entire known universe
Chef-KOdAwAri
The Shannon number, named after Claude Shannon, is a conservative lower bound (not an estimate) of the game-tree complexity of chess of 10120, based on an average of about 103 possibilities for a pair of moves consisting of a move for White followed by one for Black, and a typical game lasting about 40 such pairs of moves.
SciFiChess

One theory is that our universe is infinite in size. The observable universe universe is the portion of the universe close enough to Earth that light could travel from that point to Earth within the length of time that has elapsed since the Big Bang.

SciFiChess
gf3 wrote:

big bang is the ruy lopez of universe theories

haven't u got anything with more piazzazz

Multiverse

ryanfff

A LOT

austinmcmurtrie

Please join the official international club of Chess. com. https://www.chess.com/club/chess-com-international-1

johntromp

See https://www.chess.com/forum/view/general/on-the-number-of-chess-positions for some good estimates.

ChessOfficial2016

There are over 100,000,000 possible positions in a chess game.

DiogenesDue
ChessOfficial2016 wrote:

There are over 100,000,000 possible positions in a chess game.

You need to add 38 more zeroes to that number...

tygxc

8726713169886222032347729969256422370854716254 possible of which 8% are presumed legal
19201527561695835455154058755564594798074 sensible without excess promotions of which 8% are presumed legal

DiogenesDue
tygxc wrote:

8726713169886222032347729969256422370854716254 possible of which 8% are presumed legal
19201527561695835455154058755564594798074 sensible without excess promotions of which 8% are presumed legal

10^40ish...

Congrats, you finally realized that your 10^20 estimate was ~20 orders of magnitude off happy.png.  Welcome to a more realistic conversation.