Jane Espenson, a writer on Buffy the Vampire Slayer and Battlestar Gallactica, published an article several days ago in The New Republic on the difficulty of selling sci-fi or fantasy TV shows to networks. This is despite the fact that Harry Potter books and movies have broken money-making records with every new incarnation, Heroes was the biggest new TV show of the season, and Battlestar Gallactica is the biggest show on the SciFi network and has a rabid, cultlike fanbase. While it is interesting to hear that scifi shows are still a hard sell, Espenson's article is so slight that I'm amazed it even got published. She comes to the conclusion that a prospective scifi show needs to have a hero who discovers a secret world in which he or she is recognized for having amazing and untapped talents that nobody in the real world appreciated. While I don't doubt that that's a good archetype for a sellable scifi project, it's hardly a groundbreaking realization. It's one step up from saying that it's good to have a pretty girl in a skimpy bikini starring in your show. In fact, I don't know if anyone's bothered to do it, but I'd bet that just about every show on TV can be binned into one of ten or so archetypes-- the detective show, the dysfunctional family show, the quirky group of singles (who usually live in NYC) show, the teenage soap opera, etc. etc. The superhero show is just one of the archetypes, and all Espenson is saying is that you have a better chance of selling a show if it falls into one of the standard archetypes than if it doesn't.

I was a little disappointed by the Espenson article because for whatever weird reason I'm endlessly fascinated by the economics of Hollywood. Edward Jay Epstein wrote a book called The Big Picture about how Hollywood really makes money off of movies. The Washington Post, in their review of the book, wrote:

Each week the box-office grosses rung up by the big new movies are published, and each week it is near universally assumed, reflexively and reverentially, that they represent not merely an accurate ranking of current films but also an accurate record of how much they are making for the studios that produced them. Tommyrot, says Epstein. These seemingly huge earnings are wildly misleading, as the cost of making, distributing and showing new movies almost always far exceeds what they earn in theaters. Hollywood's highly imaginative accounting practices disguise this reality, but more to the point, theaters aren't where movies make money any more.

And Epstein writes in the book:

The main task of today's studio is to collect fees for the use of the intellectual properties they control in one form or another and then to allocate those fees among the parties -- including themselves -- who create, develop, and finance the properties. It is now essentially a service organization, a dream clearinghouse rather than a dream factory. As clearinghouses, they are very different creatures from their predecessors, and this difference is as apparent from looking at their financial reporting as it is from looking at their products.

Epstein also wrote a regular column for Slate called the The Hollywood Economist where he examined different aspects of the Hollywood money-making system, talking about, among other things, why Tom Cruise might be a wacky Scientology nut, but he's a brilliant businessman, and describing what steps you have to go through if you want to make an indie film, and telling why, until recent changes to their tax codes, Germany is an awesome place to find investors for your film. Anyway, people like me who watch movies to see something good and artistic would be well-served to never lose sight of the idea that, in a very direct way, films released in theaters are a vehicle to sell action figures in America and DVDs in Europe and Asia. If artistry manages to seep through, it is more a side effect rather than an intention of the film business.

Posted at 8:20 PM | Permalink | Comments (35) | TrackBacks (0)

I promised a follow-up to my Friday post on the Monty Hall problem, so here it is. We found on Friday that, even though the three doors from which we have to choose have equal probabilities of being a winner, after Monty reveals a door, we are twice as likely to win if we switch to the remaining door. So, basically, instead of the two remaining doors having a 50/50 probability of revealing a car, they have a 67/33 probability, with the door you chose having the 33% probability, the same probability it had when you originally chose it. So we found that, counterintuitively, the probability stays the same and does not go up even after Monty reveals a goat behind one of the other doors. But can we get even more counterintuitive? Can the probability of our door go down AFTER Monty reveals another door contains a goat. It certainly doesn't seem possible. If we have three or four or five doors to choose from, and Monty essentially reduces the number of doors that could be a winner by revealing that one contains a goat, how could the probability of our door being a winner go down? In fact, that is what Jason Rosenhouse found when he analyzed a five-door Monty Hall problem in the post I linked to on Friday. In his analysis, you start out with five doors, and you choose, for example, Door 1, and Monty reveals a goat behind Door 2. You then, unwisely as it turns out, switch to Door 3. At this point, Door 3 has a 27% (4/15) chance of being a winner. If Monty then opens Door 1, the door you originally chose, and reveals a goat, it turns out that the new probability for Door 3 to be a winner is 25%. That is, the probability has gone down even though Monty has eliminated one of the possible doors. Rosenhouse's analysis, based on a probability theorem called Bayes' theorem, is complicated and hard to follow, so let's go back to the three-door problem to see a more straightforward illustration of this weird effect. Let's now assume that the game is fixed. A trusted friend of ours who works on Monty's staff tells us that the car is most definitely not behind Door 1. So we go into the game knowing that Door 1 has zero probability of winning, and Doors 2 and 3 each have a 50% probability of winning. Our best strategy in this case is to choose Door 1, forcing Monty to reveal whether Door 2 or Door 3 contains the goat. If he opens up Door 2 and shows it has a goat, we now know that Door 3 has a 100% chance of winning, so we will switch to that door, absolutely certain that we will win the car. But let's suppose we flub things and choose Door 2 initially. Monty then opens Door 1, which we know already will not be the winner. So Monty has given us no new information, meaning that the probabilities of Doors 2 and 3 are unchanged, right? Wrong. Let's look at the scenarios, remembering that we have chosen Door 2:

1. If Door 2 is the winner (1/2 the time), then Monty will reveal the goat behind Door 1 1/2 the time and the goat behind Door 3 the other 1/2 of the time.
2. If Door 3 is the winner (1/2 the time), then Monty will always reveal the goat behind Door 1.

