Dear Uncle Colin,

This is the fifth time in the last six World Cups that Nigeria have been drawn against Argentina in the group stages. What are the odds?!

- Coincidences At FIFA? Unlikely!

Hi, CAFU, and thanks for your message!

I’m going to make some simplifying assumptions for this answer, and only partly because FIFA didn’t respond to a request for comment before press time, the major one being to assume that Argentina (who were seeded in all six tournaments) and Nigeria (who weren’t seeded in any of them) are equally likely to be in any given group.

### 1998 onwards

From 1998 onwards, 32 teams qualified for the World Cup, split into eight groups of four. That means (so far as I can tell) in each tournament, the two teams had a one-in-eight chance of meeting.

### The 1994 anomaly

The draw in 1994 was a weird one. The seedings were determined geographically, so that no two teams from the same confederation (other than Europe) could be in the same group. Two of the top seeds (guess which) were from South America; two more South American sides, Colombia and Bolivia, were in the second pot - along with Mexico (who couldn’t be drawn with USA) and three African sides (including Nigeria).

I’ve tried to get hold of a document that explains exactly how they arranged a draw that was a) guaranteed to get a valid result and b) fair - but so far, no dice.

In any event, a naive assumption would be that Nigeria had a one-in-four chance of meeting Argentina (who could only play an African side or Mexico); a properly fair draw would give slightly less, $\frac{5}{21}$, taking into account the USA/Mexico separation. Let’s go with a quarter to keep the numbers simple.

### The calculation

So, what’s the probability of Argentina and Nigeria meeting five or more times over the six tournaments?

We have to combine two distributions here: a binomial distribution for the five most recent tournaments, $\br{X \sim B\br{5,\frac{1}{8}}}$ and one for 1994 $\br{Y\sim B\br{1,\frac{1}{4}}}$.

We want $P(X+Y ≥ 5)$, which is $P(X=4 \cap Y=1) + P(X=5)$.

$P(X=4 \cap Y=1) = \br{5 \times \br{\frac{1}{8}}^4 \times \frac{7}{8}}\times \frac{1}{4}$, which is quite small - about $\frac{1}{3,750}$.

$P(X=5) = \br{\frac{1}{8}}^5$, which is smaller, about $\frac{1}{30,000}$.

Adding these gives about $\frac{1}{3360}$ (or $\frac{1}{3510}$ if you use the ‘fair’ probability), in either case about $3 \times 10^{-4}$.

That isn’t especially likely, but I wouldn’t discount coincidence as a possibility. After all, FIFA has an excellent reputation for being completely above board. Right?

- Uncle Colin