The World Cup from a Quantum Perspective

Two weeks into the Football World Cup and the group stages are over: 16 teams have gone home, leaving the top 16 teams to contend the knock-out stages. Those fans who enjoy a bet will be poring over the odds in search of a bargain—a mis-calculation on the part of the bookmakers. Is now the time to back Ronaldo for the golden boot, whilst Harry Kane dominates the headlines and sports betting markets? Will the hosts Russia continue to defy lowly pre-tournament expectations and make the semi-finals? Are France about to emerge from an unconvincing start to the tournament and blossom as clear front-runners?

But, whilst for most the sports betting markets may lead to the enhanced enjoyment of the tournament that a bet can bring (as well as the possibility of making a little money), for others they represent a window into the fascinating world of sports statistics. A fascination that can be captured by the simple question: how do they set the odds?

Suppose that a bookmaker has in their possession a list of outcome probabilities for matches between each pair of national football teams in the world and wants to predict the overall winner. There are 32768 possible ways for the tournament knock-out rounds to pan-out—a large, but not insurmountable number of iterations by modern computational standards.

However, if the bookmaker instead considers the tennis grand-slams, with 128 competitors in the first round, then there are a colossal 1.7 × 1038 permutations. Indeed, in a knock-out format there are 2n-1 permutations, where n is the number of entrants. And for those of a certain mindset, this exponentially growing space immediately raises the question of whether a quantum algorithm can yield a speed-up for the related prediction tasks.

A Tiny Cup

The immediate question which we want to answer here is, perhaps, who will win the World Cup. We will walk through the idea on the blackboard first, and then implement it as a quantum algorithm—which, hopefully, will give some insight into how and where quantum computers can outperform classical ones, for this particular way of answering the question.

Let us take a toy setup with four teams A, B, C and D;
the knockout stage starts with a game A vs. B, and C vs. D.
Whoever wins each game will play against each other, so here we have four possible final games: A vs. C, A vs. D, B vs. C, or B vs. D.
Let’s denote by p(X, Y) the probability that X wins when playing against Y.

The likelihood of A winning the cup is then simply given by

p(A, B) × ( p(C, D) × p(A, C) + p(D, C) × p(A, D) ),

i.e. the probability that A wins against B, times the probabilities of A winning against C in case C won against D, plus the probability of A winning against D in case D won.

How can we obtain the same quantity with a quantum algorithm?

First, we set up our Hilbert space so that it can represent all possible Cup scenarios.
Since we have four teams, we need a four-dimensional quantum system as our smallest storage unit—we commonly call those qudits as generalizations of a qubit, which having dimension 2 would be fit to store two teams only (we can always “embed” a qudit into a few qubits of the same dimension.

Remember: k qubits have dimension 2k, so we could also store the qudit as two qubits).
If we write |A\rangle, this simply stands for a qudit representing team A; if we write |A\rangle |B\rangle, then we have a state representing two teams.

To represent a full knockout tree, we follow the same logic: Take four qudits for the initial draw; add two qudits for the winners of the first two matches, and one qudit for the final winner.

For instance, one possible knockout scenario would be

|\text{Game 1}\rangle = \underbrace{|A\rangle |B\rangle |C\rangle |D\rangle}_\text{Initial Draw} \ \underbrace{|A\rangle |D\rangle}_\text{Finals} \ |D\rangle.

The probability associated with Game 1 is then precisely p(A, B) × p(D, C) × p(D, A).

Here is where quantum computing comes in.

Starting from an initial state |A\rangle |B\rangle |C\rangle |D\rangle, we create two new slots in a superposition over all possible match outcomes, weighted by the square-root of their probabilities (which we call q instead of p):

\begin{aligned} |\text{Step 1}\rangle = |A\rangle |B\rangle |C\rangle |D\rangle \big(\ \ &\text{q(A, B)q(C, D)} \,|A\rangle\ |C\rangle +\\ &\text{q(A, B)q(D, C)} \,|A\rangle\ |D\rangle +\\ &\text{q(B, A)q(C, D)} \,|B\rangle\ |C\rangle +\\ &\text{q(B, A)q(D, C)} \,|B\rangle\ |D\rangle\ \big). \end{aligned}

For the final round, we perform the same operation on those two last slots; e.g. we would map |A\rangle |C\rangle to a state |A\rangle |C\rangle ( q(A, C) |A\rangle + q(C, A) |C\rangle ). The final state is thus a superposition over eight possible weighted games (as we would expect).

