End and Foot

 Heuristic, Literature  21 Responses »
Mar 132007
 

Brian Doherty’s Radicals for Capitalism is a comprehensive, highly entertaining history of libertarianism with too many points of interest — Murray Rothbard’s solution to the free rider problem (“so what?”), Milton Friedman’s sterling character, The Unbearable Lightness of Being a Deontologist — to deal with in a single post. Instead I want to talk about the notes.

Radicals for Capitalism is a scholarly, though not an academic, book, and like many such books it does plenty of business in the notes. Not as much as some, like Popper’s The Open Society and Its Enemies, in which the notes are longer than the text, but enough. For instance, my friend (and frequent commenter) Jim Valliant’s book on the Brandens, The Passion of Ayn Rand’s Critics, receives a half-page treatment in the endnotes, but none in the text. Out of 2,000 notes, there are 400 or so that you want to read; the rest are simple source citations.

Doherty’s notes receive the standard treatment, which is to say the worst possible. The notes are renumbered by chapter, but each page of notes is headed, usefully, “Notes”; the chapter titles occur only on the beginning page of the notes for that chapter. To look up an endnote, then, you have to remember the number, remember the chapter number, flip to the notes section, locate the beginning page of the correct chapter, and then flip forward to the right note number, only to be disappointed most of the time with a mere source cite. (Admittedly it would be more efficient to use a bookmark, but I never have one handy, and they tend to fall out. At any rate, the necessity confesses design failure.)

Yet this is all so simple to fix. There are five rules for notes:

1. Footnotes, provided they are short and sparse, are better than endnotes. They can be consulted immediately and without effort. Obviously in a book like Doherty’s endnotes are necessary.

2. Each endnote page should be headed by the page numbers of the notes it contains, to facilitate easy flipping. For example, “Notes, pp. 537-558”; not “Notes: Chapter Seven,” or “Notes: A Stupid Chapter Title That I’ve Forgotten and Now You’re Gonna Make Me Look It Up,” or, God forbid, “Notes.”

3. Notes should not be numbered. Numbers tax the reader needlessly, especially when they reach three figures. They should be marked by a symbol in the text, something like this◊ or this•. In the back they should be referenced by the page number and the last few words of the passage that they annotate, which are the easiest things to remember.

It would be especially helpful to use two symbols, to distinguish substantive comments from simple citations, telling the reader when to flip to the back and when not to bother. I have never seen this in a scholarly book, and I wonder why.

4. The notes must be indexed. In Doherty’s book they are not. Had Jim Valliant gone looking for himself in the index, as I am assured august persons are wont to do, he would have come up empty. Why make trouble for Jim? If he merits a substantive mention, he also merits an index entry. I realize this is extra work. I expect extra work for my thirty-five bones, now marked down to $23.10, plus shipping.

5. The text should contain as little scholarly detritus as possible. Academic books often include source citations in the text, which avails the author the opportunity to look more erudite and avails the reader nothing, since if he wants to look up the source he has to consult the biblliography anyway. If the book has endnotes, that’s where the source cites belong.

A brilliant exception to this rule is Jacques Barzun’s From Dawn to Decadence, which contains no specific source cites, only an occasional parenthesis, when discussing a topic, that “the book to read is…” or “the book to browse in is…” If you are a nonagenarian and the world’s preeminent living intellectual, you can write like that. The rest of us cannot afford to be so peremptory. Still, Barzun’s asides have furthered my education, which is more than I can say for the usual uncommented bibliography.

â—ŠYes, a circle would be better. I can’t get a circle the right size using HTML character codes. Sorry.

•Yes, a larger bullet would be better. See above. I trust you get the idea.

Update: Another intransigent opponent of endnotes, Billy Beck, heard from. I thank him for his recommendation of the Zerby book, which I will look up. Kieran Healy comments. Andrew Gelman comments. James Joyner comments. Evan Hughes comments.

How To Solve It

 Heuristic  32 Responses »
Nov 222006
 

If I could require every American schoolchild of normal intelligence to read one book, it would be George Polya’s How To Solve It. (Second choice is Henry Hazlitt’s Economics in One Lesson. I keep extra copies of both books on hand to give away as necessary.) Polya was born in Hungary and taught mathematics at several European universities before ending up at Stanford. Like the authors of all the best pedagogical texts, he was a superb practitioner. Polya made important original contributions in probability theory, combinatorics, complex analysis, and other fields. He published How To Solve It in 1945; it has since sold more than a million copies. He died in 1985 at an immense age.

How To Solve It, among its other virtues, is a model of English prose style; I will let Polya himself describe what he’s up to:

The author remembers the time when he was a student himself, a somewhat ambitious student, eager to understand a little mathematics and physics. He listened to lectures, read books, tried to take in the solutions and the facts presented, but there was a question that disturbed him again and again: “Yes, the solution seems to work, it appears to be correct; but how is it possible to invent such a solution? Yes, this experiment seems to work, this appears to be a fact; but how can people discover such facts? And how could I invent or discover such things by myself?” Today the author is teaching mathematics at a university; he thinks or hopes that some of his more eager students ask similar questions and he tries to satisfy their curiosity.

Math students are regularly exhorted to “show your work,” while the great mathematicians hide theirs. Euclid’s proof that the angles of a triangle sum to 180 degrees is a masterpiece of logical thought, but however he arrived at it, it was assuredly not by the route shown in the Elements. The proofs came first, the axioms after. One can admire but not emulate. In short, what math education lacks is heuristic, and this is what Polya endeavors to supply.

