How many times can the same position repeat in chess?

Most official chess tournaments have a so-called "three-fold repetition" rule, which in short means that if the same board position is repeated three times, the game can be declared as draw. (In human tournaments declaring it as draw is optional and up to the players, even though many tournaments have an additional rule that says that if the same position repeats five times, the tournament referee can declare it a draw. In computer chess tournaments three-fold repetition is invariably declared a draw automatically.)

So the answer appears to be pretty obvious: Three times, duh.

Except it's actually not that simple. You see, there are additional rules with respect to what is considered the "same board position". Even though all the pieces might be on the exact same squares as in a previous position, for the purposes of this rule it might not be considered the "same position".

For starters, castling rights are taken into consideration when defining whether a position is the same as a previous one. Even if all the pieces are on the same squares as before, if however now one of the players has lost the castling right to either side, which he had in that previous position, it will be considered a different position.

So, how many times can the same arrangement of pieces repeat in this light? (We don't need to worry about how realistic in actual games this is. We are just pondering what the theoretical absolute maximum is.)

So, this is theoretically possible: Both players have castling rights to both sides. The same position repeats twice. Then one of the players loses one of these castling rights, after which the same position repeats an additional two times. And so on with the remaining three castling rights. Then, after the last one, the position can repeat a third time, and it will be draw.

So 2 initial times + 2 for each lost castling right (ie. times 4) + 1 final time = 11 times?

Nope! There are more conditions we haven't touched! You see, the rule also states that it has to be the same player to play, in order for it to be considered the "same" position, in the context of this rule.

Thus, we can actually repeat each of the above positions twice, except for the very last one, as long as it's the other player's turn each second time. (The last one can't be repeated with the other player having the turn, because the last one already makes it a draw immediately.)

Thus 21 times?

Still no! There's still one additional rule: Having en-passant right vs. not having it is considered yet another aspect that makes the position different.

How many times can we repeat the same position with en-passant? Only once. This is because this condition requires a pawn move, and can't be repeated with all pieces ending up on the same squares. Thus we have to start the whole sequence with having a player have en-passant right, after which it's lost, and we can proceed with the rest of the sequence above.

So the final answer is 22 times.