Estimating the Average Game Length

Unfortunately, not many pro-Quidditch matches are described in the books, so this is quite hard to estimate. On the one hand, the only game that is described in its entirety is the 1994 Quidditch World Cup final, and it was extremely short: only 15 minutes or so (4]); On the other hand, the longest game ever lasted 3 months (5]). So the average is somewhere between 15 minutes, and 130,000 minutes… great.

For the sake of our simulations, then, I will assume the average game lasts 90 minutes (assuming Rowling has soccer or rugby in mind). I will also assume that the length of a Quidditch game is exponentially distributed, since the probability of catching the Snitch remains constant (as opposed to it getting easier as the game progresses) – the Snitch’s ability to escape capture does not change (magic) and players are substituted throughout the game, so let’s assume an unchanging level of vigilance.

Together, the distribution of game lengths looks something like this:

This makes the 3 month long game very unlikely ( \(Pr(time>130,000)<10^{-600}\) ), but the 15 minute game rather probable ( \(Pr(time<15)=0.15\) ).

Estimating the Rate of Goals Scored

Throughout the books, the final scores of several Quidditch games are mentioned, but since we don’t know their lengths, it is hard to estimate the rate of goals-per-minute (GPM). Going by the 15 minute world cup, in which the final score was 170-160, or 170-10 if only counting goal points, we can see that there were 18 goals in 15 minutes, or an average of 9 goals per 15 minutes per team, or a GPM of 0.6 (or 0.07 for Bulgaria, and 1.13 for Ireland). Though this is only one game, and a very short one at that, these numbers seem close to the averages seen in basketball, where GPM are around 40/48 = 0.83 (6). This is also in line with the game’s depiction in the films as a fast sport, with many goals scored quickly and in succession.

I will also assume that this rate is constant and unchanging throughout the game, regardless of the score (or fatigue of the players, see above), and so will assume GPM has a poisson distribution, with an average of 0.8, making the GPM distribution look something like this:

R Setup

set.seed(1994)

sim_quidditch <- function(mean_game_durA, mean_game_durB, GPM_A, GPM_B,
                          snitch_points = 150, goal_points = 10,
                          n_sims = 1e6) {

  # Sim game durations and snitch Points ------------------------------------
  mean_game_dur_rateA <- 1/mean_game_durA
  mean_game_dur_rateB <- 1/mean_game_durB

  game_dursA <- rexp(n_sims, rate = mean_game_dur_rateA)
  game_dursB <- rexp(n_sims, rate = mean_game_dur_rateB)

  # Final game durations
  game_durs <- ifelse(game_dursA < game_dursB, game_dursA, game_dursB)

  # Who got the snitch (and the points)
  got_snitch <- game_dursA < game_dursB
  snitch_pointsA <- got_snitch     * snitch_points
  snitch_pointsB <- (1-got_snitch) * snitch_points



  # Sim goals and goal points -----------------------------------------------
  # Based on the length of the game
  goal_pointsA <- rpois(n_sims,GPM_A*floor(game_durs)) * goal_points
  goal_pointsB <- rpois(n_sims,GPM_B*floor(game_durs)) * goal_points


  # Find Winner -------------------------------------------------------------
  total_pointA <- snitch_pointsA + goal_pointsA
  total_pointB <- snitch_pointsB + goal_pointsB

  who_won <- rep(NA_character_,n_sims)
  who_won[total_pointA > total_pointB] <- 'A'
  who_won[total_pointA < total_pointB] <- 'B'

  list(teamA_points = total_pointA,
       teamB_points = total_pointB,
       Winner = factor(who_won),
       Got_snitch = factor(ifelse(got_snitch,'A','B')))
}

Equal Teams

Let’s start by simulating 1 million games between two teams with equal GPMs and equally good seekers.

res <- sim_quidditch(mean_game_durA = 90,  # in mins
                     mean_game_durB = 90,
                     GPM_A = 0.8,
                     GPM_B = 0.8)
with(res,{
  table(Winner) / 
       length(Winner)
})
## Winner
##        A        B 
## 0.497066 0.496060

Under these conditions, the probability of winning a game without catching the snitch is:

with(res, {
  mean(Winner[!is.na(Winner)] != Got_snitch[!is.na(Winner)])
})
## [1] 0.03747158

3.75%! Hmm, that does seem quite unfair… Though it does put Fred and George’s gambling odds at around 1:20!

How many games were determined by the snitch - i.e., had the snitch been caught at the same time by the other team, they would have won?

with(res, {
  (
    sum(teamA_points[Winner == 'A'] - 150 < teamB_points[Winner == 'A'] + 150, na.rm = T) +
      sum(teamA_points[Winner == 'B'] + 150 > teamB_points[Winner == 'B'] - 150, na.rm = T)
  ) / 1e6
})
## [1] 0.94918

94.9%! Not the best of odds…

Unequal (but matched) Teams

But just like in real sports, different teams have different strengths… Let’s see what happens when we create two teams who have an equal probability of winning, but differ in terms of their seekers and GPMs.

After some trial and error, I found that if we have a team with a seeker who catches the snitch within 67 minutes on average and a GPM of 0.71, versus a team with a seeker who catches the snitch in within 113 minutes on average but with a GPM of 0.90, both teams have a 50% chance of winning.

res <- sim_quidditch(mean_game_durA = 67,  # in mins
                     mean_game_durB = 113,
                     GPM_A = 0.71,
                     GPM_B = 0.90)

with(res,{
  table(Winner)/length(Winner)
})
## Winner
##        A        B 
## 0.497701 0.490796

So, what is the chance of winning without catching the snitch?

with(res,{
  mean(Winner[!is.na(Winner)] != Got_snitch[!is.na(Winner)])
})
## [1] 0.1213732

12.2%! Still pretty low… very disappointing. Let’s explore each team separately:

with(res,{
  table(Winner, Got_snitch)
})
##       Got_snitch
## Winner      A      B
##      A 497218    483
##      B 119494 371302

It seems that 99.9% of team A’s wins were due to their better seeker…

…but that only 24.3% of team B’s wins were achieved by their weaker seeker T!

How many games were determined by the snitch?

with(res, {
  (
    sum(teamA_points[Winner == 'A'] - 150 < teamB_points[Winner == 'A'] + 150, na.rm = T) +
      sum(teamA_points[Winner == 'B'] + 150 > teamB_points[Winner == 'B'] - 150, na.rm = T)
  ) / 1e6
})
## [1] 0.901089

90.1%!