Beggar my neighbor is a card game for two or more players. It is a deterministic zero-sum game, which makes it sound rather boring, but it's actually quite fun to play. For two to four players, one deck of cards is normally used, for more players add decks as necessary. It is not necessary to have identical decks, if there are different patterns on the backs of the cards it makes no difference to the game.

Players sit in a circle. One player is dealer, and this role passes to the left after every game. The dealer picks up the deck(s), takes out any jokers, shuffles the cards, then deals the entire pack face-down to all players in the standard fashion. The person to the left of the dealer makes the first play of the game, and play goes to the left.

A play normally consists of simply putting the topmost card from your pack face-up in the centre of the playing area (on the 'pile' - and once the first card is down, every card must be put on top of the previous one to preserve the order in which they were played). However, if one person puts down an Ace, King, Queen, or Jack the play changes, these are called 'debt' cards. If player A puts down one of the debt cards, then he or she 'has dibs on the pile'. If the card is a Jack, the next player (player B) 'owes' player A one card. If the card is a Queen, then player B owes player A two cards. If it's a king, the debt three cards, and if it's an ace, it's four cards. Player B begins to 'pay' the debt, one card at a time face up on the pile. If player B plays a debt card at any time while paying a debt (even if it's the last card of the debt), then the debt to player A is cancelled and player B now has 'dibs' on the pile. The next player must now pay player B, the debt associated with player B's debt card. (The unpaid part of the debt owed by B to A is cancelled, so a Jack *always* incurs a 1-card debt, even if it was played immediately after an ace.) This continues until a debt is fully paid-off with all non-debt cards. At that point, the player who has dibs on the pile at the time takes all the cards from the pile, unshuffled, and puts them at the back of his/her pack (face down like the rest of the pack). That player starts the next round. A player loses when his/her pack is exhausted. If the player is in the middle of paying a debt and loses, the player with dibs wins the pile as if the debt had been fully paid off. Play continues until one player has all the cards.

The outcome of the game is determined from the moment the cards are dealt, so it is deterministic, but it's still fun to play because it can go back and forth many times. Some games are short, some are quite long (even the one-deck games). If you play enough, you'll have a game (like I did) where you have only about 5 cards left in your pack, and the only debt card is a Jack. The Jack is the hardest debt to pay off, because you have the least chance to take over dibs of the pile. It just happened that my Jack appeared in the middle of paying a Queen debt, and I won the pile. I now had two debt cards… I ended up winning the game.

A very interesting question about 'beggar my neighbor' is whether there is an infinite game. It sure seems like it sometimes, especially when there are two players, the sizes of each player's pack can oscillate pretty regularly for a while. But so far, it has always ended up in a win for one player when you play long enough. There's a **lot** of different combinations a deck of cards can take, so an exhaustive search of all games would be a lot of work. By my reckoning, for a single-deck game with a fixed number of players, there are (52!) (that's 52 factorial) different combinations the deck can take. But the specific order of the 2s, 3s, 4s, … and 10s makes no difference - one could transpose any pair of cards from this set without making any difference to the outcome of the game - so that number can be divided by ((9 * 4)!) = (36!); the specific order of the jacks makes no difference so it can be divided again by (4!), same for the queens, kings and aces, so that's another (4!)^3. But still, (52! / (36! * ((4!) ^ 4))) is a large number (about (6 * (10 ^ 20)). I don't know if an infinite game is possible. I'd love to hear from someone if they have a proof that there is or isn't!

RimRod informs me that there's a very similar game called egyptian rat screw. It has an extra rule about slapping, which makes the game nondeterministic (personally, I kinda like the determinism of 'beggar my neighbor' - it makes the to and fro of the game even more surprising, which IMHO adds to the fun and excitement). Go have a look at that node - it has an example round to help you understand play.