A big round table of cards

Probability Level 1

There are 12 boys seated around a big round table. They play a game with 12 cards. Initially a boy $A_1$ has all the 12 cards with him.

Every minute, if any boy has 2 or more cards with him, he passes a card to the boy on the left, and a card to the boy on the right. The game ends when each and every boy has 1 and only 1 card with him.

How many minutes does it take for this game to end?

2(12!)

The game never ends

12!

12^2

4 solutions

Christopher Boo
Oct 21, 2014

The first player either loses 2 cards, or no net gain, or get two cards from both his left and right player. Hence the first player's card must be even, the game never ends.

Same here. Awesome problem.

Rick B - 6 years, 7 months ago

This solution is not complete; you need to mention that the only possible way the game doesn't end is when every player has exactly one card each.

Wen Z - 4 years, 9 months ago

Does the first player get one card from left and one from right simultaneously? If so please explain.

Romok Das - 3 years, 6 months ago

Yes. Because of symmetry, the guy on your left and right must have the same amount of cards. If the left one gives you a card, the right one must also give you a card.

Christopher Boo - 3 years, 6 months ago

Richard Levine
Nov 24, 2014

I agree with Christophe Boo. Eventually the boys at the round table end up with pattern 202020202020 alternating with pattern 020202020202. The game never ends.

Ayush Garg
Dec 21, 2016

Let us number the players from 1 to 12 and all cards held by a particular player with his number. For example all cards with player number 7 are numbered 7. It is obvious that at every step the sum of the 2 cards modulo 12 remains the same. Since initially the sum of the cards was 12*12= 0 mod 12, it will always be 0 mod 12 but the end state demands it to be 1+2+3+..12 mod 12 which is not 0 mod 12. Hence the end state can't be reached.

More explantions: Take $T=x 1+2x 2+3x 3+\cdots + 11x {11}+12x {12}$. (Define $x {13}=x {1}$). For each step, $(x {n-1}, x n,x {n+1})\to (x {n-1} +1, x n - 2,x {n+1} +1)$ and $(n-1)x {n-1} + nx {n} + (n+1)x {n+1} \equiv (n-1)(x {n-1}+1) + n(x {n}-2) + (n+1)(x_{n+1}+1) \pmod{12}$. Therefore $T$ is an invariance. At starting position $T=1\cdot 0 +2\cdot 0 +\cdots +11\cdot 0 + 12\cdot 12 \equiv 0 \pmod{12}$. But at finish position $T=1\cdot 1 +2\cdot 1 +\cdots +11\cdot 1 + 12\cdot 1 \equiv 6 \pmod{12}$. Hence, we can never reached finish position.

lokman gökçe - 8 months, 1 week ago

Dante Adonai
Dec 19, 2014

will the first player with two or more card it give it away then the first player shall have 0 card in hand if they do that every minute they never end because player with 2 or more card need to become 0

A big round table of cards

4 solutions

0 pending reports