Threefold draw has to be claimed by one of the players. If neither player claims, the game can go on forever, or until a TD declares the game drawn
How many times can the same position repeat?

Based on your method of computing the count, then you should also add in the ways en passant are ignored (e.g. white moves pawn 2 squares, black moves bishop, white move bishop, black moves bishop back, white moves bishop back; although the board looks the same, it isn't a repeat.).
In programming, you can just look for repeat of the FEN; in actual play, the moves usually happens consecutively.

Is there honestly nobody who knows the rules and is willing to answer the actual question being asked?
I know that in tournaments claiming draw by 3-fold repetition is optional. That's not the point. If you really want to be that nitpicky, then change the question to "how many times can the position repeat at most, before a player is allowed by the rules to draw the game by 3-fold repetition?" Please stop missing the point of the question.
And please read what I wrote. I already took en passant into account. It's right there in the second-to-last paragraph.
And yes, the correct answer is larger than 3. Please read the rules regarding 3-fold repetition.

If nobody here is either capable or willing to give a correct answer, is there some other forum where I could ask this?

I think almost everybody here is capable of answering your problem. You just have to give more incentive...

You're right according to Wikipedia, en passant and castling rights ( or not ) are classed as different positions, even if all the pieces are otherwise on the same squares.

An example game?
I'm not being confrontational here, I'm asking in complete honesty. I'm just not sure what kind of example game you want.
I suppose I could construct an artificial position and combination of moves that goes through the same piece arrangement as many times as I can possibly concoct, and check with a chess software at which point it declares it draw by 3-fold repetition.

@Djvortex, an artificial position probably wouldn't work as the software doesn't declare a draw, it has to be claimed, it could be tested though in an unrated game perhaps.

Arena declares the game a draw in 3-fold repetition cases, but its behavior in this regard seems inconsistent and, perhaps, buggy. It seems to take into account castling rights and en passant rights when deciding, but the position repeating with the other player to play seems to confuse it.

If nobody here is either capable or willing to give a correct answer, is there some other forum where I could ask this?
Make your own!

Threefold draw has to be claimed by one of the players. If neither player claims, the game can go on forever, or until a TD declares the game drawn
The TD/arbiter must claim at 5th repetition.

Djvortex, I tried it with an unrated game, losing castling rights and putting the pieces back in the same places is deemed a different position by the software so you were right about there being 11 but the en passant can only happen once so with repeat positions with the other player to move I make it 21 not 22.
Threefold draw has to be claimed by one of the players. If neither player claims, the game can go on forever, or until a TD declares the game drawn
The TD/arbiter must claim at 5th repetition.
I've never heard this rule, but even so the TD would have to see the position repeat 5 times in order to make the call. In most rated events between club players the TDs don't get involved unless someone asks or there are very few games left in the round. It's just not practical to have a TD watching every game.

I don't know about the USCF rules, but FIDE has had this rule for at least of couple of years:
9.6 |
If one or both of the following occur(s) then the game is drawn: |
9.6.1 |
the same position has appeared, as in 9.2.2 at least five times. |
The rule seems strange because I would think that one player would claim a draw on the third repetition rather than repeat the position two more times. However, there must be a reason they incorporated the rule.
Note: The arbiter doesn't have to see the repetition occur. The players either agree that the position has been repeated or one of the players has to show the arbiter that the position has been repeated.

Djvortex, I tried it with an unrated game, losing castling rights and putting the pieces back in the same places is deemed a different position by the software so you were right about there being 11 but the en passant can only happen once so with repeat positions with the other player to move I make it 21 not 22.
I suppose it's a bit unclear, and should be stated more clearly what is meant by "the same position repeats". I suppose I was counting, more precisely, how many times the same arrangement of pieces can appear on the board before the three-fold rule kicking in. This would count also the very first time that the particular arrangement appears, not just how many times that arrangement repeats (ie. appears again later).
After all, if I understand correctly, this is what the three-fold repetition rule is looking for: It's not looking for the same position to appear three times later in the game, but how many times that particular position has appeared overall (which, I suppose, if we get very technical, means that the position has repeated twice, after its first appearance).

DeirdreSkye,
I already did, I played an unrated game on here with my nephew mooly9 to check it out, and got the same "position" 5 times before I could claim a draw and I certainly could have got more, you can get the pieces on the exact same squares more than 3 times without the option to claim a draw but it's considered a different position after en passant and castling rights have gone.
Assuming the three-fold repetition rule is in effect, how many times can the same arrangement of pieces repeat on the board (as the absolute theoretical maximum)?
"Well, duh, three times. Seems pretty obvious." Except it's not that obvious, as there's more to the rule than that. Please corroborate if my math below is correct.
For starters, having castling rights to one side vs. not having it is considered a "different" position in the context of this rule, even if all the pieces are otherwise on the same squares.
Thus we can start with both players having castling rights to both sides, and have the same position first repeat 2 times, then have one player lose castling rights to one side and repeat the same position 2 times, and so on with the remaining three castling rights. Then there's one last time that the position has to repeat before it's draw by this rule.
In other words: 2 initial times + 2 times per castling right (ie. times 4) + 1 last time = 11 times.
However, there is more: The rule states that it has to be the same player to play, in order for it to be considered the same position. Therefore we can actually repeat each of those positions twice (with a different player to play each second time), except for the last one (which can't be repeated anymore because it causes the game to be immediately a draw).
Thus 2*(2+2*4)+1 = 21.
Except there's still one additional situation which causes the position to be "different": En passant rights. If two positions only differ in a player having or not having en passant rights, they will be considered "different" positions in the context of this rule. However, this situation can only be achieved once in the sequence above (because it necessitates a pawn move, which cannot be reversed).
Thus the final answer is, if I have calculated correctly, 22 times.