This question is about a game of cards, played with a standard deck without jokers.
We used to play it as kids, and called it the paying game although it probably has a real name.
It goes like this.
Divvy up the cards between all the players. First player plays a card and play continues clockwise.
If the card played is 2-10, then the next player must play a card.
If the card played is a J,Q,K,A then the next payer must count out up to 1,2,3,4 cards stopping if they play a J,Q,K,A of their own.
If they pay all the cards without finding a face card, then the previous player wins the whole pile. if they do play a face card, then the next player must pay and so on.
If a player runs out of cards, they are out and the next player round takes up playing exactly where they left off. Eg if a king is played and the next player must play 3, but only has 1 card which is not a face, then the player plays the last card and is out. The next player must now play 2.
Player with all the cards wins.
An example may help.
Andy is playing Beth
A: 2D
B: 7H
A: AS (Beth must now try to pay 4 cards)
B: 5S, 6C, QS (Andy must now try to pay 2)
A: 10D, 10S
Beth scoops the pile of cards and adds them to the bottom of her deck.
We realised after a while that this game is silly because the outcome is totally predetermined and the skill of the player does not influence it at all.
Anyway, on to the question. Is it possible for there to be an initial setup leading to a game that never ends? Can this be done for more than 2 players, and what is the maximum number of players an infinite game can involve and have no players drop out?
Originally posted by MrPhilPlayers play their cards blindly I assume.
This question is about a game of cards, played with a standard deck without jokers.
We used to play it as kids, and called it the paying game although it probably has a real name.
It goes like this.
Divvy up the cards between all the players. First player plays a card and play continues clockwise.
If the card played is 2-10, then the next player must pla ...[text shortened]... what is the maximum number of players an infinite game can involve and have no players drop out?
What happens after someone wins the pile?
Originally posted by TheMaster37Yes, you play your cards from the top of your deck without looking at them.
Players play their cards blindly I assume.
What happens after someone wins the pile?
If a player wins a pile, he adds it to the bottom of his deck and starts off a new round by playing the top card from his deck. Play then continues.
Originally posted by MrPhilAs DrPhil would say: "Hmmm, interesting..."
This question is about a game of cards, played with a standard deck without jokers.
We used to play it as kids, and called it the paying game although it probably has a real name.
It goes like this.
Divvy up the cards between all the players. First player plays a card and play continues clockwise.
If the card played is 2-10, then the next player must pla ...[text shortened]... what is the maximum number of players an infinite game can involve and have no players drop out?
Originally posted by FabianFnasI think you are describing "Beggar-My-Neighbour" or "Beat-Your-Neighbour"
As DrPhil would say: "Hmmm, interesting..."
see http://en.wikipedia.org/wiki/Beggar-My-Neighbour
Your question is not a new one!
"A longstanding question in combinatorial game theory asks whether there is a game of Beggar-My-Neighbour which goes on forever. This can happen only if the game is eventually periodic—that is, if it eventually reaches some state it has been in before. Some smaller decks of cards have infinite games, while others do not. John Conway once listed this among his anti-Hilbert problems, open questions whose pursuit should emphatically not drive the future of mathematical research."
http://en.wikipedia.org/wiki/Beggar-My-Neighbour
You are right, that does seem to be the same game.
Perhaps it being one of mathematics unsolved problems explains why there weren't more people pitching in to a solution.
On the other hand, I got as far as recognizing that it has to be periodic by myself, but couldn't get any further, so I have done as well as the greatest maths minds. Go me!
Thanks for bringing this one to a resolution
Phil.