Sunday Times Teaser 2889 – Catching the Bus

by Michael Fletcher

Published February 4 2018 (link)

I can travel to town by either the number 12 bus or the number 15 bus. The number 12 bus leaves every 12 minutes and the number 15 bus leaves every 15 minutes. The first number 15 bus leaves soon after the first number 12 bus. I arrive at the bus stop at random and catch the first bus to arrive.

If the probability of me catching the number 15 bus is 2/5 how soon does the first number 15 bus leave after the first number 12 bus?

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Brian Gladman permalink


f1, f2 = False, False

# events for bus number 12 over the first hour (in seconds)
ev15 = [(60 * x, 12) for x in range(0, 61, 12)]
# the initial time delay in seconds between buses 12 and 15
for delay in range(0, 3600):
  # events for bus number 15 over the first hour (in seconds)
  ev12 = [(60 * x + delay, 15) 
          for x in range(0, 61, 15) if 60 * x + delay <= 3600]
  
  # accumulate the waiting times for bus numbers 12 and 15
  waits = [0, 0]
  # consider the events in the first hour in time order 
  for time, bus in sorted(ev12 + ev15):
    if time:
      # accumulate the waiting times for each bus number
      waits[0 if bus == 12 else 1] += time - last_bus
    last_bus = time
  # the probabilities for bus numbers 12 and 15 3/5 and 2/5
  if 2 * waits[0] == 3 * waits[1]:
    # output the delay in minutes and seconds
    m, s = divmod(delay, 60)
    ms = f'{m} minute' + ('s' if m > 1 else '')
    ss = '.' if not s else f' {s} seconds.'
    if not s and not f2:
      print('Other solutions (first hour only):')
      f2 = True
    print(f'  Delay = {ms + ss}')
    if not f1:
      print('Other solutions:')
      f1 = True

f1, f2 = False, False

# events for bus number 12 over the first hour (in seconds)

ev15 = [(60 * x, 12) for x in range(0, 61, 12)]

# the initial time delay in seconds between buses 12 and 15

for delay in range(0, 3600):

# events for bus number 15 over the first hour (in seconds)

ev12 = [(60 * x + delay, 15)

for x in range(0, 61, 15) if 60 * x + delay <= 3600]

# accumulate the waiting times for bus numbers 12 and 15

waits = [0, 0]

# consider the events in the first hour in time order

for time, bus in sorted(ev12 + ev15):

if time:

# accumulate the waiting times for each bus number

waits[0 if bus == 12 else 1] += time - last_bus

last_bus = time

# the probabilities for bus numbers 12 and 15 3/5 and 2/5

if 2 * waits[0] == 3 * waits[1]:

# output the delay in minutes and seconds

m, s = divmod(delay, 60)

ms = f'{m} minute' + ('s' if m > 1 else '')

ss = '.' if not s else f' {s} seconds.'

if not s and not f2:

print('Other solutions (first hour only):')

f2 = True

print(f' Delay = {ms + ss}')

if not f1:

print('Other solutions:')

f1 = True

Sunday Times Teaser 2889 – Catching the Bus

by Michael Fletcher

Published February 4 2018 (link)

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