The way to write about Polya is to solve problems with his techniques. Abbas Raza at 3 Quarks Daily provided an occasion by posting fourteen moderately difficult logic problems, none requiring mathematical background. I’ve rearranged them slightly. Most of the problems are famous; you have probably seen some of them before. You may want to have at the problems first before you read my solutions and commentary on how I used Polya’s techniques to find them.

1. You are given two ropes and a lighter. This is the only equipment you can use. You are told that each of the two ropes has the following property: if you light one end of the rope, it will take exactly one hour to burn all the way to the other end. But it doesn’t have to burn at a uniform rate. In other words, half the rope may burn in the first five minutes, and then the other half would take 55 minutes. The rate at which the two ropes burn is not necessarily the same, so the second rope will also take an hour to burn from one end to the other, but may do it at some varying rate, which is not necessarily the same as the one for the first rope. Now you are asked to measure a period of 45 minutes. How will you do it?

Solution: Light the first rope at both ends, and the second at one end. When the first rope has completely burned, 30 minutes have elasped. Now light the other end of the second rope. When the second rope has completely burned, 45 minutes have elapsed.

Commentary: “If you can’t solve a problem,” Polya says, “there is an easier problem you can solve: find it.” Measuring 45 minutes may seem impossible at first, but how about 30 minutes? Thinking about 30 minutes instead, you may hit on the bright idea of lighting the rope at both ends. From there you need one more bright idea: that you need not light both ends simultaneously. Most people, including me, arrive at the second idea very quickly after thinking of the first; but I once saw an excellent problem solver find the first idea immediately and take quite a while to find the second.

2. You have 50 quarters on the table in front of you. You are blindfolded and cannot discern whether a coin is heads up or tails up by feeling it. You are told that x coins are heads up, where 0 < = x <= 50. You are asked to separate the coins into two piles in such a way that the number of heads up coins in both piles is the same at the end. You may flip any coin over as many times as you like. How will you do it?

Solution: You are given x, the number of heads. Create a subgroup of x coins. Flip them all.

Commentary: Polya asks: Did you use the whole condition? The condition here is more liberal than it looks. You need not know the number of heads in each pile. Neither must the two piles contain the same number of coins, provided the number of heads in the two piles is the same.

Polya also asks: Did you use all the data? Here we are given the total number of coins, which is doubtfully relevant, except that it is large enough to make the problem difficult. More important, we are given x, the number of coins heads up. The solution is very likely to involve flipping x coins. In fact it is a simple matter of doing just that.

Polya finally asks: Can you check the solution? Introducing suitable notation, another Pólya suggestion, yields a satisfying way to do so. x is the number of coins that are heads up; 50 – x, then, is the number of coins tails up. We divide the coins into two groups, of x and 50 – x coins. Let y be the number of heads in the x group. Then the number of heads in the 50 – x group is x – y. Now we flip all the coins in the x group. The number of heads becomes x – y. The two groups contain the same number of heads. This also demonstrates, as we suspected, that 50 is indeed irrelevant; the solution works no matter how many coins you begin with.

3. A farmer is returning from town with a dog, a chicken and some corn. He arrives at a river that he must cross, but all that is available to him is a small raft large enough to hold him and one of his three possessions. He may not leave the dog alone with the chicken, for the dog will eat it. Furthermore, he may not leave the chicken alone with the corn, for the chicken will eat it. How can he bring everything across the river safely?

Solution: Bring the chicken across. Return alone and bring the dog across. Return with the chicken, and bring the corn across. Return alone and bring the chicken across.

Commentary: This is a “hill-climbing” problem; you proceed by steps until you reach the goal. It can be difficult to solve because in hill-climbing it is natural to try to proceed directly, which gets you stuck at a local optimum of two items across the river.

Polya says, translating the Greek mathematician Pappus of Alexandria (circa 300 AD), “start from what is required and assume what is sought as already found.” Next “inquire from what antecedent the desired result could be derived.” Beginning at the end, we can see that on the farmer’s last raft trip he must bring the chicken across, because only the dog and corn can be left together safely. But for the same reason he must also bring the chicken across on his first trip. Putting this to yourself explicitly, you may eventually realize that the chicken must go back and forth and the solution will immediately present itself.

4. Late one evening, four hikers find themselves at a rope bridge spanning a wide river. The bridge is not very secure and can hold only two people at a time. Since it is quite dark, a flashlight is needed to cross the bridge and only one hiker had brought his. One of the hikers can cross the bridge in one minute, another in two minutes, another in five minutes and the fourth in ten minutes. When two people cross, they can only walk as fast as the slower of the two hikers. How can they all cross the bridge in 17 minutes? No, they cannot throw the flashlight across the river.

Solution: Two and One cross (2 minutes). One returns (3 minutes). Ten and Five cross (13 minutes). Two returns (15 minutes). Two and One cross (17 minutes).

Commentary: Polya asks: If you had a solution, what would it look like? Certainly we know that Ten and Five cannot cross more than once, or we are immediately at 20 minutes plus. But if Ten and Five cross separately we are still over 17 minutes, since there must be three other trips of at least a minute each. Therefore Ten and Five must cross together. This cannot happen at the beginning — otherwise one would have have to return — or at the end — since someone would have to return with the flashlight and would remain. Therefore they must cross in the middle. The solution appears.

