Five Thirty Eight has been discussing the World Cup and their predictions based on what I can only assume is a fancy-pants Bayesian statistical model (done in Excel, natch).

A few days ago, Nate Silver his own self wrote a post on this topic. In trying to give it a counter-intuitive hook, the post got a little too cute for its own good. The headline sums up what’s wrong, but I understand that headlines are often written separately from article proper, so we might forgive statistical silliness up top. But the same statistical silliness makes an appearance in the body. Silver writes:

For instance, the probability of correctly identifying the winners in each of the first four knockout matches — Brazil over Chile, Colombia over Uruguay, the Netherlands over Mexico and Costa Rica over Greece — was about 23 percent, or one chance in 4.3. And the chance of going 12 for 12, as the FiveThirtyEight favorites have done so far, is just one in 75.

It’s an upset, in other words, when all the favorites prevail. On average, we’d have expected three or four upsets through this point in the knockout round.

How interesting! They’re favorites, but it’s still, somehow, a (huge) upset when they prevail! What gives?

What gives is that Silver is comparing a single outcome – all favorites prevail – to every other possible outcomes. With 12 games (8 in the first round, 4 in the second), there are possible outcomes.

(As often happens to me with counting problems like this, I came up with that answer fairly quickly, then immediately doubted the assumptions that led me to this answer were correct. Specifically, I was worried that I was somehow counting second round outcomes that were ruled out by first round outcomes (e.g., Brazil winning in the second round after losing in the first round). I’m pretty sure is right, though, and we can spell out a slightly simplified version to see why.

Suppose the semi-finals feature, say, Germany vs Brazil in one game and Argentina vs Holland in the other. Call this ’round one’ and call the final ’round two.’ Using my logic, there should be possible outcomes in these three games, and indeed there are. Listing all possible combinations of winners: (1) Brazil, Argentina, Brazil; (2) Brazil, Argentina, Argentina; (3) Brazil, Holland, Brazil; (4) Brazil, Holland, Holland; (5) Germany, Argentina, Germany; (6) Germany, Argentina, Argentina; (7) Germany, Holland, Germany; (8) Germany, Holland, Holland.

We could, in theory, list out all the possibilities for the rounds with 8 and 4 games – the real rounds one and two – to get the number of possible outcomes stated above, .)

Okay, so, when Silver says that the teams that have won so far were all favorites, he was saying that the single outcome we’ve observed is the single most probable outcome. If it weren’t the most probably single outcome, then at least one team that won wouldn’t have been a favorite.

On the other hand, when he says that it’s an upset that all the favorites have won, he’s saying that the this single most probable outcome is less probable than the sum of the probabilities of all of the other possible outcomes.

Given that there are 4095 other possible outcomes, the favorites would have to have been favored to an absurd degree for the sum total of all other outcomes to be less probable.

The point being that it’s not particularly interesting to compare the single outcome of all favorites winning to every possible other outcome. It gets you a catchy headline, I guess, but it doesn’t provide any useful insight into how the tournament is playing out.