The Fifth Round Draw of the 2016/17 Emirates FA Cup was rigged.

Bold statement (literally), although that sentence probably meant nothing to anyone who doesn’t follow English Football (re: soccer) and the FA Cup in particular.

A quick introduction to the FA Cup competition, courtesy of Wikipedia (emphasis mine):

The FA Cup, known officially as The Football Association Challenge Cup, is an annual knockout association football competition in men’s domestic English football. First played during the 1871–72 season, it is the oldest association football competition in the world. For sponsorship reasons, from 2015 through to 2018 it is also known as The Emirates FA Cup.

The competition is open to any eligible club down to Levels 10 of the English football league system – all 92 professional clubs in the Premier League and the English Football League (Levels 1 to 4), and several hundred “non-league” teams in Steps 1 to 6 of the National League System (Levels 5 to 10). The tournament consists of 12 randomly drawn rounds followed by the semi-finals and the final. Entrants are not seeded, although a system of byes based on league level ensures higher ranked teams enter in later rounds – the minimum number of games needed to win the competition ranges from six to fourteen.

In the modern era, only one non-league team has ever reached the quarter finals, and teams below Level 2 have never reached the final. As a result, as well as who wins, significant focus is given to those “minnows” (smaller teams) who progress furthest, especially if they achieve an unlikely “giant-killing” victory.

It’s no secret that when it comes to the FA Cup, “giant-killing” victories are more exciting to the average viewer, and therefore better for TV ratings. Therefore, the tournament organizers are incentivized to create as many “minnow-giant” match-ups as possible. Specifically, this means matching up teams from the top level of the English football league system (more commonly known as the English Premier League, or EPL) with teams from lower levels (2nd Tier = Championship, 3rd Tier = League One, 4th Tier = League Two, 5th Tier = National League, etc.) While match-ups in the first 12 rounds of the tournament are determined using “randomly drawn” balls, it has been shown that such live draw events can be effectively rigged by cooling or freezing certain balls.

This year’s FA Cup Fifth Round Draw provided an interesting case study to test the rigging hypothesis, because out of the 16 teams going into the Fifth Round, 8 of them were from the EPL (Tier 1), while the remaining 8 were all from lower divisions. Coincidentally, the 8 EPL teams just happened to get drawn against the 8 non-EPL teams, conveniently leading to the maximum number of 8 “minnow-giant” match-ups. This result should seem suspicious even if you are not familiar with probability theory, but to illustrate just how unlikely such a result is, I will walk through the math.

In order to calculate the probability of the aforementioned result, we first need to figure out the total number of match-ups (i.e. pairs) that can be arranged among a group of 16 teams. As with most problems in mathematics, there is more than one solution, but perhaps the most intuitive one is this: Take one of the 16 teams at random. That first team can be paired up with 15 possible other teams. After a pair is made, 14 teams will remain. Again, we take one of the 14 teams at random. This team can be paired up with 13 possible other teams. By repeating this logic, we see that there are a total of 15x13x11x9x7x5x3x2x1=2,027,025 unique pairs. It turns out that mathematicians already have a function that simplifies this exact result: the double factorial (expressed as n!!). Therefore, we can generalize that for any group of n objects, the number of unique pairings is equal to (n-1)!!

To calculate the total number of ways to draw exactly 8 “minnow-giant” match-ups, we can imagine putting all 8 of the EPL teams in a line. Since we are looking to match the EPL teams one-to-one with the non-EPL teams, the question becomes: how many different ways can we line up the non-EPL teams so that they are paired up with the EPL teams? The answer to that is simply 8x7x6x5x4x3x2x1=8!=40,320. It is important to understand why we keep the order of the EPL teams unchanged while we only change the order of the non-EPL teams; otherwise, we would be grossly over-counting!

The probability of drawing exactly 8 “minnow-giant” match-ups is therefore 40,320/2,027,025=1.99%, or just a tad under 2%! To verify this, I ran a Monte Carlo simulation involving 50,000 trials, of which 961 trials ended up with exactly 8 “minnow-giant” match-ups, or 1.92%. The below table and chart also show the theoretical probabilities of drawing n “minnow-giant” match-ups, for 0 ≤ n ≤ 8. (Bonus Question: Can you convince yourself why it’s impossible to draw an odd number of “minnow-giant” pairs among a group of 16 teams?)

But wait, it gets even better. Out of the 8 non-EPL teams, 4 teams were from the Championship (2nd Tier league), 2 teams were from League One (3rd Tier), and 2 teams were from the National League (5th Tier). Arsenal, which has been sponsored by Emirates since 2006, ended up drawing Sutton United, one of only two teams (the other being Lincoln City) from the National League (5th Tier). Now, what are the chances that the team that shares a sponsor with the competition itself ends up drawing one of the two easiest (in theory) match-ups available?

The number of ways for Arsenal to draw a National League (5th Tier) team (i.e. either Sutton United or Lincoln City), without any restrictions on how the other match-ups are drawn, is 270,270. We arrive at this number by first assuming Arsenal and Sutton United are already paired off, thus leaving 14 teams reaming. The 14 teams can be paired off in 13!!=135,135 ways without restriction. We can repeat the same reasoning for an Arsenal/Lincoln City pair. Therefore, we double 135,135 to arrive at 270,270. This yields a theoretical probability of 270,270/2,027,025=13.33% (Monte Carlo resulted in 6,620/50,000=13.24%), which is almost 1 in 6. However, this is only the probability of Arsenal drawing a 5th Tier team with no other match-up restrictions. In reality, there were already 8 “minnow-giant” match-ups drawn in the first place.

Therefore, the question becomes: what is the probability that 8 “minnow-giant” match-ups are drawn AND Arsenal draws a 5th Tier team? We already know there are 40,320 possible match-ups for the first part of the requirement. Satisfying both parts of the requirement must result in a number smaller than 40,320. Think of it like this: we start off with the fact that the 8 EPL teams are matched up one-to-one with the 8 non-EPL teams. There are 2 different ways to pair Arsenal with a 5th Tier team (since there are only 2 such teams). Of the remaining teams, there are 7!=5,040 ways to pair them off such that the EPL and non-EPL teams are still matched one-to-one. Therefore, the total number of match-ups satisfying both requirements is 2×7!=10,080. This yields a theoretical probability of 10,080/2,027,025=0.50% (Monte Carlo resulted in 250/50,000=0.50%).

In conclusion, there was only a 0.50% chance that the 2016/17 Emirates FA Cup Fifth Round Draw would lead to exactly 8 “minnow-giant” match-ups AND Arsenal drawing 1 of the 2 National League (5th Tier) teams. The fact that it happened anyway suggests that the drawing process may not have been 100% random.

As always, you can find my back up here. Please note, however, that I had to change all of the Monte Carlo formulas to values and save the file as .xlsb instead of .xlsx, as the file was way too large before (71 MB).

I would also like to give credit to the Chelsea subreddit for inspiring me to explore this topic.