Math genius needed for combinatorics calculation of my card game (Shannon number-like) My name is Lych, I am a game developer from Germany.
My card game is a quite complex one and I am very curious about the question, what amount of different matches are possible in this game. Just like the Shannon number in chess. However, this will most likely be bigger than the Shannon number.
In the following, I will describe everything you need to know. Some of the instructions are already simplified to not make it harder than necessary.
After all, it is not important to get an exact result, just a rough calculation.
Please keep in mind that I am not especially gifted when it comes to math. Therefore, I do hope that my instructions are clear. If they are not, please ask me.
Thank you really much in advance! I am grateful to have the opportunity to ask people for help who can do this better than me.

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*4 players in total. After each turn, the next player clock-wise is next with their turn. Everyone has 5 cards in their hands in the beginning.
75 cards are available in total, all different, no equal ones.
(First interposed calculation that could help later: How many possible combinations of starter hand cards are possible?)


*Per turn, the player can set 2 cards from his hand to the field (choose freely from the 5 hand cards).


*Every single card on the field can do 21 different actions. In ANY order.
(That means that the order of options "1, 2, 3, 4, 5 ... 21" is a different one than "1, 3, 2, 4, 5, ... 21" and therefore another match obviously).
Also, the 2 cards can take turns in their actions as the order is completely free again.
(Therefore, the amount of possible orders with those two cards has to be an incredible high number already).


*However, the players will not always add 2 more cards per turn. In average (for this calculation), they will keep having 2 cards on the field to use, constantly. Every turn, they are destroyed and 2 new cards will be placed from the hand again, freely.


*In the beginning of each new turn, the player draws 3 more cards (don't mind because only
2 are used in the previous point. One card of them is always directly destroyed, but do not take
this into the calculation. Just act like the player only gets 2 cards per turn, but 3 are removed from the total card stack).
(Still keep in mind that every card of those 75 is unique and all possibilities of card drawings from the stack that are still available, and also all card settings from the hand to the field must be taken into the calculation).


*In total, there are 20 turns, but only 18 of them the players draw cards because then, the card stack
is empty (55/3 = 18 roughly. 55 because in the beginning, 20 cards are already drawn).
In the last 2 turns, they don't draw but still play as before with 2 cards on the field.
-> How many different matches are possible in this game?
Closing words: As I said, thank you very much for your effort to everyone who helps me with this. If anything should be not clear, please ask. After all, it is enough if the result is an estimation and not exact.
~ Lych
 A: Here is a rough approximation, and you should consider it as a lower bound more than anything else. I will not look at hand size of the players. My assumptions will be very basic. Players go in order and play 2 cards each. Each card has 21 different actions, and any permutation of the actions will make a different game. When all the cards are played, the game ends, but not before that. I will suppose the order in which a player plays his 2 cards does not matter. If there are any assumptions you think very important to change let me know. I will approximate using lower bounds, so that this indeed gives a lower bound.
The number of different ways we can choose 37 packs of 2 cards and one pack of one card, and then have any order in between the 37 packs of two cards is
$$P=\frac{75!\cdot37!}{(2!)^{37}}\approx 2.48 \cdot 10^{141}.$$
The number of different ways we can have any permutation of the actions of a card is $21!$. There are 75 different cards that will be played, and so we have about
$$C_1=(21!)^{75} \approx 1.33 \cdot 10^{1478}.$$
Multiplying these two numbers together gives us our first approximation for the number of games :
$$A_1=P \cdot C_1 \approx 3.29 \cdot 10^{1619}.$$
If instead we consider the case where the actions on the cards during a players turn can be done in any order, then we have that
$$C_2=(42!)^{37} \cdot 21!\approx 1.48 \cdot 10^{1912}$$, which in turn leads the the second approximation :
$$A_2=P\cdot C_2 \approx 3.67 \cdot 10^{2053}.$$
I hope this answers your question, and let me know if any assumptions should be changed!
