p = 0.5

In the middle of a fascinating article on South Africa’s preparations for the World Cup, R. W. Johnson makes the following offhand remark:

Any minute now the usual groaning will be heard from teams which claim that they, uniquely, have been drawn in a ‘group of death’. What is the point, one might ask, in groaning about a random draw? Well, the trouble starts there, for the draw is not entirely random. In practice, seven teams are seeded, according to how well they’ve been doing in international matches, along with an eighth team, the host nation, whose passage into the second round is thus made easier – on paper. The draw depends on which balls rise to the top of the jar and thus get plucked out first; but it’s rumoured that certain balls get heated in an oven before a draw, thus guaranteeing that they will bubble to the top. The weakest two teams aside from South Africa and North Korea are South Korea and New Zealand. The odds are, of course, heavily against any two or more of these bottom four finding themselves in the same group. If they do, we will have to be deeply suspicious of the draw.

This got me wondering. What is the probability that the bottom four teams will actually end up in different groups?

Given the rules as stated above, eight of the teams (including South Africa) start in eight different groups. There are 24 slots remaining. Now let’s assign the next three low-ranking teams. The first has a 21/24 chance of being in one of the seven groups that does not have South Africa; the next has a 18/23 chance of being in one of the six remaining groups, and the next has a 15/22 chance of being in one of the five remaining. Combining these, the probability that the bottom four teams are in four different groups is 1-(21/24)*(18/23)*(15/22) = 0.53. (Unless I did the calculation wrong. Such things happen.)

So, no, I don’t think that if two of these teams happen to find themselves in the same group, that “we will have to be deeply suspicious of the draw.”

P.S. The 53% event happened: the four bottom-ranked teams are in different brackets. So we can breathe a sigh of relief.

8 thoughts on “p = 0.5

  1. Actually, there was a 3/8 chance that North Korea, South Korea, or New Zealand would have been paired in the same group with South Africa (Group A, which always has the host). There were 4 pots, with a group being made up of a draw from each pot. Pot 1 had SA and the 7 best teams. Pot 2 had Asia, North America, and Oceania. North Korea, South Korea, and New Zeland where all in this pot, so they had 0% of being grouped together, but a 3/8 chance of being with SA.

    Pot 3 had Africa and South America and Pot 4 the remaining teams from Europe. They create these pots so that there is not a group with four European teams or groups with more than one member from same region (except Europe, who gets 13 slots, forcing some groups to have two teams from Europe).

    If you use Nate Silver's Rankings (http://espn.go.com/soccer/spi/rankings), South Africa has a rough draw. They got the 2nd best team from Pot 2 (Mexico), the third best team from Pot 3 (Uruguay), and the 2nd best team from Pot 4 (France). Considering the average within-pot ranking should be 4.5 for each group, South Africa got the short end of the straw.

  2. Would people be more accepting of the draw if it was done according to the last time a country won the world cup, with the host automtically seeded eighth, and non-world cup winners seeded randomly?

    So Italy would be seeded #1, then France #2, then I think Brazil and so on. Would this seem fairer than the current system?

  3. Ed, Brazil won in 2002 (not France).
    But Rutter is right.
    And using Nate Silver's Ranking you will see that Brazil's group is the death group.
    As a Brazilian, I am not very happy, but still confident in our team – we are the best :).

  4. Did you see the several good articles Nate Silver wrote about the World Cup draw, a couple on 538 and a couple on ESPN I think? Good stuff.

  5. You did get the calculations wrong. You inserted "1-" into the calculations for no reason.
    Hence the event that happened was actually a 47% event, rather than a 53% event.

  6. No, Prof. Gelman. It's not at all a picky question of decimal places. Your reasoning is very badly flawed.

    You correctly go through the parlay where all the teams are in different groups. That probability has 47%. But, you get confused about whether you're calculating the different-groups probability or its opposite. So, instead of the correct answer of 47%, you took 1-47% to get the wrong answer of 53%.

    With the correct number of 47% instead of 53%, your case would have been even stronger.

    Paul Epstein

Comments are closed.