Polya also asks: Do you know a related problem? This problem bears an interesting reciprocal relationship to Problem 3, of the dog, chicken, and corn. There we infer the procedure from the first and last trips; here we infer it from the trip in the middle.

5. You have four chains. Each chain has three links in it. Although it is difficult to cut the links, you wish to make a loop with all 12 links. What is the smallest number of cuts you must make to accomplish this task?

Solution: Three cuts. You cut all three links of a single chain and use them to connect the other three together.

Commentary: Cutting one link in each of the four chains will obviously do the job, but that’s not interesting enough to be the right answer. Can we do better?

Pólya suggests enumerating the solution space, when possible; or guessing, to put it bluntly. How many different ways can we cut three links? Well, we can cut one from each of three chains: that won’t work. We can cut two from one chain: that doesn’t help either. Or we can cut all three from a single chain… aha!

6. Before you lie three closed boxes. They are labeled “Blue Jellybeans”, “Red Jellybeans”, and “Blue & Red Jellybeans”. In fact, all the boxes are filled with jellybeans. One with just blue, one with just red and one with both blue and red. However, all the boxes are incorrectly labeled. You may reach into one box and pull out only one jellybean. Which box should you select from to correctly label the boxes?

Solution: Choose from the box labeled Blue & Red.

Commentary: Another good guessing problem. The solution is obvious. It is functionally equivalent to choose from the Blue or Red box, and the problem stipulates a single answer, which must be Blue and Red.

All that is left is the reasoning. Suppose you choose a red jellybean. Then you know the Blue & Red box should be labeled Red, and that, since the other two boxes are also mislabeled, that the Red box must be Blue and the Blue box must be Blue and Red.

Guess first, reason later: it works more often than you’d think.

7. Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have two children,” she replied. What is the probability that both of her children are girls?

Solution: There are four prior possibilities for the sex distribution of her two children: boy-boy, girl-girl, boy-girl, and girl-boy. We’ve seen a girl, so boy-boy is out. Of the three remaining possibilities, once you’ve revealed a girl, a boy remains in two of them. Therefore the probability that the other child is a girl, P(G) = 1/3.

Commentary: The difficulty here is less in finding the answer than in believing it. As with the Monty Hall Problem, many people deny that the solution is true, and they have distinguished company. (The solution depends subtly on the precise wording with which the problem is given; this comment thread has an extensive discussion, which is beyond the scope of this discussion.)

Polya asks: Can you draw a diagram? No, but you can model the problem experimentally. Dump a bunch of coins on the table and pair them up randomly. Remove all the tail/tail pairs. Now tabulate the results for the rest of the pairs. They will be tail/head approximately 2/3 of the time.

8. A glass of water with a single ice cube sits on a table. When the ice has completely melted, will the level of the water have increased, decreased or remain unchanged?

Solution: The water level sinks, because ice has lower specific gravity than water.

Commentary: Polya asks: Have you seen this problem before? You have. It’s the famous problem Archimedes solved in his bath. A king asked Archimedes to determine if a crown he owned was pure gold, without melting it down. Archimedes stepped into his bath, watched the water rise, and ran naked into the street, shouting “Eureka!” Maybe not. At any rate, he realized that his body displaced an equivalent volume of water, and he could measure the volume of any irregular object the same way, by submerging it.

Once Archimedes determined the volume of the crown, he simply weighed it against a lump of gold of identical volume. Gold is denser than silver, so if the crown was lighter, it had been adulterated. Water is denser than ice, so the water level sinks as the ice melts.

Abbas Raza, after setting this problem, got it wrong. The floating cube does not “displace its own weight in water”; it displaces its own volume in water. Had he regarded Polya’s advice to check the solution, by melting a few ice cubes in a glass of water, he would have spared himself some embarrassment. (See the update for who’s embarrassed now.)

9. You are given eight coins and told that one of them is counterfeit. The counterfeit one is slightly heavier than the other seven. Otherwise, the coins look identical. Using a simple balance scale, can you determine which coin is counterfeit using the scale only twice?

Solution: Weigh three against three. If they are equal then the counterfeit is one of two coins and it’s easy. If not, then the counterfeit is one of three coins. Take two of the three and weigh them against each other. Whichever is heavier is the counterfeit, or, if they’re equal, the third is counterfeit.

Commentary: This problem would be far easier if it were given with nine coins instead of eight. The same solution applies, but since three divides nine evenly, and two does not, you would immediately think to weigh three against three. With eight coins the opposite is true. You think of weighing four against four, and it may be some time before you disentangle yourself.

10. There are two gallon containers. One is filled with water and the other is filled with wine. Three ounces of the wine are poured into the water container. Then, three ounces from the water container are poured into the wine. Now that each container has a gallon of liquid, which is greater: the amount of water in the wine container or the amount of wine in the water container?

Solution: The water in the wine is equal to the wine in the water.

Commentary: This problem, like Problem 2, is overspecified. In fact almost all of the given data — how much liquid is in each container, the mixing sequence — is irrelevant. It matters only that the two containers begin and end with equal amounts of liquid. Polya asks: Did you use all the data? Here that question gets you into trouble.

But Polya also says that sometimes the general problem is easier to solve. (In computer science the general problem is always easier to solve.) He has caught grief for this remark, and the example he gives is somewhat artificial, but here it bears out. The specifics make the problem confusing.

