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I try to solve some exercises from olympiads and I have difficulties with this one:

Consider a round table with 20 people. One of these players receive a book and chooses one of his neighbors and passes the book to him (with probability 1/2). The next player again chooses one of his neighbors (each with prob 1/2) and passes the book. The game ends until at least everyone received the book one time.

Every person has of course a probability for being the last one getting the book. Which player of the group has the highest prob. reiceiving the book as the last player?

Intuitively I would say the player which sits opposite to the player that receives the book first (Lets call him Player 1 and the opposite player is Player 11).

I first tried to analyze the situation for 4 players. Assume at the bottom is Player 1, right to him Player 2, opposite of Player 1 is Player 3 and left to Player 1 is Player 4.

I assume Player 1 receives the book first:

=> Player 2 as the last one: 1->4->3->2 Player 3 as the last one: 1->4->1->2->3 or 1->2->1->4->3 Player 4as the last one: 1->2->3->4

Therefore I would say Player 3 has the prob. which attains the maximum over the group. This is exactly the opposite player of Player 1.

How can I use my argument for the case of 4 Players to prove it for 20 Players?

Montaigne
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  • If you don't get help here, another place to ask about former olympiad-type problems is: http://www.artofproblemsolving.com/Forum/ – GEdgar Mar 11 '13 at 14:16
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    This might help http://math.stackexchange.com/questions/116446/random-walk-on-n-cycle – Jean-Sébastien Mar 11 '13 at 14:44
  • @Jean: Great, I was reminded of that but didn't know how to find it :-) – joriki Mar 11 '13 at 14:48
  • @joriki That was me being lazy to type the argument, googling for it instead – Jean-Sébastien Mar 11 '13 at 14:49
  • @joriki There should be a search function inside our past answers – Jean-Sébastien Mar 11 '13 at 14:50
  • @Jean: There is; you can enter your user number in the search field (preceded by "user:", in your case "user:31493"). – joriki Mar 11 '13 at 14:51
  • That is good to know! – Jean-Sébastien Mar 11 '13 at 14:52
  • But a random walk on n cycle just give you the probability for receiving a specific value. In my case I do not want to know when a certain player receives the book, instead I want to analyze the case where each player gets the book at least one time and find the people whose prob. attains the maximum over the group. – Montaigne Mar 11 '13 at 14:52
  • @Montaigne: Please take some time to digest what people have written and linked to; the answers are all there. At least consider the possibility that you've misunderstood the answers before jumping to the conclustion that we've all misunderstood the question. – joriki Mar 11 '13 at 14:55
  • Thanks for the link. I made an error reasoning. – Montaigne Mar 11 '13 at 14:57

1 Answers1

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You may want to calculate the probability of expanding the current "territory" (set of people that have held the book) in either the forward or backward direction, as a function of the size of the territory.

When the territory has size $1$, forward and backward both have probability $1/2$. When the territory has size $2$, forward has probability $1/2+1/8+\ldots=2/3$ and backward has probability $1/3$. So expanding as $123$ or $143$ (leaving $4$ or $2$ as the last to hold the book, respectively) have probability $1/2\cdot 2/3=1/3$, while expanding as $124$ or $142$ (leaving $3$ as the last to hold the book) have probability $1/6$ each, for total probability $1/3$.

In other words, in the $n=4$ case, all players but player $1$ are, in fact, equally likely to hold the book last.

mjqxxxx
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  • I am not sure you interpreted my question in the correct way. Each person has a certain probability of being the last to reiceive the book; find the persons whose probability attains the maximum over the group.

    Which consequences would your calculations have for the case n=20?

     – Montaigne Mar 11 '13 at 14:48
  • @Montaigne: I don't see why you think this answer doesn't address that question. It shows that in the case $n=4$ this probability is the same for all. If you follow the link Jean-Sébastian provided, you'll find that it's in fact always the same for all, independent of $n$. – joriki Mar 11 '13 at 14:53
  • Ok I think I get it now, thank you for your help – Montaigne Mar 11 '13 at 14:56