How many times can the same position repeat?

I'll try to answer OP's question (as I re-stated in post # 62). I think the answer is significantly higher than the answers posted here so far.  But first I need clarification of the rules: if the only difference between 2 positions is that 2 same-colored knights (or bishops, or rooks) have switched places, is the position considered identical with regards to the repetition rule?  Without searching the USCF or FIDE rule books, I would guess that the 2 positions would be considered identical, i.e, we don't distinguish between the king's knight and the queen's knight.

Assuming the above is correct, then there are 3 general ways that 2 "optically identical" positions can differ:

1. Whether it's white's or black's turn (2 distinct states)
2. Castling options (4 distinct states per color: O-O only, O-O-O only, both, or neither)
3. Availability of en passant (2 distinct states)

Any set of "optically identical" positions are not repetitions if they differ in any of the ways listed above. But note that the en passant move must be available in the initial position corresponding to the first "optically identical" position (since no pawn moves are allowed subsequently). So the en passant option only adds one "optically identical" position (so we will add 1, not multiply by 2 when calculating the possible combinations of differences).

To achieve the maximum number of "optically identical" positions before actual repetition, we need to start with full castling rights available fro both sides, and a pending en passant move already present.

The number of uniquely different "optically identical" positions should then be:

(turn states) x (W castling states) x (B castling states) + (alternative en passant state)

= (2 x 4 x 4) + 1 = 33

Now add 1 more position since after all of the merely "optically identical" positions are exhausted, the last such position will occur 1 more time before 3-fold repetition finally applies.

Answer = 33 + 1 = 34.

I won't be surprised if anyone finds mistakes in my reasoning, and I welcome corrections.

EDIT: it just occurred to me that I'm over-counting the number of possible castling states.  I can't have some positions where white has only O-O available (but not O-O-O), then others (in the same game) where white has only O-O-O available.  Once O-O or O-O-O becomes unavailable, it can't be resurrected.

I did it the old fashioned way with the starting position.  I made moves until the computer flashed draw, three-fold repetition.  The maximum number of "repeats" was 11.  I estimated that if I could have found a way for Black to move first, the maximum number of "repeats" would have been 13.  This is consistent with the OP's estimate minus the double counting of castling opportunities. 

Yes, it's an academic exercise, of course. I believe it's extraordinarily rare that in an actual real game the same position would repeat even just 4 times before the three-fold repetition rule kicks in, not to talk about the whopping 22 times!

However, it's very interesting to ruminate on what the theoretical absolute upper limit is, considering all the rules pertaining 3-fold repetition. I don't think many people would have guessed that it's as much as 22 times.

Yeah, you can only count four repeats of the piece arrangement for each one of the four castling rights (two for positions where white is to move, and two for ones where black is to move). That's 16. Then prior to that we can have 4 repeats (2 for each side to move), bringing it up to 20. Then we can have that one initial en-passant repeat, bringing it to 21. Then the position can finally repeat one last time and the 3-fold repetition rule finally kicks in, for a total of 22.

@DjVortex, we're almost in perfect agreement.  I would revise my original calculation to have 3 possible castling states for each color: can castle long or short, can castle only one way (long or short but not both), or can't castle at all.  Then the maximum possible number of "optically identical" positions should be:

(2 x 3 x 3) + 1 = 19

And then this last "optically identical" repeats (in the true sense) one time *before* the 3-fold repetition rule applies:

19 + 1 = 20

The repetition rule could then be correctly invoked on the 21st occurance of the said "optically identical" position.  Or would the answer be 20 since the repetition rule must be invoked just before the 3rd repition appears on the board? (so that the opponent doesn't just instantly bang out a move to invalidate the claim)

You are counting incorrectly. You are assuming that each player has three possible "castling states", and that we go through each possible combination of them (ie. 3x3). But that's not possible. You can't count one player's "castling states" as many times as the other player, because that would mean repeating the same "castling states" for the same player many times. They can only happen once.

It's easier to think of there being 4 castling rights on the board, and repeating the same piece arrangement twice for each side, for each of those 4 castling rights. That would be 2x2x4=16 (ie. for each of the four castling rights we can have the same piece arrangement twice, for the same player to play, and again twice for the other player to play).

Prior to this, ie. when all four castling rights are still existent, we can repeat the same piece arrangement twice for each player to play, ie 4. prior repetitions. Then we add even prior to that the en passant situation, and finally at the very end of the whole thing one last time and the rule kicks in.

@DjVortex, I need to think about what you wrote regarding castling states.  But doesn't the en pasant move need to occur first, since it changes the pawn structure?

Your math is wrong because you are considering White and Black totally separately!

21 times is the maximum number that it can be repeated without 3-fold, then the 22nd occurrence must force 3-fold.

