How Many Unique Games of Commander Can Be Played?

[Khursed]

TL;DR Summary

Card game.

I play magic the gathering in a format called commander.

Each player start with a deck of card that has 99 cards in it, for simplicities sake, I'll give very simple rules.

Each card is different, we have 4 players, each shuffling their deck of cards and drawing 7.

Roll for who goes first, then draw 1 each turn and pass clockwise.

My real question was how many different games of magic the gathering of commander can 4 different people with 4 different deck play, but I'm going slightly insane trying to figure it out.

My neophyte guess is that it's going to be an insanely huge number, since my bare starting point assuming 99 different card is 9x^155 or so different deck arrangement. Draw 7, with 3 other people doing the same. Not taking into account there are over 27,000 different cards of magic. Let's be realist and say the playable pool is 5000. and for this exercise only 4 decks of 99 plus the one card that is always revealed.

The way I see it, it should be possible to calculate the initial board state on the first turn, but I can not imagine any way to calculate the odds for any future turns because of card interactions as the turn order go on. Well maybe a few?

I was wondering about it, because while playing magic with friends, we were discussing very large numbers, and I was asked what I'd do with Graham's number of years to live and I said "play commander". So now I wonder how entertained I'll be. Either by how repetitive or varied it's going to be....

 

[jedishrfu]

This is not an easy question to answer without more detailed knowledge of the cards and the game rules.

For a simpler problem you could look into how mathematicians approach combinatorics with a standard 52 count card deck. There is a lot of work done on calculating the number of winning hands and probabilities of getting those hands as an example.

Here some stats that I found for MTG:

https://www.dicebreaker.com/games/magic-the-gathering-game/how-to/mtg-colour-combinations-explained#:~:text=A deck with a single,and five four-colour combinations.

Here’s a discussion on using math to make decisions in MTG:

https://articles.starcitygames.com/...very current,binomial coefficient” in no time!

 

[Khursed]

Thank you for your answer. I must admit I am at a loss on how to figure it all out. The game is very complex and there are so many different cards and then the rules are a nightmare because basically the game has like 8-900 rules that the cards constantly break or tweak in a way or five.

Games could go on forever. So we'd have to rule them out. Make rules for exceptions etc...

 

[jedishrfu]

The basic notions of combinatorics comes from determining the number of words a given set of letters can produce:

As an example, if there are 5 letters in the word and there are 5 letters that can be used:

$5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 5^5$

but if there is a comstraint like no letter can appear twice then:

$5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5!$

If instead they say they say you can use 26 letters then

$26 \cdot 26 \cdot 26 \cdot 26 \cdot 26 = 26^5$ possibilities for 5 letter words using the same letter twice

and if they add the constraint that a letter can't be used again:

$26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 = 26! / (26-5)! = 26! / 21!$

Anyway hopefully you get the idea of how you determine the counts.

One other thing to consider is the whether order is important which requires you to divide by n! where n is the number of letters in a word.

 

[Hornbein]

Then there's taking x cards from a set of n differing cards. Of the order doesn't matter then there are n!/(x!(n-x)! ways to do this. It's called "n chose x" or "binomial coefficients".

N = 5000
So 4*[(N chose 7) * (N-7)!/(N-99)!]^4 for initial setup. Maybe * 5000 for "the card that is always revealed."