I was at my lunch table today and was trying to come up with a card game, and here is what I came up with:
Let there be a standard deck of $52$ cards called $\mathbb{D}$, with four suits: spades, clubs, diamonds, hearts. Each card represents a number from $1$ (ace) to $13$ (king). There are three players; each player is dealt $5$ cards, and one card is placed on the table. Players go in an arbitrary but constant order, playing one card at a time.
The stack, we'll call it $S$, has a value starting with the first card placed on the table. Each suite has one binary operation ($\cdot$) defined by it, such that when a player plays the card $b\in \mathrm{suite}$, the current value of $S$ changes to the value $a \cdot b$, where $a$ was the previous value of $S$.
The operations are $\beta_1$ (spades), $\beta_2$ (clubs), $\rho_1$ (diamonds), $\rho_2$ (hearts); such that $$ \begin{align} \beta_1 &: \mathbb{N}^2/\{13\} \rightarrow \mathbb{N}/\{13\} &| (a,b) \rightarrow a + b \mod(13)\\ \beta_2 &: \mathbb{N}^2/\{13\} \rightarrow \mathbb{N}/\{13\} &| (a,b) \rightarrow a \times b \mod(13)\\ \rho_1 &: \mathbb{N}^2/\{13\} \rightarrow \mathbb{N}/\{13\} &| (a,b) \rightarrow | a - b | \mod(13)\\ \rho_2 &: \mathbb{N}^2/\{13\} \rightarrow \mathbb{N}/\{13\} &| (a,b) \rightarrow \lfloor \frac{a}{b} \rfloor \mod(13)\\ \end{align}$$
The game is played such that each player is trying to make $S$ equal the identity element of the suit $b$, and if this is achieved then the player does not draw a card from the deck, however, if the player fails to make $S$ equal the identity of $b$'s suit, then the player draws 2 cards from the deck. This process stops when a player reaches $14$ cards, as which point they forgo drawing anymore cards, and have to do a calculus problem designed to make them lose 2 turns, meanwhile, the other players continue gameplay. Should the deck be exhausted, the cards of stack $S$ will be shuffled and returned to the deck, however, $S$ retains its value. The first person to get rid of all of their cards wins.
My question is simply: is this game winnable in a finite amount of turns?
I have no idea how to even approach this problem, but (since I designed the game) I do know some of its algebraic properties:
Each color (whose operations are either $\beta$ or $\rho$) is meant to form a ring, with the first one being an abelian group, and the second being a monoid. I have not proved this to be the case, and it very well may not be, but I just thought of the operations off of the top of my head.
Edit: As noted in the answers, $\rho$ does not form a ring, and from what I can tell it does not preserve any algebraic structure. However I can not think of any two operations on $\mathbb{N}/ \{13\mathbb{N}\}$ that form a ring, and so I am not really sure if there is any operations that would preserve any algebraic structure. Either way, this discussion of algebraic structure doesn't bring me closer to a solution.
As for winning in a finite number of steps, the shuffling seems like an obvious weak spot. If a sequence of cards results in a reshuffle that just duplicates the same in winnable situation at the end then conceivably the bad shuffle could reappear arbitrarily many times. I think it would be more interesting to put a finite cap on this and ask how often players can win given a certain sequence of cards.
As for the algebraic properties, the $\beta$'s certainly make a ring, but the $\rho$'s don't. Neither of those two are associative operations. For example, $\rho_1(1,\rho_1(2,3))=0$ but $\rho_1(\rho_1(1,2),3)=2$. $\rho_2$ is not even defined on all of $\Bbb Z_{13}\times \Bbb Z_{13}$.
I would also be pretty amazed if the one distributed over the other. I imagine that it's easy to find a counterexample. If you want t study the algebraic operations as a single structure, then you'll have to relax your expectations to something more general than rings.