What you aren't considering is that you have to consider White and Black castling rights together.  You can't bring castling rights back to life, so it's not all permutations, as the edit in post 64 indicated, but you have to think of it like this:

1. En Passant.  This will only apply for 1 move, so this can only occur once!  That's 1.

2. Now, instead of saying White can castle both ways, one way, and no ways, and same with Black, think of them TOGETHER!  You have:

• Both sides can castle either direction (that's 4 occurrences, twice with Black to move, twice with White to move, added to the one prior is now 5 occurrences of the position, no 3-fold)
• ONE side can castle EITHER direction and the OTHER can ONLY CASTLE ONE WAY.  For example, White moves one of his Rooks.  Don't do the same with Black in parallel.  Now do 4 more occurrences, we are at 9.
• Now, it doesn't matter if Black loses a way to castle or White loses his other way (You can force either side to be on the move as the 4 occurrences could be, to move, W-W-B-B or W-B-B-W, so you can make it end with the same player to move as started.  But for the sake here, let's say Black is to move and moves a Rook.  Now White has 1 way to castle and so does Black.  4 more times and we are now at 13.
• Now one side loses all rights of castling by moving either the other Rook or the King.  The opposite side still has 1 way to castle.  4 more occurrences makes 17.
• Now lose all castling rights for the other player, so that neither side can ever castle again, and you get 4 more occurrences.  21.

Only now, the next time that position occurs, no matter who is to move, it WILL be 3-fold, and that will be the 22nd occurrence.  So it is indeed 22, not 20 or 11 or 34 (post 64) or any other numbers thrown out there!

Actually, there are 5 states of castling rights, no matter which order they lose their rights.  There are 2 possible paths:

Path 1:

- Both can castle both ways

- One can castle one way and the other both

- Both can castle one way

- One can castle one way and the other not at all

- Neither side can castle.

5 times 4 plus the en passant scenario at the beginning makes 21 non-3-fold repeats.  22nd is 3-fold.

The other possible path is, if the 2-time repeat goes something like W - B - B - W:

- Both sides can castle both ways

- One side can castle only one way but the other can still castle both

- One side can't castle at all and the other can still castle both

- One side cant' castle at all and the other can still castle one way

- Neither side can castle at all.

Here's the challenge.  So far we have several theories about the number of times that a position can be duplicated.  However, they are just theories.  Can anyone demonstrate on a chessboard that it's possible to have 21 (or whatever) duplicate positions before three-fold repetition occurs?   Without experimental evidence supporting your theory, it's just an interesting idea.

ThrillerFan

You list five options and multiply by four.  I'm assuming that you're considering positions where White moving first can be transposed to Black moving first.

The problem is that the castling options can only be multiplied by 2.  Once the castling privilege has been lost, it cannot be restored by transposing who moves first.

Uhm, to answer this and your post 74, it has already been answered!  SEE POST 24!  I prove it there already!  I demonstrate a case of 21 occurrences with no 3-fold repetition, and then state that the next time the position occurs, no matter who it is to move, it will at that point be 3-fold, which will be the 22nd occurrence.

Smyslov fan, the question he was asking, talking now in Lehman Terms for you, was how many times the same arrangement of the pieces can occur on the board, MAXIMUM, before 3-fold occurs.  For example, if you start with a White Rook on a1, White King on e1, White Rook on h1, White Knight on a4, White pawn on d5, Black Rook on a8, Black King on e8, Black Rook on h8, Black Knight on h5, and Black pawn on e5, where Black's last move was e7-e5, then you can PHYSICALLY HAVE A WHITE ROOK ON a1, WHITE KING ON e1, WHITE ROOK ON h1, WHITE KNIGHT ON a4, WHITE PAWN ON d5, BLACK ROOK ON a8, BLACK KING ON e8, BLACK ROOK ON H8, BLACK KNIGHT ON h5, AND BLACK PAWN ON e5 FOR A GRAND TOTAL OF 21 TIMES WITH NO 3-FOLD REPETITION.  THE 22ND TIME THAT THESE 10 PIECES LAY EXACTLY ON THESE 10 SQUARES, 3-FOLD REPETITION WILL OCCUR NO MATTER WHAT!

SHEESH!  Stop trying to take your stupid technicalities to purposely mis-understand what the OP is asking for!

ThrillerFan

You might want to check your example.  I put it into my chess program (Stockfish 8) and got three-fold repetition on move 6 (Note:  it was one of the maneuvering positions rather than the subject position).

I have seen this position over and over again, in almost every single gamei have played:

Here's a simplistic, unrealistic example of what I was talking about in Post 75.  The position has been duplicated 10 times, but only eight can be attributed to castling.  Now that it is Black's turn to move, neither side has the castling privilege, so you cannot gain eight more duplications by using the castling maneuver.

Here's an example, and it looks like my initial calculation over-counted the number of distinct castling situations.  I get only 11 different "optically identical" positions, without true repetitions in the sense of the 3-fold repetition rule.  The 2nd diagram below shows the position that will occur optically 11 times, before it starts repeating in the actual sense of the 3-fold repetition rule.  The sentence fragment "B lost O-O" means that black lost the ability castle short.

This example indicates that the maximum number of instances of an "optical position" that can appear before 3-fold repetition occurs is 12 (including 1 actual repetition at the end).