First, I'm going to call the 3 piles to be deciphered Deck 3, Deck 2, Deck 1, to correspond with the number of deciphered cards required for the "phrase."

I like it. I'll change it to read this way, too.

As I understand it, we slide the top card (I'll call it the "slider") down letter-by-letter until the hidden card's title is completely revealed. That revealed ("deciphered) title card is then placed beside that deck until the required number of cards for a "phrase" (3, 2, or 1)

The "slider" card is placed aside. This makes the next card that you just 'revealed' to be the new slider, and so on. This makes the motion of slide/revealing always a discovery/mystery, and very fluid if you're working on a long phrase.

When the phrase is fully deciphered, the player who completed it takes the slider card from that deck into his discard pile (to be shuffled and drawn as needed for our 3-card hand to begin each turn.)

When the phrase is fully deciphered, the play takes all previous slider cards that have been set aside. In the case of the three-card phrase it would be all three slider cards. In the case of "Deck 1" (pawn, court, and The Excuse Deck) you would just collect the slider card each time you successfully decipher a one-card phrase.

If a phrase is partly deciphered, say only 2 of the 3 required cards are beside Deck 3, it is possible that the opponent may be able to claim the cards by finishing the reveal of the 3rd. Is this correct? So player A might have revealed 3 of the four letters of the card's title, for example, but he has no cards left in his hand and his turn is over. Now Player B, on his turn, could reveal the final letter, and thus claim the phrase card(s) next to the deck. Is this correct?

That is all correct.

I don't quite understand that "overshoot the phrase" business. If we play a 3, for example, but there are only 2 unrevealed letters on a card we're working on, what card(s) does the opponent get? Just the card that was overshot, or that card plus any other phrase cards which are next to that deck?

If you 'overshoot the phrase' by any amount, all the cards that were set aside for that phrase go to your opponent's discard. For the case of Deck 1, it would just be the current slider card. For the case of Deck 3 it would be all three slider cards.

Over on the BGG forum you suggested examples of play. I'll add them here.

Thanks again.

First, I'm going to call the 3 piles to be deciphered Deck 3, Deck 2, Deck 1, to correspond with the number of deciphered cards required for the "phrase."

As I understand it, we slide the top card (I'll call it the "slider") down letter-by-letter until the hidden card's title is completely revealed. That revealed ("deciphered) title card is then placed beside that deck until the required number of cards for a "phrase" (3, 2, or 1), When the phrase is fully deciphered, the player who completed it takes the slider card from that deck into his discard pile (to be shuffled and drawn as needed for our 3-card hand to begin each turn.)

If a phrase is partly deciphered, say only 2 of the 3 required cards are beside Deck 3, it is possible that the opponent may be able to claim the cards by finishing the reveal of the 3rd. Is this correct?

So player A might have revealed 3 of the four letters of the card's title, for example, but he has no cards left in his hand and his turn is over. Now Player B, on his turn, could reveal the final letter, and thus claim the phrase card(s) next to the deck. Is this correct?

I don't quite understand that "overshoot the phrase" business. If we play a 3, for example, but there are only 2 unrevealed letters on a card we're working on, what card(s) does the opponent get? Just the card that was overshot, or that card plus any other phrase cards which are next to that deck?

Sorry if I'm a little thick. I blame it on my age.

So it seems like card ranks only matter for the tiebreaker, which (as you note) doesn't really extend the strategic space of the game. Are ties common enough that there even needs to be a tiebreaker? If there does, you could just say that if there is a tie on points then the winner is the player who won the top-most Ace. If there was a tie on that Ace, then the player who won the second-highest Ace. And so on.

You might take a look at Colour Bazaar, which is a similar game.

1. What happens if a player ends with suit chips for suits that only have an Ace, no other cards?

2. We often ended play with players having uneven amounts of suit chips. If you have players A, B, C, and D, and player B ends the round by placing the last card and taking a chip, players A and B have one more chip each than players C and D. Is this by design?

When you sell a card to gain more tokens, instead of discarding it, you play it as a developed property on your opponent's side. You can choose which district to play it on, but it must match suit as in the regular game. You'll want to put most of the cards in one district and write that one off as a loss, but sometimes the suits won't let you.

If you want to sell a card that can't be placed with either suit you may discard it.

You only run through the deck once since there are no (or very few) discards. Scoring is the same as usual.

To make the game harder:
If one of the suits on your discard is one of your automa opponent's crown suits, you must place it using that suit if possible. If not, you can place it using the other suit. This mimics the two-player game where your opponent is more likely to be able play cards with suits from their crown cards. It makes the game more difficult because it limits where you can play the opponent's cards.

These are the core elements of the game. You may be able to work out the rules with these elements in mind. It is hard to start on rules if you have no idea what the game is about.

Taxation is used to deter players from hoarding goods (otherwise playing deeds were inefficient).

It's OK. I've fixed it.

Given that I changed it from LaTeX to wiki mark-up with a couple of quick and dirty regular expressions, I'm surprised there weren't more stray brackets.