# Shithead Loops

Is the game “Shithead” guaranteed to terminate?

Shithead is a card game, played with a poker deck without jokers (52 cards), with an arbitrary number of players. I may write down the rules another time. Will every match of this game end, or is it possible to play indefinitely?

For convenience let *n* be the number of players. For an
“actual” game, there need to be at least two players, but one
could also imagine playing alone, trying to
get rid of all their cards. Because the number of cards and the
number of places where cards can be are finite, the game has
(for fixed * n *) only a finite number of possible states.
Therefore, if a match goes on indefinitely, it must repeat itself.

I’ll say, a loop is *forced*, if independent of the players’
strategies, the match will not end.

In a loop, each player that remains in the game and plays cards, must at some point take up cards. Otherwise they would win eventually. In a loop, no card can leave the game (through bombs), otherwise it’d be impossible to reach the same state again.

Corollary: In a forced loop, there can be no 10s in a player’s hand and in a not-forced loop, players must never play 10s or cause quadruples.

A single player (*n* = 1) is in a forced loop, if they have
a 7 and a 9 (or other card >9) in their hand, and the stack is
empty. No matter which card they play, the next card will not be
playable. So they have to take up the stack, getting back into
their initial position.

In a loop, the stored cards on the table can never be played. They can only be “used” to prevent a player from winning, but they must never be played. But they can’t simply be ignored.

At each point in the loop, the active player must (before making their turn) have at least two cards or be unable to play a card. Otherwise they would win and end the loop.

If all players have the same kind of card, the game always terminates. Therefore in a loop there must always be at least two kinds of cards.

There is a non-forced loop, found by experimentation: Consider two players both with a 5 and an 8, and an empty stack. The first player plays 8, skipping the second player. Unable to play again, the first player takes up the stack, giving them the same hand as before. Now the second player has to move. The situation is now symmetric and can go on forever.

This extends to 4 players:

Player | Hand |
---|---|

1 | 5, 8 |

2 | 5 |

3 | 5 |

4 | 5, 8 |

Every player that plays a card, plays their eight. This creates a non-forced loop.

With 5 players:

Player | Hand |
---|---|

1 | 5, 8 |

2 | 5, 8 |

3 | 5 |

4 | 5, 8 |

5 | 4, 8 |

Of course, the cards 4, 5, 6 and 7 can be used interchangeably here.

It seems impossible to use this strategy to create a loop for 3 players, because if the 3rd player takes the stack, the first is moving again, so they will need infinitely many 8s.

## Open Questions

- Does such an 8-based loop exist for all
*n*>3? (Until there are too many players and too few 4, 5, 6, 7, 8 cards in the deck) - Are there “other” loops, where players play other kinds of cards as well?
- Are there forced loops?