The Post I've Been Wanting to Write

I've been trying to figure out the right angle on a post for a few weeks, one that talks about the idea that the real problem with the American economy is that we aren't consuming enough. Until we kick-start consumptions, the idea goes, the economy will remain sluggish. (The idea is poppycock, by the way.)

Anyway, I've been pre-empted by Kevin Williamson, who has explicated it better than I could have.

You Didn't Ask Me But I'm Going To Tell You Anyway

I have what I guess is a pretty un-nuanced, old-fashioned view of Don't Ask Don't Tell. Basically, the purpose of the military is to bring violence upon our enemies. It's not a tool of social policy. If having gays serve openly in the military enhances its ability to do violence, then I'm all for it. If not having them serve openly works better, then I'm all for that. If DADT is the best policy, then... well, you get the picture.

It's possible to take that "best for the military" thing too far, of course. I wouldn't support a policy of executing every tenth man in a platoon that screws up, even if that could be proved, via scientific study, to enhance military effectiveness. But our policy with regard to gays in the military has never been anything like that draconian. If discovered, they are discharged; that's all.

Being gay might be a personal choice, or it might be inherent. There's a big debate about this, and in some cases it might matter which is right. (Personally I suspect it's a bit of both, in different proportions for different people.) In the case of military policy, I don't see how it matters one bit. If it's a choice, then that's one choice that's denied to you, like wearing shorts and sandals on parade, or wearing a nosering and a mohawk. If it's inherent (which is a broader term than "genetic" - it might be inherent due to upbringing), then so are lots of other things that keep people from serving in the military. If you have flat feet, you cannot serve. Why? Because it detracts from military effectiveness.

Which of these policies actually is best for the military is not a question I am at all qualified to answer. But it should be the only question being asked. Whether it enhances social justice or something is irrelevant.

A Death Tax Tale

Once upon a time there was a company. This company was founded during the Great Depression, and survived that turmoil and the World War that followed it. In the 1960s it was bought by a pair of entrepreneurs who thought they could make it even better. And they did. Under their ownership it prospered. Eventually one of these owners died and his family's share was bought back by the company. That was costly, but the company survived it. The company survived other hardships as well: legal troubles, financial market struggles, moves, new products.

The company has been run by this family of entrepreneurs for nearly fifty years now. And after all of this history, it has been forced to sell itself. Why? Has the company been poorly run? Is it drowning in debt and has to find a partner to help pay its creditors?

No. The company generates millions of dollars in profits a year. It has no debt.

What is happening is simple: the second of the pair who acquired the company is aging, and so is his wife. At some point - maybe tomorrow, maybe ten years from now - they will be dead, and the estate tax will kick in. And if that happened, the company would basically be out of business, because while it is a cash cow, it doesn't have the funds to buy back enough stock to allow the family to pay those taxes. Furthermore, such buybacks would radically change the capital structure of the company, with the family possibly losing control. And all for no good reason.

Naturally, the company wants to avoid this outcome. So it is being acquired. The family and the minority shareholders get their payoffs now, and the company will be absorbed by its new partner. While the company has been lucky - the partner has pledged to run the company as a stand-alone operation, and not mess with its corporate operations or culture - they will install a new CEO, and inevitably some changes will be coming. Employees are nervous, and rightly so.

This is the sort of unintended, and injurious, consequence that Republicans talk about when they attack the "death tax." To make matters worse, in this case at least, the government won't even collect the tax. The acquisition will trigger some capital-gains taxable events (I assume - it may even manage to avoid those depending on the specifics of the deal), but those will be timed to coincide with current lower capital-gains tax rates. The 55% estate tax that will go into effect in 2011 (barring new legislation to prevent the sunsetting of the 2001 Bush tax cuts) will collect zero revenues from the company. It's hard to find a purer example of the Laffer curve at work.

Stupid Koran Tricks

So there's this dimwit (whose name I won't honor by putting in print) in Florida who wants to burn a Koran tomorrow. Stupid.

But what's nearly as stupid is the press coverage of it. Years ago, networks figured out that when some drunken sports fan jumps onto the field to make an ass of himself, the thing to do is not to broadcast his antics to millions of viewers across the nation. You studiously don't film it, talk about something else for a few minutes while security takes care of it, then get back to the game.

I suppose it's harder to do that here. This Florida pastor isn't doing anything illegal, so the police won't be hogtying him and marching him off to the drunk tank. And he can make as much noise as he wants to on the Internet and wherever else he can. But why help him by giving him free publicity? I'll do my part by not providing a name or link, although that's spitting in the wind considering that his name, church, and probably blood type are already in all media outlets.

It's kind of sad when the sports media is wiser than the rest of the media.

On Same-Sex Marriage

I've written before that I find arguments both for and against same-sex marriage to be about equally inane. That leaves me, net-net, against codifying the practice into law, because as a conservative I favor a millennia-old tradition over a newfangled idea unless significant evidence can persuade me otherwise.

National Review's recent editorial "The Case For Marriage" is the best argument I've read to date bolstering my default position. Is there a similarly persuasive case in favor of same-sex marriage? If so, I haven't read it - and as I've said, it would have to be even more persuasive to swing me against tradition.

Why is Labor Day a "patriotic" holiday?

