A World Cup conundrum

My friends and I were discussing England’s world cup fortunes at lunch the other day, and there was a strong feeling that our boys would have no chance against France, should that end up being the final. I wanted to check the odds for this, but there is of course not a market since the teams in the final are not yet confirmed. I had a feeling that you could piece it together from existing markets, and so I present to you the following problem.

Note: After trying this out myself I asked Claude Fable, and it got it very wrong!

A problem

Polymarket gives world cup winner probabilities for France, England, Spain and Argentina as

P(F) = 0.39, \quad P(E) = 0.23, \quad P(S) = 0.21, \quad P(A) = 0.17,

with semifinals France v Spain and England v Argentina priced at

P(F \text{ beats } S) = 0.60, \qquad P(E \text{ beats } A) = 0.55.

Estimate the probability that France beat England in the final.

A solution

Note: this reflects the process of how I personally tried to work it out, so there is definitely a more simple/concise way to write it!

Let $m_{fe}, m_{fa}, m_{se}, m_{sa}$ denote the probabilities that France beat England, France beat Argentina, Spain beat England, and Spain beat Argentina in a potential final. Each team wins the tournament by winning its semifinal and then beating whichever opponent emerges from the other semifinal, which can be either of the teams in the other side of the bracket. So we can write, for each team, the probability of a tournament win as follows:

0.6\bigl(0.55m_{fe} + 0.45m_{fa}\bigr) = P(F) = 0.39 \tag{F}

0.55\bigl(0.6(1-m_{fe}) + 0.4(1-m_{se})\bigr) = P(E) = 0.23 \tag{E}

0.4\bigl(0.55m_{se} + 0.45m_{sa}\bigr)= P(S) = 0.21 \tag{S}

0.45\bigl(0.6(1-m_{fa}) + 0.4(1-m_{sa})\bigr) = P(A) = 0.17 \tag{A}

We can try to solve this set of simultaneous equations by rewriting as $A\mathbf{m} = \mathbf{b}$ , with $\mathbf{m} = (m_{fe}, m_{fa}, m_{se}, m_{sa})^{\top}$ and constants moved to the right:

\begin{pmatrix} 0.33 & 0.27 & 0 & 0\\ -0.33 & 0 & -0.22 & 0\\ 0 & 0 & 0.22 & 0.18\\ 0 & -0.27 & 0 & -0.18 \end{pmatrix} \begin{pmatrix} m_{fe}\\ m_{fa}\\ m_{se}\\ m_{sa}\end{pmatrix}=\begin{pmatrix} 0.39\\ -0.32\\ 0.21\\ -0.28 \end{pmatrix}

where $0.33 = 0.6 \times 0.55$ is the probability the final is France v England, etc. Unfortunately there is a bit of an issue here though, because if you look closely, each matchup probability appears once as $m$ (winner’s equation) and once as $1-m$ (loser’s), so every column of $A$ has one $+c$ and one $-c$ , and so the rows sum to zero, which implies they are not linearly independent, and so there is not one unique solution.

Oh dear! This is where I got a bit stuck. There are are actually a range of solutions to this problem which are consistent with the probabilities. This kind of makes sense if you think about it. Say France get to the final, they could be rubbish against England (🤞) and great against Argentina, and that would result in the same probability of them winning the tournament as if they were just quite good against both. So how do we move forward?

We have to introduce an additional assumption to the model to eliminate one of the unknowns. We can do this in a number of different ways, but to me it makes sense to assume France’s relative difficulty against England versus Argentina matches the market’s view of the England–Argentina matchup.¹ We can write this as

\frac{m_{fa}}{1-m_{fa}} = \frac{0.55}{0.45}\cdot\frac{m_{fe}}{1-m_{fe}}

If you sub this into the France equation from above, you can work out $m_{fe}=0.63.$ We don’t now even need to use the rest of the equations unless we want to work out the other probabilities, in which case we can just sub in our answer.

This is related to the assumption in the Bradley-Terry model ↩︎

A World Cup conundrum

2026/07/14

A problem

A solution