Of course if you solve the general problem then you have, by definition, not used all the data. Sometimes one procedure works; sometimes its opposite.

11. Other than the North Pole, where on this planet is it possible to walk one mile due south, one mile due east and one mile due north and end up exactly where you began?

Solution: Polya gives this exact problem in How To Solve It in its more famous form, in which a bear does the walking and the problem is what color is the bear. I will quote his solution, if only to demonstrate how comprehensive his thinking is next to mine:

You think that the bear was white and the point P is the North Pole? Can you prove that this is correct? As it was more or less understood, we idealize the question. We regard the globe as exactly spherical and the bear as a moving material point. This point, moving due south or due north, describes an arc of a meridian and it describes an arc of a parallel circle (parallel to the equator) when it moves due east. We have to distinguish two cases.

1. If the bear returns to the point P along a meridian different from the one along which he left P, P is necessarily the North Pole. In fact the only other point of the globe in which two meridians meet is the South Pole, but the bear could leave this pole only in moving northward.

2. The bear could return to the point P along the same meridian he left P if, when walking one mile due east, he describes a parallel circle exactly n times, where n may be 1, 2, 3… In this case P is not the North Pole, but a point on a parallel circle very close to the South Pole (the perimeter of which, expressed in miles, is slightly inferior to 2Ï€ + 1/n).

Commentary: Before solving the problem Polya offers the following hints:

What is the unknown? The color of a bear — but how could we find the color of a bear from mathematical data? What is given? A geometrical situation — but it seems self-contradictory: how could the bear, after walking three miles in the manner described, return to his starting point?

12. I was visiting a friend one evening and remembered that he had three daughters. I asked him how old they were. “The product of their ages is 72,” he answered. I asked, “Is there anything else you can tell me?” “Yes,” he replied, “the sum of their ages is equal to the number of my house.” I stepped outside to see what the house number was. Upon returning inside, I said to my host, “I’m sorry, but I still can’t figure out their ages.” He responded apologetically, “I’m sorry. I forgot to mention that my oldest daughter likes strawberry shortcake.” With this information, I was able to determine all of their ages. How old is each daughter?

Solution: The factors of 72 can be combined into three factors with identical sums only one way: 6, 6, and 2; and 3, 3, and 8, both of which sum to 14. “My oldest daughter likes strawberry shortcake” implies that there is one daughter who is older than the other two. (This isn’t quite sound, since two of the daughters could be, say, 6 and 1 month and 6 and 11 months, and even twins are not precisely the same age; but, as Polya would put it, we idealize the question, as it is more or less understood.) Therefore 3, 3, and 8 are the ages.

Commentary: Polya might suggest introducing suitable notation. Let the ages of the three daughters be x, y, z. There must be a uniqely oldest daughter, so x > y >=z. Let S be the sum of their ages.

We have:
x * y * z = 72
x + y + z = S

Now we enumerate x, y, and z, looking for those with non-unique sums. Since the prime factors of 72 are (2^3) * (3^2), the job is pretty simple. The solution suggests itself shortly.

13. The surface of a distant planet is covered with water except for one small island on the planet’s equator. On this island is an airport with a fleet of identical planes. One pilot has a mission to fly around the planet along its equator and return to the island. The problem is that each plane only has enough fuel to fly a plane half way around the planet. Fortunately, each plane can be refueled by any other plane midair. Assuming that refuelings can happen instantaneously and all the planes fly at the same speed, what is the smallest number of planes needed for this mission?

Solution: Three planes. Send out all three, flying clockwise. At 45 degrees each plane has burned a quarter of its fuel. Plane 1 gives a quarter of its remaining fuel each to Plane 2 and Plane 3 and uses its remaining quarter-tank to return to base. Planes 2 and 3, now both full, continue to 90 degrees. Plane 2 gives Plane 3 one-half of its fuel and uses its remaining half-tank to return to base. Plane 3 continues to 270 degrees. When it reaches 180 degrees, Planes 1 and 2, having refueled at base (Plane 2 will have just returned by then), fly out counter-clockwise, using the same procedure.

Commentary: Polya says, first be sure you understand the problem. Abbas Raza specified, in reply to a reader’s query, that the planes may not fly suicide missions. Oddly, if they were permitted to, the answer would still be three, although two of them would plunge into the drink. But then the problem would not be interesting.

14. You find yourself in a room with three light switches. In a room upstairs stands a single lamp with a single light bulb on a table. One of the switches controls that lamp, whereas the other two switches do nothing at all. It is your task to determine which of the three switches controls the light upstairs. The catch: once you go upstairs to the room with the lamp, you may not return to the room with the switches. There is no way to see if the lamp is lit without entering the room upstairs. How do you do it?

Solution: You turn one on. You turn a second one on, wait a minute, then turn it off. Then you go upstairs and see if the bulb is off, on, or warm.

Commentary: Here the question that is so effective for Problem 12 — could you restate the problem? — can lead you astray. Introducing notation will probably also steer you wrong. The solution depends on the physical characteristics of the problem elements, and different, more abstract language may cause you to miss it. (This is why many mathematicians hate this problem.) But that’s why it’s called heuristic, as Pólya explains:

You should ask no question, make no suggestion, indiscriminately, following some rigid habit. Be prepared for various questions and suggestions and use your judgment. You are doing a hard and exciting problem; the step you are going to try next should be prompted by an attentive and open-minded consideration of the problem before you….