On Independence Day we celebrate our founding as a nation - if we're going to have a flag-waving holiday, that's the logical choice. On Memorial Day we celebrate our fallen warriors - so that one makes sense, too. We have two days on which we celebrate specific American heroes - Presidents' Day and Martin Luther King Day - but those evince at most mild patriotic displays. Thanksgiving is traditionally American, but centers more around family than country. New Year's Day is a strictly secular, global holiday; finally, Easter and Christmas are religious, global holidays.

Other than minor holidays like Veteran's Day and Columbus Day, which most of us don't get anyway, that leaves only Labor Day. In my experience, Labor Day is third behind only Independence Day and Memorial Day in patriotic displays. In my town, for example, we have a fireworks display. I see kids wearing American flag clothing in the park.

But Labor Day is not especially American, and the part of it that is has not much to do with founding American values. Labor Day commemorates the bloody breaking of a strike in 1894. Its associations are with the International Labor movement and it is essentially the American version of May Day, which is the international (not American) socialist holiday (and also commemorates a massacre during a 19th-century strike). Labor Day was created as a conciliatory gesture toward the growing labor movement.

So I don't quite understand the fireworks and the flags. USA! USA! We're unionized!

Math Corner

John Derbyshire has another good one up in his August Diary. It's another probability question, this time about cards:

I have an ordinary deck of 52 playing cards. I shuffle it thoroughly. What is the probability that not one card is in its original position?

As often happens, the way to approach this is to look at the conjugate question: what is the probability that at least one card is in its original position? Suppose, for example, that one card is in its original position - we'll call this a stationary card, because it didn't move after shuffling. There are C(52,1) ways to pick this one card, and the other 51 cards can be arranged arbitrarily, so there are 51! ways to arrange them. However, we've done some double-counting here, because some of those 51! include arrangements that have a second stationary card.

It's worth going over this point in some detail, because this is an argument we're going to come back to again. To see what's happening here, it's useful to reduce the number of cards. So let's say there are only 4 cards, numbered 1, 2, 3 and 4. There are, of course, 4! = 24 possible arrangements of these cards. Let's look at all the arrangements with the "1" card in its original position:

1,2,3,4
1,2,4,3
1,3,2,4
1,3,4,2
1,4,2,3
1,4,3,2

Also, let's look at all the arrangements with the "2" card in its original position:

1,2,3,4
1,2,4,3
3,2,1,4
3,2,4,1
4,2,1,3
4,2,3,1

Notice something? The lists aren't distinct. Those first two entries are common to both lists. What we've done is double-count arrangements that contain at least two stationary cards.

We can subtract those back out pretty easily: there are C(52,2) ways to pick two cards, and then 50! arrangements of the other 50 cards. So we'll subtract C(52,2)50! arrangements.

But wait! We've removed too much, because both of the previous sets included arrangements that had at least three stationary cards. Going back to the 4-card example, take a look at the arrangement "1,2,3,4". We originally double-counted it, but then we double-removed it. So to count those arrangements we have to add back in arrangements with at least three stationary cards, and that's C(52,3)49!.

It should be no surprise at this point that this pattern continues. Now we've double counted arrangements with four stationary cards, so we subtract C(52,4)48! of those, at which point we need to add back the ones with five - there are C(52,5)47! of them - and so on. We end up with this many arrangements:

C(52,1)51! - C(52,2)50! + C(52,3)49! - C(52,4)48! + ... + C(52,51)1! - C(52,52)0!

There are 52! total arrangements of 52 cards, so to get the probability, we divide the above expression by 52!. This simplifies down to:

1/1! - 1/2! + 1/3! - 1/4! + ... + 1/51! - 1/52!

But this is the probability of having at least one stationary card, and we wanted the probability of having zero stationary cards, which is:

1 - 1/1! + 1/2! - 1/3! + 1/4! + ... - 1/51! + 1/52!

Reasoning the same way you can see that if you had n cards, the probability would be the first n+1 terms of this series. (It's an interesting fact that the above expression is the first n+1 terms of something called the Taylor series for 1/e, where e is the base of natural logarithms that you may dimly remember from high school or college. For more than 8 or so cards, the difference between the actual probability and 1/e is very small: less than 0.01%.)

Let's look at Derbyshire's second (related) problem:

I have a deck of n cards, numbered from 1 to n. I shuffle the deck thoroughly. Then I turn the cards over one by one. If the k-th card I turn over bears the number k, call that a "match." What is the probability that after going through the whole deck I shall have tallied m matches, where m is some number in the range from zero to n?

Based on the work we did before, this isn't hard at all.

Let's write the number of arrangements of n cards that have m matches (what I earlier called "stationary cards") as A(n,m). Then the probability p(n,m) that Derbyshire seeks is just A(n,m)/n!. Furthermore, we can reason about A(n,m) as follows: Suppose we select m cards and call them the matches. There are C(n,m) ways to make this selection. For each selection, there are A(n-m,0) arrangements of the remaining cards that have no matches. Any combination of a selection of m matches along with an arrangement of n-m cards that have no matches is equivalent to an arrangement of n cards with m matches. So A(n,m) = A(n-m,0)C(n,m).

Furthermore, since p(n,m) = A(n,m)/n!, we can write p(n,m) = A(n-m,0)C(n,m)/n! = p(n-m,0)C(n,m)(n-m)!/n! = p(n-m,0)/m!. Since our solution to the first problem gave us p(n,0) for every n, we now know how to calculate p(n,m) for every n and m.