Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose you can make infinitely many copies of yourself. Each of them starts his/her life in a Hilberts Hotel, where each room is labeled by an element in the free group with two generators, and structured as the Cayley graph of the group. (All the room have the same size, so in particular this hotel is not embedded in $\mathbb{R}^3$!) In the beginning, each clone have 1£. If they all cooperate, they can get richer exponentially fast: If they all give all their money to the neighbor in the direction of $e$, then everyone except the person in $e$ will receive money from three persons, so after n transactions he will have $3^n$£ (and $e$ will receive money from 4 persons, so he will be even richer).

Question: Suppose instead that the rooms were unlabeled. You can decide on a strategy before being copied, and you are allowed to use randomness in this strategy. However, all the copies will be identical, so all of them will think that they are the original "you". Each of the copies can send money and information to each of their four neighbors once each day. Is there a strategy that will make each of them rich exponentially fast?

Comment: If only one of the copies thought he/she was the original you, you could solve the problem: The real you is consider to be $e$. The first day he/she tells his/her neighbors. The next day the neightbors send 2/3 of their money to $e$ and tell their neighbors that "the original you" is in this direction, and so on. With this strategy, each copy become rich exponentialy fast, although it will take some time ("distance to $e$" +3 days) before it starts.

I originally asked the question on my blog, here.

share|improve this question
    
Suggested alternative get-rich-quick scheme in connection with Hilbert's hotel: mathoverflow.net/questions/22033/… –  Joel David Hamkins Feb 13 '12 at 0:22
1  
Incidentally, JDH's answer there is a viable candidate for a second "reversal" badge. –  S. Carnahan Feb 13 '12 at 1:56
    
Now he got it :) –  Sune Jakobsen Feb 13 '12 at 16:38
2  
Hey, thanks, you guys, I'm flattered! Shall I pin my shiny new gold badge on my lapel? Perhaps at the next seminar meeting... ;-) –  Joel David Hamkins Feb 13 '12 at 17:00
add comment

2 Answers 2

up vote 18 down vote accepted

No, you cannot get rich with identical copies on the unlabeled tree. This is a special case of the Mass Transport Principle - take a look at the book of Lyons and Peres, chapter 8.

share|improve this answer
3  
The book is available online, and the relevant theorem is on page 283 in the book (page 293 in the pdf-file). –  Sune Jakobsen Feb 12 '12 at 18:38
add comment

If the distance matters, one could ask his neighbors what's the distance between them and their neighbors, to understand what the actual position of $e$ is; then is possible to send money as you pointed out.
Well, this is probably cheating on the definition of "unlabeled" copies...

share|improve this answer
1  
Well, all the distances might be the same. As I said, the hotel is not embedded in R^3 ;) –  Sune Jakobsen Feb 27 '12 at 8:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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