How many chess games are possible?

https://win-vector.com/2026/01/27/how-many-chess-games-are-possible/

by jmount • 6 months ago

85 46 6 months ago

tromp3• 6 months ago

For the chess problem we propose the estimate number_of_typical_games ~ typical_number_of_options_per_movetypical_number_of_moves_per_game. This equation is subjective, in that it isn’t yet justified beyond our opinion that it might be a good estimate.

This applies to most if not all games. In our paper "A googolplex of Go games" [1], we write

"Estimates on the number of ‘practical’ n × n games take the form b^l where b and l are estimates on the number of choices per turn (branching factor) and game length, respectively. A reasonable and minimally-arbitrary upper bound sets b = l = n^2, while for a lower bound, values of b = n and l = (2/3)n^2 seem both reasonable and not too arbitrary. This gives us bounds for the ill-defined number P19 of ‘practical’ 19x19 games of 10^306 < P19 < 10^924 Wikipedia’s page on Game complexity[5] combines a somewhat high estimate of b = 250 with an unreasonably low estime of l = 150 to arrive at a not unreasonable 10^360 games."

Our final estimate was that it is plausible that there are on the order of 10^151 possible short games of chess.

I'm curious how many arbitrary length games are possible. Of course the length is limited to 17697 plies [3] due to Fide's 75-move rule. But constructing a huge class of games in which every one is probably legal remains a large challenge; much larger than in Go where move legality is much easier to determine.

The main result of our paper is on arbitrarily long Go games, of which we prove there are over 10^10^100.

[1] https://matthieuw.github.io/go-games-number/AGoogolplexOfGoG...

[2] https://en.wikipedia.org/wiki/Game_complexity#Complexities_o...

[3] https://tom7.org/chess/longest.pdf

jmount2• 6 months ago

Nice stuff, thanks for sharing that.

I remember from a lot of combinatorial problems (like cutting up space with hyper-planes or calculating VC dimension) that one sees what looks like exponential growth until you have a number of items equal to the effective dimension of the system and then things start to look polynomial.

BTW: I was going through some of your lambda calculus write-ups a while ago. Really great stuff that I very much enjoyed.

qsort3• 6 months ago

I wonder if/how that interacts with the new draw rule. (For the uninitiated: the formal rule to adjudicate games as draws automatically or on time is that the game is a draw if there exists no sequence of moves that could lead to checkmate. Interestingly, although this has almost no strategic implications, it means that... it's almost impossible to write a program to detect draws that's technically correct. A similar corner case is draws in Magic the Gathering, which is literally undecidable in general.)

Sesse__5• 6 months ago

Is that a new rule? I was under the impression that it had been the case for a very long time that if you went out on time but there was no possible sequence of moves leading to checkmating you, it was a draw instead. (Meaning, of course, that having more pieces could be a disadvantage in such situations, which feels a bit unfair. E.g., KvKB is a draw, but KPvKB can lead to a mate if both sides cooperate, and thus would be a time loss for white even if black would never win in practical play.)

Someone4• 6 months ago

the game is a draw if there exists no sequence of moves that could lead to checkmate. Interestingly, although this has almost no strategic implications, it means that... it's almost impossible to write a program to detect draws that's technically correct.

I don’t see what makes that technically difficult. The number of possible positions is finite, so just enumerate the game tree and check whether it contains a checkmate situation.

I also don’t see why it would be almost impossible in practice. Aren’t the only weird situations ones where there are pawns that could be promoted to queens if they weren’t blocked by other pawns, and those pawns prevent all other pieces on the board from taking pawns and from checkmating the king?

lxgr2• 6 months ago

For 7 remaining pieces or less, there are actually tablebases of all possible positions showing whether there's a possible win or loss for either side: https://en.wikipedia.org/wiki/Endgame_tablebase

Someone4• 6 months ago

I do not see how that’s a good estimate. For example, take a game length of, on average, 4 and a branching factor of 10. That gives an estimate of 10,000.

Chances are there are games of lengths 3 and 5, too. With that branching factor, there are 1,000, respectively 100,000 of those, for a total of 111,000. That’s over ten times as many games as estimated.

The larger the spread in game length towards games that are larger than average, the more the proposed estimate underestimates the actual number.

tromp3• 6 months ago

That’s over ten times as many games as estimated.

That's still a pretty good estimate of an exponentially large quantity; the exponent being off by only 1. With these estimates you cannot hope to do better than estimating the exponent.

GMoromisato• 6 months ago

One thing I always wondered is how many moves, on average, do you have to play before reaching a position that has never before seen on Earth?

Or maybe the question should be what percent of games reach a position that has never before been seen?

tromp3• 6 months ago

I think that the average chess game played between humans contributes between 20 and 40 new positions (note that a 30 move chess games has 60 plies).

bdamm• 6 months ago

You'd probably need to make a determination of the skill of the players. A very strong player vs a novice could be scholar's mate most of the time.

reassess_blind• 6 months ago

Yes, the stronger the players, the more often they will both go deeper into established theoretical lines that have been played before.

dpc050505• 6 months ago

A very strong player would show the novice the scholar's mate once and then move on to hanging tactics and pieces on purpose so that the novice starts seeing things, probably leading to positions that are a lot more rare.

recursivecaveat• 6 months ago

Apparently ~75% of the positions in the lichess database (as of 6 years ago) have only been seen once ever. Average game length is 30-40 moves, so for the completely average player it would be like 10+ moves I suppose. The stronger the players the longer it will take: I found some comments suggesting 20+ for high level players.

matusp• 6 months ago

I don't think the math is correct here. The 25% of positions that have been seen more than once represent more than 25% of the occurrences. Even if all of them would be seen only twice, you should already see them in 40% occurences.

squidbeak• 6 months ago

It depends totally on the opening. You can be out of book and database far quicker than that for offbeat stuff, or in book far longer for popular openings.

Another distinction needs to be made between positions seen and positions played. Almost every viable position will have been seen in preparation well beyond 10 moves. But seeing them on the board is rarer.

RA_Fisher• 6 months ago

Infinite. :) Chess is strictly unbounded.

Mordisquitos• 6 months ago

That was also my intuition. Unless there's a rule that can stop two immortal but dumb-as-bricks players from indefinitely cycling through the same non-capturing moves surely the answer is 'infinity'.

eulgro• 6 months ago

Well there is. The three/five fold rule. And 50 moves rule.

DSMan1952762• 6 months ago

It depends what rules you're using, but there are the three-fold repetition and 50-move rules which allow a player to force the game to end in a draw. The catch is they both require one of the players to claim a draw under the rule, otherwise they can keep playing.

There is additionally the 75-move rule where the the game is forced to be over without either player claiming the rule (the arbiter just ends the game), that rule would give an upper bound regardless of the players knowledge of the rules.

erehweb• 6 months ago

This link https://wismuth.com/chess/longest-game.html from the article talks about the 2014 changes (75-move rule and draw by 5-fold repetition) that make it no longer infinite.

paulpauper• 6 months ago

meh. I think it would have been more interesting had the author discussed more granular estimates. Mathematicians have narrowed it down more by considering the properties of the pieces and bijections.

LegionMammal9782• 6 months ago

Assuming you're referring to [0], that's a statistical estimate of valid chess positions (based on clever methods of uniform position sampling + fast validity testing), not valid chess games (based on estimating branching factors for very long games).

[0] https://github.com/tromp/ChessPositionRanking

jonas_kgomo• 6 months ago

I watched a movie a few days ago and they basically said there are more states in the game of chess than atoms in the universe? https://www.youtube.com/watch?v=xfMQ7hzyFW4

adonovan• 6 months ago

Sure, but in combinatorics the number of atoms in the universe (say 1e80) is not a large number. For example, the factorial of 59 is larger. If you own 30 pairs of shoes, there are factorial(60) ways to arrange the individual shoes in a sequence.

BobaFloutist2• 6 months ago

To be fair, that's atoms in the observable universe.

The total size of the universe is unknown, and could (and likely does) have way more atoms than that.

Actually, that's a fun thought: assuming homogenuity of matter between the observable and unobservable universe, how much bigger would the unobservable universe need to be to render some of these claims no longer true?

Because you're right to point out that factorials grow absurdly quickly. It's entirely possible my caveat straight up doesn't matter.

Edit: Ok, I'm seeing Wikipedia has a (disputed) estimate for the diameter of the total universe as 10^10^10^128 megaparsecs. Then, radius cubed should be 1/2(10^10^10^128^3)=1/210^10^10^131, as opposed to the radius of the observable universe being a nice, clean 14 billion parsecs = 1410^3 megaparsecs, making the radius cubed 1410^4 megaparsecs. I don't think I have a big enough calculator for this, but for fun, let's say 128^3 is roughly 2,000,000. Then we can rewrite T, the relative volume of the total universe, as 1/210^10^10^2*(10^ 6). I guess if we call 14 close enough to 10, then our density is 10^80/10^6=10^74 atoms for every pi megaparsecs cubed.

Going off the heuristic that n!<n^n, and the total universe can trivially produce (10^10)^(10^10), we would need to rearrange >10^10 objects just to even start to think about the number of (megaparsecs cubed)/pi it might have, let alone the 10^74 those each have.

We might not have enough decks of cards for this one.

(Feel free to criticize/tear down my math or logic anywhere in this one, it's very much off the cuff and I'm sure I made at least as many egregious errors in computing exponents as I did computations. No math class I've taken yet really prepares you to handle exponents raised four deep.)

How many chess games are possible?

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