And if you are inclined to be a pedant and must rely on some rule learn this one: Always use your own brains first.

Update: On Problem 8, as Adam points out in the comments, Abbas Raza is right and I am wrong. The best correct explanation is here. Polya does not say, but should, that if you insist on solving problems in public you do so at your peril. I will leave up my own foolishness as a lesson in hubris.

I Am Irrational (I Think)

 Heuristic  6 Responses »
Aug 262006
 

I’m a betting man, and yesterday I was offered a betting proposition. The U.S. Bridge Championships are going on now, and the great Nickell team, which has won the event eight years running, has a bye to the semifinals. My friend Justin Lall, a bridge pro, offered me 6:1 odds on $50 on the field: in other words, he would pay off if any team but Nickell won the event.

Ordinarily I would accept happily. 6:1 is very long odds, and no matter how good Nickell is they still have to win two matches against excellent teams. Except Justin informed me that he was getting 6.5:1 on the same 50 bucks from someone else. So taking the bet gives him a freeroll: plus $25 if Nickell loses, break-even otherwise.

I refused the bet, which, from a strictly economic point of view, is irrational. If I like the odds, then I like them. Why should I care if Justin is using me to hedge his risk?

Anxiety mostly — anxiety, first, about one’s place in the dominance hierarchy. One hates to be a pawn in someone else’s game, Es to his Ich, a means to his end. A moment’s reflection will convince you of the idiocy of this attitude, on which several moral philosophies, like Kant’s and Martin Buber’s, have been erected. Regardless of who initiates the transaction, Justin is just as much a means to my own end — obtaining a bet against Nickell at favorable odds — as I am to his of laying off his risk. Hasn’t he also earned a transaction fee for having done the work of negotiating the bet in the first place and then offering it, at a small profit, to me? I regard the philosophies as foolish and atavistic yet, in this case at least, persist in the attitude. If you want an instance of the dictionary definition of “irrational,” this will serve.

Also involved is a related, slightly different form of anxiety which, for lack of a fancy psychological term, I will call shopping anxiety. Mencken defined Puritanism as the haunting fear that someone, somewhere may be happy: shopping anxiety is the haunting fear that someone, somewhere got a better deal. It is not clear to me why it should detract from someone’s pleasure in his new 56-inch plasma TV to discover that his neighbor bought the same model for $200 less. Neither is it clear why it annoyed me that Justin found a better bet than I had, especially since I hadn’t been out looking. But it did.

Finally there is the fact that Justin is a bridge pro. He knows and has played with members of the Nickell team. He is, in short, far more competent than I to evaluate the odds, and he would rather freeroll than eat the risk. Perhaps the bet isn’t as good as I thought it was. This conceivably sound reason, I am sure, influenced me far less than the stupid ones.

Nickell, as I write, has a huge deficit late in its semifinal match. Who’s sorry now?

Update: Despite a furious comeback, Nickell loses. I’m out 300 bones.

All the Best People

 Culture, Heuristic, Poetry  9 Responses »
Mar 122004
 

All rock critics like Elvis Costello because all rock critics look like Elvis Costello.
–David Lee Roth

Were you a grade-school liberal like me? Anyone who isn’t a socialist at 10 has no heart, anyone who still is at 20 has no brains. I grew up in New York’s legendarily Republican Dutchess County, of which Gore Vidal remarked, after a losing run for assemblyman, “Every four years the natives crawl out of their holes and vote for William McKinley.” Maybe so; what they don’t crawl out of their holes to do is vote for Gore Vidal. Dutchess was FDR’s home county, and he never came close to carrying it in four tries. A straw poll of my 6th grade class revealed that I was the only kid who supported McGovern. What I lacked in numbers I made up in energy, plastering McGovern posters all over the walls of the elementary school. My Nixon hatred confirmed, by junior high I knew all the Watergate players, not just the big boys like Haldeman and Ehrlichman but the whole supporting cast — McCord, Segretti, Egil Krogh, right down to Frank Wills, the security guard at the Watergate Hotel who blew the whole thing open. My chess club adjourned early one sultry night in July 1974 to tune in Nixon’s resignation speech, which I watched with undisguised, not to say lip-smacking, relish.

I understood no more of politics than my Nixonite classmates did. I hated Nixon because my parents hated Nixon and I was too young to have learned to hate my parents; I liked my parents. But I was plenty old enough to hate my classmates, and took a none-too-secret pleasure in the fact that my politics differed from theirs. They were nothing more than a way to be superior.

In high school I refused to listen to Led Zeppelin and Pink Floyd: that shit was for the heads who wore cutoff jean jackets and smoked in the parking lot. I went in instead for Devo, Talking Heads, Sex Pistols, Clash, a few deservedly forgotten groups like the Fabulous Poodles (“Mirror Star” anyone?), and of course, as Professor Lee Roth would have predicted, Elvis Costello. Later on, when my ex-stoner buddies sat me down with the headphones and forced me to listen carefully to Zep and Floyd, I was astonished to discover that it was good, really good, and that my own tastes at the time had held up spottily by comparison. The jean jacket boys were right, and I was wrong. It bothered me, as it would bother anyone. Only after several years of conscientious deprogramming could I listen to these bands without prejudice.

When I started to read poetry I stayed away from Keats and Shelley and Christina Rossetti: that shit was for the girls who liked rainbows and ponies, not that I had anything against rainbows or ponies, just the girls who liked them, who wouldn’t go out with me anyway. Even now I can’t read any Keats besides the Grecian Urn, am notoriously unfair to Shelley, and can admire one or two poems by Rossetti only from a discreet distance.

I have spilled my share of pixels here defending objective values in art. Some art is good, some bad, and confusing them is like thinking that the earth is flat or that there’s a fortune to be made in buying real estate with no money down. I am very far from recanting but I have nagging doubts. Elsewhere, discussing public and private reading, I instanced someone whose favorite song is “Desperado” because it happened to playing when he kissed a girl at the junior high school dance. The example is tendentious; in truth most “private readings” are far more subtle and insidious. You admire someone, and he plays you music, and shows you pictures, and lends you books. You admire the exhibits, but to what extent can that be disentangled from your admiration of the exhibitor, if at all? I have a taste for poems about the relationship between the abstract and the particular; on what grounds can I claim it is any more universal or important a theme than the tribulations of love or the inevitability of the grave?

I exhibit certain poems here, and convince certain readers who might not see them otherwise that they are good. But you will never share my tastes exactly unless you’re exactly like me, and God forbid. Yvor Winters, as any steady reader here knows, is my favorite critic. Do I like him on the merits, such as they are, or do I like him because, of all poetry critics, he’s most like me? David Lee Roth might know. I don’t.

(Update: Rick Coencas comments. George Wallace points out that it was Egil Krogh, not Emil as I originally had it; I would have known that in 8th grade. Eddie Thomas has some especially interesting remarks. Eloise of Spit Bull comments. Jeff Ward comments.)

Restricted Choice

 Heuristic  21 Responses »
Feb 252004
 

Eddie Thomas, a better philosopher than he is a statistician, poses the following problem. He is interviewing five candidates for two jobs. Each candidate’s chance of receiving a job offer, a priori, is 40%. After interviewing four candidates Eddie wants to offer a job to the best of the four to protect against his taking another job elsewhere. He is puzzled because it now looks like the final candidate has a 25% chance of a job offer, being one in four remaining, while his chances should be unaffected, remaining at 40%.

This is a type of restricted choice problem. The classic illustration of true restricted choice is the old game show Let’s Make a Deal. Monty Hall shows the player three doors, behind one of which is the grand prize, a Hawaiian vacation or a brand new Cadillac Eldorado. The player chooses one door. Monty then reveals another, behind which the prize is not hidden, and asks the player if he wants to switch. The player should always switch. His chance of choosing the grand prize in the first place was 1 in 3. If he switches, it is 2 in 3, because Monty’s choice of which door to open has been restricted by the choice the player already made. Many people don’t believe this even after it has been explained to them, but it’s true, and can be verified easily by experiment if you doubt it.

Here, on the other hand, because of restricted choice, the probabilities only appear to change. To receive a job offer you must be one of the best two of five candidates. Consider it from the point of view of the first four candidates. Each one has a 25% chance of receiving an offer after four interviews, and a 15% chance (.75 X .20) of receiving an offer after five interviews, for a grand total of 40%, as you would expect. So one way to look at it is that the remaining candidate must also have a probability of receiving an offer of 40% for the probabilities to add up to 200% (two job offers).

Since this will satisfy no one, least of all Eddie, I’ll try another approach. To receive an offer he has to be better than the three remaining candidates. However, his three remaining competitors are not randomly chosen; they have already failed to finish first among the first four. Choice has been restricted. Each one of the three, not having received the first offer, has a far less than 40% chance remaining of securing an offer. In fact, they now have half that chance; for they can only finish second, at best, among the five, while the fifth candidate can still finish either second or first. Therefore their remaining chance is half of their original 40%, since one of the two offers has been closed to them, or 20%, rather than 25%. The fifth candidate still has a 40% chance.

God of the Machine: all probability, all the time!

Time Won’t Tell

 Baseball, Heuristic  16 Responses »
Feb 212004
 

Five years ago, after the 1999 season, a fellow fantasy league baseball owner and I fell into an argument about Roger Clemens. Clemens was 37 years old. In 1998 he had a brilliant season with Toronto, winning the pitching triple crown — ERA, wins, and strikeouts — and his fifth Cy Young Award. In 1999, his first year with the Yankees, he slipped considerably, finishing 14-10 with an ERA higher than league average for the only time since his rookie season. His walks and hits were up, his strikeouts were down, and my friend was sure he was washed. He argued that Clemens had thrown a tremendous number of innings, that old pitchers rarely rebound from a bad season, and that loss of control, in particular, is a sign of decline. I argued that Clemens is a classic power pitcher, a type that tends to hold up very well, that his strikeout ratio was still very high, that his walks weren’t up all that much, and that his diminished effectiveness was largely traceable to giving up more hits, which is mostly luck.

Of course Clemens rebounded vigorously in 2000 and won yet another Cy Young in 2001. He turned out not be finished by a long shot, and still isn’t. Does this mean I won the argument? It does not. Had Clemens hurt his arm in 2000 and retired, would my friend have won the argument? He would not.

Chamberlain wasn’t wrong about “peace in our time” in 1938 because the history books tell us Hitler overran Europe anyway. He was wrong because his judgment of Hitler’s character, based on the available information in 1938, was foolish; because, to put it in probabilistic terms, he assigned a high probability to an event — Hitler settling for Czechloslovakia — that was in reality close to an engineering zero. He would still have been wrong if Hitler had decided to postpone the war for several years or not to fight it at all.

“Time will tell who’s right” is a staple of the barroom pedant. Of course it will do no such thing: time is deaf, blind, and especially, mute. Yet it is given voice on blogs all the time; here’s Richard Bennett in Radley Balko’s comments section: “Regarding the Iraq War, your position was what it was and history will be the judge.” It’s not an especially egregious instance, just one I happened to notice.

Now you can take this too far. If your best-laid predictions consistently fail to materialize, perhaps your analyses are not so shrewd as you think they are. You might just be missing something. Or not. But this should be an opportunity for reflection, not for keeping score.

We fumble in the twilight, arguing about an uncertain future with incomplete knowledge. Arguments over the future are simply differences over what Bayesian probability to assign the event. There is a respectable opposing school, frequentism, which holds that Bayesian probability does not exist, and that it makes no sense to speak of probabilities of unique events; but it has lost ground steadily for the last fifty years, and if it is right then most of us spend a great deal of time talking about nothing at all. Like Lord Keynes, one of the earliest of the Bayesian theorists, we are all Bayesians now.

This, for argument, is good news and bad news. The good news is that history won’t prove your opponent out. The bad news is that it won’t prove you out either. You thrash your differences out now or not at all. Then how do you know who won the argument? You don’t. Argument scores like gymnastics or diving, not football. It will never, for this reason, be a very popular American indoor sport.

Chicken Little

 Culture, Heuristic  12 Responses »
Nov 092003
 

How full of ourselves we bloggers grow:

Some might conclude from the above that, because I reject the solutions that [Steven] Den Beste and [Victor David] Hanson offer, that I’m implying that something more dire be done to “solve” this problem. I am not. Frankly, personally, I am increasingly resigned to the fact that these problems are without solution, to the point that I’m that close to simply giving up, mothballing this site, and accepting that yes, we’re watching Western Civilization self-destruct before our very eyes and there is nothing to be done about it… I’ll probably end my life in a Death Camp of Tolerance for expressing “divisive” views and making “insensitive” remarks.

Thank God for stalwart conservative bloggers! You might think that manning the barricades against the imminent fall of Western Civilization is a lonely job. You would be wrong; the barricades are crowded with Chicken Littles of all parties, although the smoke from all the shooting prevents them from seeing each other. For some of these brave soldiers Western Civ has already fallen and its revival is the consummation devoutly to be wished. The early Objectivists used to say of Atlas Shrugged, “if this book sells 50,000 copies, the culture is cooked.” Several million copies later, well, here we are.

The sky is always falling. The “new philosophy” was putting “all in doubt” in the 17th century (Donne); “Chaos and dread Night” were descending in the 18th (Pope); “the demons [of unreason] were let loose upon the land” in the 19th (Robert Bridges). Today’s featured blogger, one Porphyrogenitus, has found that it is impossible to persuade people with reason who deride reason itself. ‘Twas ever thus, dude. Derrida and Foucault are pretty small beer compared to Hume’s attack on induction, or Bishop Berkeley’s on the evidence of the senses.

Too many bloggers confuse civilization, or culture, with Zeitgeist, which is white noise. Culture does not consist, and never did, of what is taught in college, or what appears on television or in the newspapers. It is an underground stream, the product of a few dozen of the most intelligent people of each generation, and it always appears sounder retrospectively because time takes out the trash. It is opaque not only to statistical analysis but to all but the most acute critics of the time: there is too much to sort through, and it is too easy to read in the light of the pressing issues of the day. Edmund Wilson ventured in 1935 to guess which contemporary poets would survive, a fool’s errand, and came up with Edna St. Vincent Millay (OK, he was married to her) and several other people you haven’t heard of, for excellent reason. He found Frost dull and ignored Crane, Stevens, and Williams altogether. The point here isn’t that Wilson was a dummy — far from it — but that the state of the real culture, except from a very long vantage point, is extremely difficult to discern.

Is Western Civilization on the verge of destruction? I doubt it, but I don’t know, and neither do you. Ask me in a couple hundred years.

(Update: Marvin Long comments. Julie Neidlinger comments. l8r comments.)

The Linear Fallacy

 Business, Heuristic  6 Responses »
Oct 042003
 

(I had a pleasant holiday from you, dear readers, and, I trust, you from me as well. Now let’s get down to it boppers.)

Toxicologists say that the dose is the poison, and Americans could save themselves millions of dollars if they only understood what that means.

Everything on earth, from arsenic to mother’s milk, is toxic if ingested in sufficient quantity. If we graph the lifetime dose on the x-axis and the chance of resulting loathesomeness on the y (what’s with me and the graphs lately?), we wind up with the risk curve. For your quotidian poisons like cigarettes, red meat, and smog, the risk begins at zero and stays very close to it until a certain dose is reached, at which point the curve inflects and the risk begins to increase quite radically. Not all risk curves have this shape, of course. For highly toxic substances like sarin or finely-ground anthrax it inverts. The risk escalates very rapidly and then flattens at the top of the graph after a certain exposure, at which point you die.

