Sunday Times Teaser 2484

by Andrew Skidmore

Published: 2 May 2010 (link)

Josh put all the snooker balls except the cue into a bag. He and I then played an imaginary frame. At each person’s turn, he drew a ball. If it was red, he scored a point, discarded the red and drew again. If that was a “colour” (not red), he scored the appropriate number of points, returned it to the bag, then tried again for a red. If a player drew the wrong ball, it was returned to the bag and his break was over. On one break, the first four balls drawn earned me points; at that stage, the chance of this happening was one in N, N being the number of points I’d got from the break so far.

Which two colours had I drawn?

3 Comments Leave one →

Brian Gladman permalink


# Four balls means two reds and two colours.  If the number of reds at
# the start of the break is r, the probability of this is:
# 
#    P = {r / (r + 6)}.{6 / (r + 5)}.{(r - 1) / (r + 5)}.{6 / (r + 4)}
# 
# hence the odds are:
# 
#    odds = 1 / P = (r + 6).(r + 5)^2.(r + 4) / {36.r.(r - 1)}

from itertools import combinations

# the colours for the scores between 2 and 7
colours = {2:'yellow', 3:'green', 4: 'brown', 5:'blue', 6:'pink', 7:'black'}

# the number of reds at the start of the break
for r in range(2, 16):
  odds, rem = divmod((r + 6) * (r + 5) ** 2 * (r + 4), 36 * r * (r - 1))
  # the minimum (maximum) score is 2 reds with 2 yellows (blacks)  
  if not rem and 4 <= odds <= 16:
    for c in combinations(colours, 2):
      if 2 + sum(c) == odds:
        col = [colours[x] for x in c]
        print('The colours are {} and {}.'.format(*col))

# Four balls means two reds and two colours. If the number of reds at

# the start of the break is r, the probability of this is:

# P = {r / (r + 6)}.{6 / (r + 5)}.{(r - 1) / (r + 5)}.{6 / (r + 4)}

# hence the odds are:

# odds = 1 / P = (r + 6).(r + 5)^2.(r + 4) / {36.r.(r - 1)}

from itertools import combinations

# the colours for the scores between 2 and 7

colours = {2:'yellow', 3:'green', 4: 'brown', 5:'blue', 6:'pink', 7:'black'}

# the number of reds at the start of the break

for r in range(2, 16):

odds, rem = divmod((r + 6) * (r + 5) ** 2 * (r + 4), 36 * r * (r - 1))

# the minimum (maximum) score is 2 reds with 2 yellows (blacks)

if not rem and 4 <= odds <= 16:

for c in combinations(colours, 2):

if 2 + sum(c) == odds:

col = [colours[x] for x in c]

print('The colours are {} and {}.'.format(*col))

Frits permalink

@Brian, your first Sunday Times Teaser?

As you use combinations() you assume that the coloured balls have a different colour.

Reply
- BRG permalink
  
  Yes, in my view the wording indicates that there are two different colours.
  Otherwise the wording would be “which colours had I drawn”, i.e. not
  explicitly asking for two.
  
  Reply

Sunday Times Teaser 2484

by Andrew Skidmore

Published: 2 May 2010 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta