The shortest number of moves that a game of chess can have is 2, as far as I know:

  • White moves pawn from f2 to f3, black moves pawn from e7 to e5
  • White moves pawn from g2 to g4, black moves queen from d8 to h4. Checkmate.

Which results in this situation: End game

There might be more games which end after 2 rounds, but as far as I know there is no game with fewer rounds / moves.

How many moves do the longest games take?

I thought I have read that the number is about 8000 moves.

A finite maximum number exists, because of the fifty-move rule and threefold repetition and I assume that players claim draw by those rules if possible.

Please link also to the source of your information!

edit: Jacob Schlather mentioned a blogpost with this information

[...] The Belgrade Marathon was a contest between Ivan Nicolic and Goran Arsovic that lasted over 20 hours and ended in a draw after 269 moves due to the so-called “50 Move Rule”, [...]

Source: The Longest Possible Chess Game

So the longest game that was actually played took at least 269 moves. Later he explains how he comes to 5,870 possible chess moves.

  • $\begingroup$ There's some pretty good work on this blogpost with a further discussion in the comments. There seems to be some contention based on what exactly 50 move rule you pick. It seems to around 5950 or 5900 or 5850... $\endgroup$
    – JSchlather
    Sep 11, 2012 at 6:35

3 Answers 3


There is no upper bound on the length of a legal chess game. All that is required is for the game to reach a position where the threefold repetition rule would apply, and for neither player to claim the draw. The same position could recur over and over, and so long as neither player claims the draw, the game can continue indefinitely.

  • $\begingroup$ To the downvoter - the question was edited after I answered it. This answer was correct when I posted it, even though it's incorrect now! $\endgroup$
    – user22805
    Sep 11, 2012 at 7:55
  • $\begingroup$ Yes, we must make some (weird) additional assumptions. A player having the opportunity to apply the 50move rule either thinks he can force win - but then he must make a pawn/capture move because otherwise the opponent cuold claim draw in his next move. Or he thinks he is about to lose - and will happily claim draw. The only case is where the game is "objectively" draw and we may assume that the payer will then rather end this boring game. :) $\endgroup$ Sep 11, 2012 at 7:57
  • $\begingroup$ This does not take into account the five fold repetition rule. $\endgroup$
    – PyRulez
    Oct 2, 2018 at 1:14

Longest game that can be played with neither player having the opportunity to claim a draw under the 50 move rule (or any other rule): 5898 moves.

Longest game that can be played under current FIDE rules without being automatically drawn: 8847.5 moves.

Explanation: Each pawn can make 6 moves on the way to promotion. To do so without capturing each other, 8 pieces have to be sacrificed along the way (to open up the files for them to pass each other), leaving 22 pieces after all the promotions (plus the two kings). Those 22 pieces can all be captured, but the last capture ends the game (as King vs King is an automatic draw based upon insufficient material to bring about checkmate). Being forced to occasionally (four times) switch the side moving the pawns/making the captures (first to open up all the files for the pawn moves, and later to finish capturing all the pieces) means that four half-moves--two moves--are lost. (Only a minor amount of cleverness is needed during the stalling moves to insure that there is no possible claim of draw due to threefold repetition of position.)

Most internet sources I have found (for instance, Sonny_E's comment on the blogpost mentioned in the question, and many sites he lists) show that this leads to the following calculation:

2 sides * (6 pawn moves to promotion * 8 pawns + 8 captures of opposing promoted pawns + 3 captures of other opposing pieces than the four that need to be captured by pawns) * 50 move rule - 4 times that the side that makes pawn moves or captures switches * (1/2 move) = 5898.

It might seem (it did to me) that the calculation should use 49.5 moves instead of 50, to avoid allowing each player to claim a draw immediately before every capture or pawn move, since 50 full moves will have passed since the last such event. However, the 50 in this calculation includes 49.5 stalling moves, and then adds the pawn move/capture for a total of 50.

White can win on his 5898th move (having one piece left), but if he does not, Black must finally be able to claim a draw on his 5898th move: he can capture White's last piece (if possible) and force a draw base on insufficient material to mate, or his move can leave White's piece alone, letting him claim the draw based on the 50 move rule, the first time that rule could be invoked the entire game.

But the players may not want to claim draws at all, or even entertain the possibility of drawing, and so intentionally avoid making claims and risking draws. FIDE recently added a 75 move rule (9.6b) that automatically draws a game without recent pawn move or capture regardless of either player's claiming a draw (or even being made aware of it). Doing the same calculation with 74.5 moves in between captures/pawn moves gives 8848 moves, the longest possible chess game under current FIDE rules. Black's 8848th move would automatically end the game based on the 75 move rule (or on the insufficient material rule, since Black could only avoid that rule by capturing White's final piece, leaving King vs King). Hence the longest possible non-drawn game would involve White checkmating Black on his 8848th move (or either player resigning just before Black's 8848th for that matter, since it is that move that causes the draw to happen, and resignations can still be offered even if the next move would necessarily result in a draw).

  • $\begingroup$ I'm quite certain that Sonny_E's analysis is incorrect; the longest game has just three changes of control, resulting in 5898.5 moves using the 50-move rule, or 8848.5 moves using the 75-move rule. See for example chess.com/forum/view/general/… $\endgroup$
    – Deedlit
    Aug 24, 2015 at 15:00

Each of the 16 pawns can move at most 6 times and there are 30 captures possible. Therefore $(16\cdot6+30)\cdot 50=6300$ is a rough upper bound (for example, not all pawns can make it to the opposite line without sometimes capturing - which would mean that sometimes pawn move and capture occur together).

  • $\begingroup$ Foiled again! Good observation - should your 50 be 49 - the draw can be claimed on the 50th move. Note also that the 50 move draw and threefold repetition have to be claimed and are not automatic - they have been missed on the board ... $\endgroup$ Sep 11, 2012 at 6:31
  • $\begingroup$ To get a non-rough upper bound, one also notes that there can be 50 moves before the first pawn move or capture. $\:$ $\endgroup$
    – user57159
    Sep 11, 2012 at 6:31
  • $\begingroup$ Yes, but after the 30th capture the situation will be K-K and thus draw anyway. $\endgroup$ Sep 11, 2012 at 6:33
  • 2
    $\begingroup$ @DavidWallace - the length is then infinite by only moving knights, which is not an interesting question. It could be reformulated as a game between players/computers which always claim when they can. $\endgroup$ Sep 11, 2012 at 7:29
  • 2
    $\begingroup$ @moose: We assume that in each block of consecutive block of 50 moves at least one of the following events happens: a) a pawn move b) a capture. There are a total of 30 captures possible and a total of 16*6 pawn moves possible. That's waht I wrote. Of course, in order for a pawn to make more than 4 moves, it or its opposing pawn must either make a capture (which decreaeses the limit as it is pawn+capture at the same time) or his opposing pawn gets captured prematurely (i.e. it is deprived of at least 2 moves). Thererfore 16*6+30 can be lowered considerably. $\endgroup$ Sep 11, 2012 at 7:52

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