So you can tell me who wins the World Cup?

Yes. Or well, probably. We find out by measuring the rightmost qudit.
As we know, the probability of obtaining a certain measurement outcome, say A, will then be determined by the square of the weights in front of the measured state; since we put in the square-roots initially we recover the original probabilities. Neat!

And since there are two possible game trees that lead to a victory of A, we have to sum them up—and we get precisely the probability we calculated by hand above. This means the team that is most likely to win will be the most likely measurement outcome.

So what about the World Cup? We have 16 teams; one team can thus be stored in four qubits. The knockout tree has 31 vertices, and a naive implementation can be done on a quantum computer with 124 qubits. Of course we are only a bit naive, so we can simulate this quantum computer on a classical one and obtain the following winning probabilities:

0.194 Brazil
0.168 Spain
0.119 France
0.092 Belgium
0.082 Argentina
0.075 England
0.049 Croatia
0.041 Colombia
0.04 Portugal
0.032 Uruguay
0.031 Russia
0.022 Switzerland
0.019 Denmark
0.018 Sweden
0.012 Mexico
0.006 Japan

It is worth noting that all operations we described can be implemented efficiently with a quantum computer, and the number of required qubits is quite small; for the four teams, we could get away with seven qudits, or fourteen qubits (and we could even save some, by ignoring the first four qudits which are always the same).

So for this particular algorithm there is an exponential speedup over its non-probabilistic classical counterpart: as mentioned, one would have to iterate over all trees; tedious for the World Cup, practically impossible for tennis. However…

Classical vs. Quantum

Does using a quantum algorithm give us a speedup for this task? Here, the answer is no; one could obtain similar results in comparable time using probabilistic methods, for instance, by doing Monte Carlo sampling.

But there are several interesting related questions that we could ask for which there might be a quantum advantage.

For some team A, we can easily create a state that has all game trees in superposition that lead to a victory of A—even weighting them using their respective probabilities.
Given this state as a resource, we can think of questions like “which game tree is most likely, given that we fix A and B as semifinalists”, or “which team should A play in the knockout stages to maximize the probability that B wins the tournament”.

Or, more controversially: can we optimize the winning chances for some team by rearranging the initial draw?

Some questions like these lend themselves to applying Grover search, for which there is a known speedup over classical computers. To inquire deeper into the utility of quantum algorithms, we need to invent the right kind of question to ask of this state.

Let us think of one more toy example. Being part physicists, we assume cows are spheres—so we might as well also assume that if A is likely to win a match against B, it always wins—even if the probability is only 51%. Let’s call this exciting sport “deterministic football”. For a set of teams playing a tournament of deterministic football, does there exist a winning initial draw for every team?

This becomes an especially interesting question in cases where there is a non-trivial cyclic relation between the teams’ abilities, a simple example being: A always beats B, B always beats C, and C always beats A. For example, if this problem turns out to be NP-hard, then it would be reasonable to expect that the quadratic improvement achieved by quantum search is the best we can hope for in using a quantum algorithm for the task of finding a winning initial draw for a chosen team—at least for deterministic football (phew).

To the finals and beyond

World Cup time is an exciting time: whatever the question, we are essentially dealing with binary trees, and making predictions can be translated into finding partitions or cuts that satisfy certain properties defined through a function of the edge weights (here the pairwise winning probabilities). We hope this quantum take on classical bookmaking might point us in the direction of new and interesting applications for quantum algorithms.

Hopefully a good bargain!

(Written with Steven Herbert and Sathyawageeswar Subramanian)

4 thoughts on “The World Cup from a Quantum Perspective

    • Indeed, this is based on the sixteen teams that were part of the initial draw. For the final eight, the statistics are:
      0.244 Brazil
      0.219 France
      0.147 England
      0.113 Belgium
      0.099 Croatia
      0.080 Russia
      0.060 Uruguay
      0.037 Sweden

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