Calculating each of the probabilities, we can rewrite the list like this:

1. Door 2 is the winner, and Monty will reveal the goat behind Door 1 1/4 of the time.
2. Door 2 is the winner, and Monty will reveal the goat behind Door 3 1/4 of the time.
3. Door 3 is the winner, and Monty will reveal the goat behind Door 1 1/2 of the time.


So, numbers 1 and 3 in the above list correspond to the situation where you choose Door 2 and Monty opens Door 1. We see that for those list items, it is twice as likely that Door 3 is the winner than that Door 2 is the winner. In other words, Door 3 has a 2/3 chance of being a winner and Door 2 has a 1/3 chance of being a winner. But remember that Door 2 initially had a 1/2 chance of being a winner. So, amazingly, the probability of Door 2 being a winner went down when Monty revealed the goat behind Door 1. The reason for this is that Monty gave us a lot of information by not choosing to reveal a goat behind Door 3. If Doors 1 and 3 both contain goats and Door 2 is the winner, then Monty will be equally likely to open either door. The fact that he chose Door 1 tells you that there is a good possibility that he avoided Door 3 because it is the winner. In effect, Monty raised the probability of Door 3 by opening Door 1. Since only Doors 2 and 3 had any probability of winning, if the probability of Door 3 goes up, the probability of Door 2 must go down. Even though we analyzed the simple 3-door problem, it will always be the case that, when the doors have unequal probabilities, Monty will lower the probability of your chosen door by opening a low probability door.

Posted at 7:00 PM | Permalink | Comments (0) | TrackBacks (0)

This post on EvolutionBlog brought back to mind one of the classic non-intuitive mathematics problems, the Monty Hall problem. The problem goes like this-- you are on a Let's Make a Deal-style game show, where you have three doors to choose from. One of the doors has a car behind it, and the other two have goats. There is an equal 1/3 probability for the car to be behind each of the doors, and the game show host, Monty Hall, knows which door contains the car. You choose Door 1. Monty Hall, wanting to maximize suspense, reveals that Door 2 contains a goat, and asks if you would like to switch to Door 3. Intuition tells you that there is nothing gained by switching doors, since each door is equally likely to contain the car. But in fact, your probability of winning by sticking with Door 1 is 1/3, while your probability of winning by switching to Door 3 is 2/3. How can that be? If each of the doors started off with equal probability of being a winner, how can the doors suddenly change their probabilities mid-game? There are a couple ways to view the situation. Probably the most straightforward is that your door has a probability of winning of 1/3, while the other two doors have a combined probability of winning of 2/3. When Monty reveals that a goat is behind Door 2, that door suddenly has a probability of winning of 0, meaning that now Door 3 has a probability of winning of 2/3. Another way to look at it is in terms of information. Monty knows which door contains the car, and he can share that information with you in various ways. When you choose a door, you restrict Monty from opening the door, so you restrict him from revealing any information (directly, at least) about the door you have chosen. So the information he can give you is about the other two doors. To better illustrate the situation, imagine that there are four doors, each with a 1/4 probability of containing the car. But before you choose a door, Monty tells you that Doors 1 and 2 both contain goats. We are assuming that Monty is always honest with us, so what he has done is told us that Doors 3 and 4 now have a 1/2 probability of containing the car. But let's say that we know Monty will never tell us anything about Door 4. All we know is that at the beginning of the game each of the doors has an equal probability of being a winner. So, if he tells us before we choose that he knows that, among Doors 1, 2, and 3, Doors 1 and 2 contain goats, we know that Door 3 now has a 3/4 probability of containing the car. This is because he will never reveal anything about Door 4 to us, but the fact that he DIDN'T tell us that Door 3 contains a goat tells us volumes. Since we assume that Monty is randomly choosing which goat doors to reveal to us, there is very low probability that Monty would randomly not tell us that Door 3 contains a goat if it did, but much better probability that Monty would avoid telling us about Door 3 because it contains a car. Once again, he will never say anthing about the door we chose, Door 4 in this case. If you're thoroughly confused by my analysis, you can fairly easily go through the possibilities in the 3-door problem and convince yourself that you're better off switching. Assume you choose Door 1, which is a winner 1/3 of the time, then the possibilities are:

1. If Door 1 is a winner (1/3 of the time), then Monty will reveal the goat behind Door 2 1/2 the time and the goat behind Door 3 the other 1/2 of the time.
2. If Door 2 is a winner (1/3 of the time), then Monty will always reveal the goat behind Door 3.
3. If Door 3 is a winner (1/3 of the time), then Monty will always reveal the goat behind Door 2.

We can calculate the various probabilities, and rewrite the list like this:

1. Door 1 will be a winner and Monty will reveal Door 2 1/6 of the time.
2. Door 1 will be a winner and Monty will reveal Door 3 1/6 of the time.
3. Door 2 will be a winner and Monty will reveal Door 3 1/3 of the time.
4. Door 3 will be a winner and Monty will reveal Door 2 1/3 of the time.

So, if you've chosen Door 1 and Monty chooses Door 2 (possibilities 1 and 4 in the above list), then we can see that there is twice the probability that Door 3 will be the winner than that Door 1 will be the winner. Similarly, if you choose Door 1 and Monty chooses Door 3 (possibilities 2 and 3 in the list), you will have twice the chance of winning by switching to Door 2. More on this tomorrow.

Posted at 8:27 AM | Permalink | Comments (23) | TrackBacks (0)