Finite threats always lose?

While driving down the freeway tonight a strange application of backward induction popped into my mind. Here’s the intuition:

Suppose Albert, a terrorist, has a nuclear device. But Albert isn’t a very good terrorist. He doesn’t particularly want to kill anyone, he’s indifferent between killing and not-killing, so he decides to use his nuke as a threat to make himself immune from the law. One day, Albert’s doing 100 in a school zone, and he gets pulled over by Bob, a police officer. Their conversation goes, in relevant part, as follows:

Albert: “I have a concealed nuclear weapon somewhere in a major population center. If you write me a ticket, I’ll detonate it.”

Bob: “Your threat isn’t credible. After you detonate your nuclear device, you will have nothing left to do to me, but there will be more rounds to play. Indeed, there will be as many rounds as I damn well please. And I’ll use them to lock you in a cell and beat you every day for your natural life.”

Albert, fuming, signs the ticket and drives home, well within the speed limit.

A week later, Albert is speeding again, and Bob pulls him over again.

Albert: “Ha-ha! I built a second nuke and hid it in a second city! Now, you really can’t write me a ticket or beat me, because I can detonate one nuke and still have a second in reserve.”

Bob: “Nice try. But we’ve already established that you can’t detonate your last nuke, so it’s not a real threat. And that means I can beat you with impunity after you detonate your next-to-last nuke. That, in turn, means you can’t detonate that one either, by the same reasoning.”

Once again, Albert signs his speeding ticket and drives ever so slowly home. He sits around for a while, fuming, and contemplating buying more uranium from Russia. It slowly dawns on him… Bob’s last point can be iterated to any finite number of nukes. Which means that Bob’s little infinitely repeatable threat can just beat an arbitrarily large number of Albert’s big finite threats. (Then, of course, he starts to think of doomsday machines and other commitment devices, but let’s leave that off.)

Tempted to formalize this to see if the intuition really makes sense; don’t really have the time, though, plus someone at the Rand corporation doubtless already whipped this up in the 50’s.


4 Responses to “Finite threats always lose?”

  1. Andrew Stevens Says:

    Similar to the unexpected hanging paradox, no?

  2. Kenny Says:

    There’s a paradox that’s been discussed in the literature with a similar structure. I forget what it’s called, but it goes like this:

    Joe is convicted of a crime, and is sentenced to be executed next week on a day when he’s not expecting it. He reasons that he can’t be executed on Friday because, when he finds that he hasn’t been executed by Thursday night, he’ll know that he’ll be executed on Friday, so it won’t be a surprise. But since he can’t be executed on Friday, if he isn’t executed on Wednesday, he’ll know that the execution will be Thursday, so it can’t be Thursday either. And so forth. So the sentence cannot be carried out.

  3. Paul Gowder Says:

    Oh yeah — I guess they are pretty similar, aren’t they? Thanks guys.

  4. Andrew Stevens Says:

    Kenny, you forgot to mention the punchline. After working all this out and confident that he can’t be executed when he isn’t expecting it, Joe, to his complete surprise, is executed on Wednesday.

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