So I've been wondering recently, How many unique games of Carcassonne are there? Turns out quite a lot, here's my quick calculations:
Basic Deck has 72 tiles, but the start tile is always the same. This leaves 71 tiles to be ordered in quite a few different ways.
First you have to choose one of the 71 tiles, then 70, then 69.....
So to start of with there's 71x70x69 different orderings of tiles. Which is 71! which is huge.
But we should remove the instances of tiles which are duplicates, like the 9 and 8 roads.
Let's just remind ourselves of all the tiles:
I'm going to assume the tiles with Pennants are different.
I won't bore you with all the detail but take a look at "Theorem 3 - Permutations of Different Kinds of Objects" from this page:
http://www.intmath.com/counting-probability/3-permutations.php If you're interested, it's pretty obvious.
So the total number of games is:
....
Drum Roll
....
Total Number of Tile Sequences for Carcassonne = 71! / (8! * 9! * 4! * 1! * 2! * 4! * 3! * 3! * 3! * 3! * 5! * 3! * 2! * (1! * 2!) *(2! * 3!) *(2! * 3!) * (2! * 1!) *(3! * 1!) * 1!)
Let's simply this. 1! = 1 and 2! = 2 and 3! = 6
Total Number of Tile Sequences for Carcassonne = 71! / (8! * 9! * 24 * 2 * 24 * 6 * 6 * 6 *6 *5! *6 *2 *2 *12 * 12 *2 *6)
...
Total Number of Tile Sequences for Carcassonne = 71! / (8! * 9! * 5! * 24^2 * 2^4 * 6^6 *12^2)
... Google Tells me:
Total Number of Tile Sequences for Carcassonne = 7.8232304x10^78Let's give this some context - It's approximately this big:
7800000000000000000000000000000000000000000000000000000000000000000000000000000I'll try and give this some order of magnitude:
Population of Earth: 7x10^9
Age of the Universe in seconds: 4.32 × 10^17
Number of observable Stars in the Universe~: 7X10^22
Number of Atoms in a Human Body: 7x10^27
(
http://www.physicsoftheuniverse.com/numbers.html)
So, if you multiple those numbers together, it's a bit like counting all the games of Carcassonne there could be. If for every second the Universe has existed, you could count every atom, of every person on a planet the same size as Earth, around every Star you'd have counted:
1.47X10^78
So you'd need to do it ~five times.
And that's just the number of different ways the tiles come out of the bag!I've not considered tile or meeple placement.