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I have been developing a card game called "Infinity", which involves a unique play mechanic based on card interactions. In this game, each card displays a set of symbols, and players match cards based on these symbols under specific conditions.

Here is a picture of a game against the computer: Infinity_unique_card_game_with_potentially_infinitely_many_cards

Gameplay Rules:

  • The game is played with a deck of c unique cards distributed among p players.
  • Each player begins with n cards and plays clockwise.
  • A "hit" occurs during a player's turn when the card they play matches exactly one differing symbol compared to the last card on the stack called "pot". If a player has a hit, he collects all the cards in the pot, but his played card stays in the pot. (Additionally to make it more dynamic, the player who has a hit can reverse the direction of the game if he wishes.).
  • If all players have no cards on their hand, the last player who did a hit collects the remaining cards from the pot.
  • The player with the most collected cards, wins the game.

Number theoretic interpretation: Each card has a number $a$ and the symbols correspond bijectively to the prime factorization of $a$: Symbols of $a$

$$S_a = \{ p^k \mid 1 \le k \le v_p(a), p\mid a \}$$

Then a hit between $a,b$ occurs iff:

$$\max\{a/b,b/a\} \text{ is a prime number}$$

This leads to an interesting question regarding the probability of a "hit" occurring:

Question: Given p players and a deck of c cards numbered from $1$ to $c$, what is the minimum number of cards n each player must randomly draw to guarantee that at least one "hit" will occur during the game? ( I will assume that as soon as a player has the possibility to do a hit with the pot, he or she will do so. For otherwise if the players play on intent to avoid hits, then there is the sequence by @PeterTaylor which avoids hits: $2,3,4,5,6,\ldots$ since $m,m+1$ have $\gcd(m,m+1)=1$ and so $\max\{m/(m+1),(m+1)/m\}$ is not a prime number.)

This problem seems to intersect elements of combinatorics and probability theory and number theory. Any insights or related mathematical principles that could help in determining or estimating this minimum n would be greatly appreciated.

This game is based on some hobby research I did two years ago:

Why is this bipartite graph a partial cube, if it is?

Example of sequence of graphs which satisfy the Riemann hypothesis or the prime number theorem?

Largest eigenvalue of a Laplacian matrix to lower bound the prime counting function?

https://math.stackexchange.com/questions/4559171/is-this-a-t-1-topological-space-and-what-other-properties-or-name-does-this

Edit: Alternatively I would also be interested in the following question: Suppose there are two players each with a set of cards $1,2,3,\ldots,c-1,c$ and in the pot is the starting card $1$. For instance we might have $c=33$. Is the game fair or is there a strategy with which the starting / the other player can always win? ( I am not sure how the wording is in game theory, so excuse my language it does not fit.) The second edited question has also been asked at MSE in case it is not research related: https://math.stackexchange.com/questions/4904197/two-questions-around-some-new-card-game-based-on-prime-factorization .

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    $\begingroup$ If I've understood correctly, you cannot guarantee a hit unless all cards are drawn. $1$ forms a hit with every other card, but otherwise the sequence $2, 3, 4, 5, \ldots$ has no hits. $\endgroup$ Commented Apr 23 at 7:25
  • $\begingroup$ @PeterTaylor: Thank you for your comment. $1$ forms a hit only with the prime numbers. I am not sure what you mean by the sequence $2,3,4,5,\cdots$ has no hits? $\endgroup$ Commented Apr 23 at 7:33
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    $\begingroup$ @PeterTaylor $2$ and $4$ form a hit because $\max\{2/4,4/2\}=2$ is a prime number. $\endgroup$ Commented Apr 23 at 7:51
  • $\begingroup$ True, it's possible to get no hits even if all cards are played. If played in sequence, the last card on the stack when $4$ is played will be $3$ and that's not a hit. $\endgroup$ Commented Apr 23 at 7:58
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    $\begingroup$ In that case it's certainly not as simple. There is a third scenario which is even trickier, and that's to introduce game theory: a player might choose not to make a legal hit in the hope of making a bigger hit later with the same card. $\endgroup$ Commented Apr 23 at 8:34

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Here is a rephrasing that may help: the graph $G$ has vertices $[c]$ and an edge between two vertices if one is a prime multiple of the other. Then the game is for each player to take some vertices, name one of theirs at a time, and a hit occurs if two consecutive vertices are adjacent. If the hands are open as in the example shown, both players can see which hits are possible. The two-player strategy will depend on the card distribution as well as order. For example, if only one hit is possible, the second player will be able to win by waiting. If the distributions are $L=\{2,3,5\}$, $R=\{4,6,9\}$, then $L$ will be able to win playing first or second.

If $p\geq 5$, you can get a lower bound on $n$ by setting up a distribution so that no hit is possible. Give card $k$ to player $P_i$ if the number of prime divisors of $k$, with multiplicity, $\Omega(k)$ is $i\pmod 5$, and arrange the players in the cyclic order $P_1, P_4, P_2, P_5, P_3$. This partitions the graph so that there are no edges between consecutive players. If $c$ is large, $[\Omega(k)]$ is normally distributed, which I think means the $\bmod 5$ distribution is expected to be uniform (someone correct me here), but that's probably irrelevant if you are planning to play with less than $10000$ cards.

Because of the random distribution of starting cards, it will be difficult to say if the two player game with $c=33$ is fair. Playing on the graph (with open hands) makes it more obvious which cards can make many hits, and which can not make a hit after the first turn (and so are useful for passing a turn). You may want to modify your game to remove these.

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  • $\begingroup$ Thanks for your insight in this question. $\endgroup$ Commented May 1 at 18:47

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