3
$\begingroup$

2014 card players sit around a big table. One of the players begins with 2014 cards on his hand, and the other have none. The rules for the game are:

  • Every minute shall every player, who has 2 cards or more, give one card to the persons next to him.

  • The game is finished when everybody has exactly one card.

I imagine that the table has a form as a rectangle. At the end of the table there sits one person - one of them would be the person with 2014 cards. At the other sides there will be 1006 persons. enter image description here

I discovered that by $t=1$, $t=3$, $t=6$, $t=10$, $t=15$, $t=\frac{n(n+1)}{2}$ that will be $n$ persons on each side of the tables with one card, and the persons with most cards would have $2014-2n$ cards. Because there are 1006 persons of each side of the table (the long side) they would have 1 cards after $t=\frac{1006(1006+1)}{2}=506521$ minuts. Then we haveenter image description here

But if that is true then they ever will have one card each, because there always will be one person with 2 cards. Is that correct?

$\endgroup$
  • $\begingroup$ 2014 card players sit around a big table. In the beginning one has cards, the others - no. So why 1006, not 2013? $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 12:27
  • $\begingroup$ Sorry, I didn't get your point. What do you mean? $\endgroup$ – Nick Podowalski Jan 1 '15 at 12:31
  • $\begingroup$ From the first sentence of your question we have 2014 players. So why you are considering only $1+1006$ of them? $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 12:33
  • $\begingroup$ I said that there will be one person on each of the short sides of the tables. There wil also be 1006 persons on each sides of the long sides of the table. Then we have $1+1+1006+1006 = 2014$ persons. $\endgroup$ – Nick Podowalski Jan 1 '15 at 12:39
  • $\begingroup$ Oh, I see. But then 2 from your second picture "moves" through the upper part to the place before 0, and then the game ends... $\endgroup$ – Przemysław Scherwentke Jan 1 '15 at 12:45
9
$\begingroup$

Yes, the game will never finish.

Consider the number of cards the even-numbered players have. At the start it's zero, and in the end it ought to be 1007. But we always pass an even number of cards between the even- and odd-numbered players.

Hence it can never finish.

$\endgroup$
0
$\begingroup$

Lord_Farin's answer can be extended to all cases where there are $n$ players and cards and $n$ is even but not a multiple of $4$

Try this for all even $n$:

  • Every time players who start with all the cards pass cards, they pass an even number (the same left and right)

  • By the reflective symmetry of a round table and the passing rules, every time players who start with all the cards receive cards, they receive an even number (the same left and right)

  • So when the numbers of people and cards are even, players who start with all the cards (an even number) always have an even number of cards, and so can never have exactly $1$

So it is impossible for all even $n$, including $2014$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.