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*. 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*, then each player doubles their amount of money on the table, and the die is thrown again. The doubling repeats any number of times, 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 pot evenly between them, concluding the round.

That's really all, but there is a final little twist: If the round ends after only one roll, without any doubling, the pot will consist of an odd number (three) of coins. The two winners get their coins back, but instead of using smaller change to split the third coin, the tradition is that the loser buys the next round of beer using that coin.

The game is

*zero-sum*, meaning no money enters or leaves (unless we regard the twist as a proper part of the game). The game is also

*symmetric*in the sense that all three players have the same status with respect to the rules. These are what we might call properties of fairness: What somebody wins, somebody else must have lost, and 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 those probabilities somehow 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 2 coins and A gets half of that. 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 2 coins with the same probability, and we start to see a pattern: Player A loses 2 coins of there is a

*Double*and then an

*A*, and wins 2 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 4 coins, we have to win after 3 doubles or lose after 2, both of which have probability $1/24$. Winning or losing 8 coins requires winning after 4 doubles or losing after 3, probability $1/48$ each, 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. Perhaps we shouldn't be that surprised in view of that "fairness" we just discussed.

But wait, we forgot about winning half a coin! That scenario didn't match up with any scenario where we lose half a coin, because we never do! So it seems that a given player, say A, will win half a coin with probability $1/3$ (when the first roll is a

*B*or a

*C*), and break even in the remaining cases. So perhaps after all it's when we include that final twist as a proper part of the game that it becomes "fair", and our wins and losses exactly balance out.

But then the question is: Where does all that beer come from, if 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. That's how well you know where you are when you're on tour with LiTHe Blås :)

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