Sunday Times Teaser 2692 – Runners Up

by Robin Nayler

Teams in our local football league play once each week at 15:00 on Saturdays, playing each other twice in a season. Teams earn three points for a win and one for a draw. All teams played their last match on the same day.

Last season, the top team won two thirds of their matches; we were the runners-up with one point less even though we had won more matches. The top team had also lost twice as many matches as we had drawn.

How many games did the top team win, draw and lose?

One Comment Leave one →

brian gladman permalink


# For M teams playing each other twice in a season, there will be
# M * (M - 1) games with each team playing a total of 2 * (M - 1)
# games. M must also be even since all teams play on the last day
# of the season. The number of games each team plays is divisible
# by three and must be less than or equal to 52.
for games_per_team in range(6, 53, 6):
  # the number of teams must be even
  if (games_per_team // 2 + 1) % 2:
    continue
  top_wins = 2 * games_per_team // 3
  # since the top team has at least one less win than us and yet
  # beats us, it must win at least 6 draws more than we do - but
  # it must also have at least 2 losses
  for top_draws in range(6, games_per_team // 3 - 1):
    # calculate the top team's points
    top_points = 3 * top_wins + top_draws  
    # but our team wins more matches (but not all of them)
    for our_wins in range(top_wins + 1, games_per_team):
      # we can draw none, some or all our remaining matches
      for our_draws in range(games_per_team - our_wins + 1):
        # check that we draw more than once, that the top team
        # loses twice as often and has one more point than us
        top_losses = games_per_team // 3 - top_draws
        if (3 * our_wins + our_draws == top_points - 1 and
              top_losses == 2 * our_draws and our_draws > 1):
          print('{} wins, {} draws and {} losses'
                .format(top_wins, top_draws, top_losses))

# For M teams playing each other twice in a season, there will be

# M * (M - 1) games with each team playing a total of 2 * (M - 1)

# games. M must also be even since all teams play on the last day

# of the season. The number of games each team plays is divisible

# by three and must be less than or equal to 52.

for games_per_team in range(6, 53, 6):

# the number of teams must be even

if (games_per_team // 2 + 1) % 2:

continue

top_wins = 2 * games_per_team // 3

# since the top team has at least one less win than us and yet

# beats us, it must win at least 6 draws more than we do - but

# it must also have at least 2 losses

for top_draws in range(6, games_per_team // 3 - 1):

# calculate the top team's points

top_points = 3 * top_wins + top_draws

# but our team wins more matches (but not all of them)

for our_wins in range(top_wins + 1, games_per_team):

# we can draw none, some or all our remaining matches

for our_draws in range(games_per_team - our_wins + 1):

# check that we draw more than once, that the top team

# loses twice as often and has one more point than us

top_losses = games_per_team // 3 - top_draws

if (3 * our_wins + our_draws == top_points - 1 and

top_losses == 2 * our_draws and our_draws > 1):

print('{} wins, {} draws and {} losses'

.format(top_wins, top_draws, top_losses))

I my first attempt I didn’t find a unique answer for this one but I didn’t spend much time on it as I don’t much like football teasers. But Peter Hayes has pointed out that my description (now amended) missed one aspect of the original text that can be interpreted to imply that the runner up draws more than once. This interpretation can hence be used to resolve the ambiguity.

Sunday Times Teaser 2692 – Runners Up

by Robin Nayler

Leave a comment to brian gladman Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta