Saturday, March 16, 2013

A fun game-theoretic paradox

You observe Fred and Sally playing a game repeatedly. The game works as follows. Fred spins a spinner counterclockwise on a circular board that is so calibrated that its final position is uniformly distributed. The board has some angles labeled in degrees, counterclockwise, but the payoff for Sally depends solely on the exact final position of the spinner (so if it is between markings, the exact position between markings matters). You don't know how the payoff depends on the final posiiton of the spinner. But now you notice an oddity. Each time, right after the spinner stops, Sally is not allowed to see where it stopped, but Fred asks her: "Do you want me to rotate the spinner by an extra square-root-of-two degrees counterclockwise?" And Sally says: "Of course."

You ask Sally: "Are you superstitious or trying to make more work for Fred or have you peeked at where the spinner landed?" And Sally says: "Not at all. The payoff structure for the game gives me a good reason to ask for that extra counterclockwise rotation if the chance is offered."

And now you're really puzzled. You know Sally is really smart, and yet you wonder how it could be that an extra fixed rotation on top of what is already a uniformly distributed rotation of the spinner could make any difference.

It turns out that it could. Suppose that the game works as follows. Let S be the set of positions defined by the angles r, 2r, 3r, ... in degrees (with wrapping around, of course, so that 361 degrees defines the same angle as 1 degree), where r is the square root of two. If the pointer ends in S, Sally gets an infinite payoff (the idea of using an infinite payoff to make zero-probability outcomes relevant is Alan Hajek's)—maybe she gets to go to heaven. Otherwise, she gets nothing. In this setup, Sally has a good reason to ask for the extra rotation of r degrees. For if after Fred's initial spin, the pointer is already in S, it will still be in S after that extra rotation (nr+r=(n+1)r). But if Fred's initial spin put the pointer at zero, then the extra rotation by r will put the pointer at r, which is in S. Thus, if Fred's initial spin is not at zero, Sally loses nothing by asking for the extra rotation, and if it is at zero, Sally gains an infinite payoff by asking. Since it could be at zero (though the probability of that is, well, zero), we have domination. If Fred allowed it, Sally would have even better reason to ask for a rotation of 2r degrees, and an even better reason to ask for a rotation of 10100r degrees.

Of course, Sally is all but certain not to get anything. But infinite utilities call for desperate measures, and it costs Sally nothing to agree. (We could suppose it delays the next game. That can be taken care of by ensuring successive games are always at fixed times.)


Alexander R Pruss said...

I am now thinking that the case provides a counterexample to domination.

Alexander R Pruss said...

Update: My paper arguing that this is a counterexample to domination and regularity has just been accepted by Analysis.