The curve, however, is always a curve, never a straight line. Have you heard that every cigarette you smoke cuts five minutes, or eight, or ten, off your life? This is the linear fallacy in full flower. Smokers’ diseases like emphysema and lung cancer concentrate overwhelmingly in the heaviest, longest-term smokers. People who smoke for a few years have scarcely higher mortality than people who never smoke at all. (You kids bear this in mind when you’re thinking of lighting up.) The first cigarette you smoke probably does you no harm at all. The 150,000th — pack a day for twenty years — may, like W.C. Fields’ Fatal Glass of Beer, be the one that does you in. The dose is the poison.

It gets worse. An intense dose over a short period is generally far more toxic than the same dose spread out over a lifetime. Risk varies radically not only with the lifetime dose, but also with its rate, which renders extrapolation effectively impossible. Animal tests classically deal with this fact by ignoring it. Suppose you want to determine the long-term risks of swilling pistachio nuts and maraschino cherries, which contain Red Dye No. 3. Time’s a-wasting, and you don’t have 50 years to conduct your research. Instead you stuff a bunch of gerbils with a whole lot of Red Dye No. 3 over a few weeks or months and see what happens. If a few gerbils get cancer, you extrapolate, bury your reliance on the linear model in a couple of footnotes, and voilà! a new carcinogen. Politicians and journalists thunder against the unacceptable risks to maraschino cherry addicts, the Delaney Clause is invoked, Red Dye No. 3 is banned, a new, slightly less attractive red dye replaces it, and the cycle begins anew.

Good-sized industries have sprung up to exploit the linear fallacy. The EPA tells gullible homeowners to shell out a couple grand to a radon-removal outfit if their radon level in their water is more than 4 pCi/I (picocuries per liter). Turns out that exposure at that level for 20 years increases the lifetime risk of cancer by less than 1%, unless you also smoke, which bumps it up to a ghastly 3% or so. Mind you, this is not increased mortality, but increased risk of cancer. Since the lifetime risk for cancer is in the 25% range, we’re discussing, in terms of overall mortality, something less than 0.3%. Save your money and try to stay out of automobiles instead.

The asbestos boys make the radon boys look positively public-spirited. Asbestos is dangerous if you spend your life working with it; non-smoking asbestos workers have cancer rates about five times the general non-smoking population. Asbestos is essentially harmless when it’s minding its own business insulating pipes. In 1985 the British epidemiologists Doll and Peto estimated the annual lung cancer risk from such exposure to be 1 in 1,000,000; other reputable estimates are similar. Yet the Asbestos Hazard Emergency Response Act, passed in 1987, mandated asbestos removal for 45,000 public schools, many with airborne asbestos concentrations no higher than the outdoors. When you remove asbestos improperly you stir it up and increase the exposure, and since removing asbestos properly is extremely expensive the incentive to do it improperly is immense. $100 billion or so later, overall asbestos risk is probably higher than it ever was. Lead paint, Alar, DDT: the song remains the same.

So there’s good news and bad news. The bad news is you’re going to die of something. The good news is that it almost surely won’t be an exotic environmental poison.

A Brief Topology of Junk Science

 Business, Heuristic  10 Responses »
Jul 292003
 

You might be a junk scientist if:

You prophesy disaster in some remote future. It is safest to choose a date in the long run, when we’re all dead, but even the less judicious have little to worry about: by the time D-Day rolls around people will have forgotten what you said. If someone does happen to remember, you can either issue a new report revising your predictions, or, as a last gasp, maintain that you were right in general, even as your every specific prediction has been falsified. Or both.

You deal in poorly-understood, multi-causal phenomena, traditional playgrounds for the scientific crank. Cancer and climatology are especially popular.

You have trouble with extrapolation, like Ralph Hingson of the Boston University School of Health, who concluded that college drinking causes 1,400 deaths annually, by taking the total number of alcohol-related deaths among people 18-24 and multiplying by, uh, the percentage of them in college. (Note that this is supposed to establish that college drinking is especially dangerous.) Social science: it’s easy!

You are famous for work outside your field, like Paul Ehrlich, a bug man best-known for speculation on overpopulation and global cooling (yes, cooling); Barry Commoner, the cancer-biologist-cum-nuclear-testing-authority-cum-geneticist; Rachel Carson, an expert on chemicals by virtue of her master’s in marine biology; and Stephen Jay Gould, another genetics authority, trained in paleontology. The press, notwithstanding, can be relied on to refer to you as “Dr.,” “Ph.D.,” or “distinguished scientist.”

You do a lot of testifying for plaintiffs in class-action suits. Extra credit if this is how you make your living.

Spontaneous Order

 Heuristic  10 Responses »
Mar 192003
 

One of my favorite examples of the Hayekian concept of “spontaneous order” is stairway traffic. In the subway at rush hour, when people are trying to get up and down the stairs in a hurry, two lines always form between the guardrails, and they are always on the right. Any idiot who tries to plow through on the left is forced to the right by the sheer mass of the traffic. The escalators, two bodies wide, work the same way. The stationary riders stay to the right, and the walkers to the left, the passing lane, as it were.

Hayek explains far better than I ever could why such rules arise. But why this particular rule? I theorize that it’s because in America we drive on the right and pass on the left. This hypothesis is easily tested: in England or Japan or any number of other countries, where they drive on the left and pass on the right, do they walk the opposite way we do? If so, that would suggest that spontaneous rules are formed by analogy with preexisting rules. If not, it’s time for a new hypothesis. Can any readers enlighten me on this score?

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