Perpetuum mobile drinking game

Let's imagine we're in a pub in some sort of Pippi Långstrump universe where (1) money comes in the form of physical coins that you can move around on a table, (2) you never really run out of them, and (3) you can buy a round of beer for a single coin.

Three people called A, B and C play a game: There is a six-sided die where three of the sides are labeled A, B, and C, and the other three are labeled Double+1. At the start of the game, each of the three players puts one coin on the table, and the die is thrown. If the side that comes up says Double+1, then each player doubles and adds one to their amount of money on the table, and the die is thrown again. The doubling repeats any number of times, leading to the players putting 3, 7, 15 etc coins on the table, until one of the sides labeled with a player comes up. Whenever the die shows A, B, or C, that player loses, and the other two split the loser's pot between them. Traditionally the odd coin is used for buying the next round of beer. 

This game is symmetric in the sense that all three players have the same status with respect to the rules: Whatever somebody else wins or loses, you could have won or lost with exactly the same probability. Let's calculate the probabilities of winning or losing a certain amount, and see if they reflect this fairness. 

First let's find the probability that a player, say A, loses exactly one coin. This happens if A loses without any doubling, and the probability is $1/6$ since it happens precisely when the first roll of the die is an A. In order for A to instead win one coin, there has to be one double and the second roll must be a B or a C, so that somebody else loses three coins and A gets one of them. The probability of starting with a double is $1/2$, and the probability of the second roll being a B or a C is $1/3$, so the probability of winning one coin is $1/6$ too.

It turns out that we also win or lose 3 coins with the same probability, and we start to see a pattern: Player A loses 3 coins of there is a double and then an A, and wins 3 coins if there are two doubles and then a B or a C. Both these scenarios have probability $1/12$. And to win or lose 7 coins, we have to win after 3 doubles or lose after 2, both of which have probability $1/24$, and so on. Each of the scenarios has probability exactly half of the previous one. So the game has the property that winning $X$ coins is exactly as likely as losing $X$ coins. 

That seems fair.

But wait, where does all the beer come from? Doesn't it seem that in the long run, nobody is paying? 

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I discussed this sort of thing and related stuff in 2007 with the German probabilist Nina Gantert and my colleagues Jeff Steif and Olle Häggström. Thanks to Jeff who asked me about it a few days ago, because I had almost forgotten. I can't recall how it started, but I remember that when we met, Nina said she was from Münster, and I told her that I had been there and visited the Panzermuseum. Strangely she had never heard of it, and it turns out there isn't one. The Panzermuseum is in the small town Munster just south of Hamburg, some 200 km or so from the city of Münster.