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Cards work as they would in a physical version, with fixed fronts (type) and backs  (position) this remains the same for every game. The deck is shuffled each time.

There are:

13 Trees, 10 Houses, 9  Beeches, 9  Waves, 5  Mountains, 5  Churches, 3  Boats

There are an even number of each of the position backs.

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Does this imply, that not every field has the identical probability to offer e.g. a boat? There are only 3 positions (back) containing a boat?

If yes, knowing all the cards and thereby possible locations of certain types will drastically raise your outcome.

Or isn't a certain front sticked to a certain back, meaning in fact you have to separate decks: one for the types and one for the position and their combination is random.

For each of the 81 cells of the grid the possibility of it being used for a specific card is not identical but it is reasonably close. 

Each card has a position on the back side of it. It is not possible for the back and front of a specific card to appear at the same time in the game, just like a real card. 

If you only look at the back of the 54 cards each cell in the grid has 9 total chances to be used (three for each the row, column and section. 3 x 3 = 9). So any single specific card takes nine of those cells and reduces the opportunities of them being used to 8 (as those options are on the back of that card). So the opportunities of a cell being used for a specific card is 8/54 for 9 of the cells and 9/54 for the other 72 cells.

This is all further complicated by multiple cards having the same card type on their front, which tweaks the odds again.

The reason there is only a single deck of cards with fronts and backs, is that it would reduce the manufacturing cost in a physical game, which is what I was modelling. Two decks would allow the cells possibility of being used for a specific card equal.

At least I think that's right.