Steven Landsburg, who is supposedly a “hardcore libertarian,” is on the verge of surpassing Bryan Caplan as the most wrong-headed libertarian economist on my RSS reading list. Landsburg’s upside-down view of the world has led me to issue the following posts:
- “Landsburg Is Half-Right“
- “Rawls Meets Bentham“
- “The Case of the Purblind Economist“
- “Take Landsburg’s Money“
Now comes Landsburg with yet another tour through the land of tortured logic: “Another Rationality Test.” There, Landsburg sets out the following problem:
Suppose you’ve somehow found yourself in a game of Russian Roulette. Russian roulette is not, perhaps, the most rational of games to be playing in the first place, so let’s suppose you’ve been forced to play.
Question 1: At the moment, there are two bullets in the six-shooter pointed at your head. How much would you pay to remove both bullets and play with an empty chamber?
Question 2: At the moment, there are four bullets in the six-shooter. How much would you pay to remove one of them and play with a half-full chamber?
In case it’s hard for you to come up with specific numbers, let’s ask a simpler question:
The Big Question: Which would you pay more for — the right to remove two bullets out of two, or the right to remove one bullet out of four?
The question is to be answered on the assumption that you have no heirs you care about, so money has no value to you after you’re dead.
If you think the right answer is to pay more for the right to remove two bullets out of two, you’re wrong, according to Landsburg. Why? Well, Landsburg — with a train of “logic” that reminds me of Lou Costello’s “proof” that 7 x 13 = 28 — “proves” that removing two of four bullets has the same value as removing two of two bullets. Here’s how Landsburg does it:
[T]hink about four questions:
Question A: You’re playing with a six-shooter that contains two bullets. How much would you pay to remove them both? (This is the same as Question 1.)
Question B: You’re playing with a three-shooter that contains one bullet. How much would you pay to remove that bullet?
Question C: There’s a 50% chance you’ll be summarily executed and a 50% chance you’ll be forced to play Russian roulette with a three-shooter containing one bullet. How much would you pay to remove that bullet?
Question D: You’re playing with a six-shooter that contains four bullets. How much would you pay to remove one of them? (This is the same as Question 2.)
Now here comes the argument:
- In Questions A and B you are facing a 1/3 chance of death, and in each case you are offered the opportunity to escape that chance of death completely. Therefore they’re really the same question and they should have the same answer.
- In Question C, half the time you’re dead anyway. The other half the time you’re right back in Question B. So surely questions C and B should have the same answer.
- In Question D, there are three bullets that aren’t for sale. 50% of the time, one of those bullets will come up and you’re dead. The other 50% of the time, you’re playing Russian roulette with the three remaining chambers, one of which contains a bullet. Therefore Question D is exactly like Question C, and these questions should have the same answer.
Okay, then. If Questions A and B should have the same answer, and Questions B and C should have the same answer, and Questions C and D should have the same answer — then surely Questions A and D should have the same answer! But these, of course, are exactly the two questions we started with.
In case the trick isn’t obvious on first reading — and it wasn’t to me — here’s what Landsburg does.
He starts by positing a situation in which removing two of two bullets (Question A/Question 1) and removing one of one bullet (Question B) have the same value, and therefore should be worth the same amount to the hypothetical, involuntary player of Russian Roulette. Why should they have the same value? Because, given the player’s unstated but implicit objective — which is to survive Russian Roulette, and nothing else — he stands to improve his chance of surviving by 1/3 in both cases. Probabilistically, removing two of two bullets from a six-chamber gun is the same as removing one of one bullet from a three-chamber gun. In both instances, the chance of being shot goes from one-third to zero.
The point of the preceding exercise isn’t to get the reader to think about probabilities; it’s to get the reader (a) to assume that Landsburg’s “proof” is on the up-and-up, while (b) planting the idea that removing one of one bullets from three chambers is just as good as removing two of two bullets from six chambers. The reader is being set up to fall for a real whopper, to which I’ll come.
The next step (Question C) is to juxtapose the threat of dying by three-shooter with an irrelevant but equally probable threat. The 50% probability of summary execution is irrelevant, in the context of the problem at hand, because there’s nothing the player can do about it. What the player can do is to influence his probability of surviving Russian Roulette, which increases by one-third if he buys the single bullet that’s in the three-shooter.
The point of the preceding exercise is to reinforce further the reader’s confidence in Landsburg’s “proof,” while planting the idea that it’s legitimate to ignore a 50% threat of certain extinction. I use the word “confidence” because the setup of this “proof” is like the setup of a confidence game. The reader, like the victim of a con game, is now ripe for plucking.
Landsburg’s answer to Question D (Question 2) does the trick. It parallels Question C, but it doesn’t have the same import with respect to the player’s objective: surviving a game of Russian Roulette. Landsburg waves away 50% of the threat to the player’s existence (the three bullets that the player can’t buy), even though the three bullets (one of which may come up 50% of the time) are part of the game of Russian Roulette, unlike the summary execution of Question C. Landsburg, having dismissed the threat posed by the three bullets as somehow irrelevant, then focuses on the remaining bullet. Because this bullet can be in one of three chambers (the three not occupied by the other bullets), it seems that the player can increase by one-third his chance of surviving Russian Roulette if he buys one bullet. Thus, by Landburg’s “logic,” buying one of four bullets has the same value to the player as buying two of two bullets.
If you swallow that one, I’d like to sell you a bridge.
The fact that the player can’t buy three of the four bullets posited in Question D/Question 2, has no bearing on the probability that he survives Russian Roulette, which is his objective. With four bullets in the gun, his chance of survival is one-third. If he pays to have one bullet removed, the gun still has three bullets in it. Thus his chance of survival increases by one-sixth (3/6 [1/2] – 2/6 [1/3] = 1/6). Clearly, the value of removing one of four bullets isn’t the same as the value of removing two of two bullets, which increases his chance of survival by 1/3.
If you’re not convinced, let’s return to the original problem and restate it in a way that doesn’t distort the essential question: If the player’s objective is to survive a game of Russian Roulette, would he pay more for a six-gun with no bullets in it (Question 1) or a six-gun with three bullets in it (Question 2)? If you say that he’d pay the same amount for either gun, your name is Steven Landsburg.