Sunday Times Teaser 3030 – Pot Success

by John Owen

Sunday October 18 2020 (link)

In snooker, pot success (PS) is the percentage of how many pot attempts have been successful in that match (eg, 19 pots from 40 attempts gives a PS of 47.5). In a recent match, my PS was a whole number at one point. I then potted several balls in a row to finish a frame, after which my improved PS was still a whole number. At the beginning of the next frame, I potted the same number of balls in a row, and my PS was still a whole number. I missed the next pot, my last shot in the match, and, remarkably, my PS was still a whole number.

If I told you how many balls I potted during the match, you would be able to work out those various whole-number PS values.

How many balls did I pot?

3 Comments Leave one →

Brian Gladman permalink


sols = []
# the total number of pot attempts in the match 
for tpa in range(4, 50):
  # the number of balls potted at the end of
  # frame one and at the start of frame two
  for nir in range(2, tpa // 2):
    # the number of pot attempts at the first checkpoint
    pats = tpa - (2 * nir + 1)
    
    # the number of pots at this point
    for pots in range(1, pats + 1):
    
      # and hence the pot success 
      p1, r = divmod(100 * pots, pats)
      if r:
        continue
      
      # the pot success at the end of frame one (larger)
      p2, r = divmod(100 * (pots + nir), pats + nir)
      if r or p2 <= p1:
        continue
      
      # the pot success after the initial run in frame two
      p3, r = divmod(100 * (pots + 2 * nir), pats + 2 * nir)
      if r:
        continue
    
      # the pot success at the end of the match
      p4, r = divmod(100 * (pots + 2 * nir), pats + 2 * nir + 1)
      if r:
        continue

      sols.append((tpa, nir, pots + 2 * nir, (p1, p2, p3, p4)))

# find solutions that are unique given the total number of pots
unique = partition_unique(sols, lambda x: x[2], lambda x: x[3])[0]
for tpa, nir, tpots, p4 in unique:
  print(f"{tpots} pots in {tpa} attempts (2 runs of {nir}) -> pot successes {p4}")

sols = []

# the total number of pot attempts in the match

for tpa in range(4, 50):

# the number of balls potted at the end of

# frame one and at the start of frame two

for nir in range(2, tpa // 2):

# the number of pot attempts at the first checkpoint

pats = tpa - (2 * nir + 1)

# the number of pots at this point

for pots in range(1, pats + 1):

# and hence the pot success

p1, r = divmod(100 * pots, pats)

if r:

continue

# the pot success at the end of frame one (larger)

p2, r = divmod(100 * (pots + nir), pats + nir)

if r or p2 <= p1:

continue

# the pot success after the initial run in frame two

p3, r = divmod(100 * (pots + 2 * nir), pats + 2 * nir)

if r:

continue

# the pot success at the end of the match

p4, r = divmod(100 * (pots + 2 * nir), pats + 2 * nir + 1)

if r:

continue

sols.append((tpa, nir, pots + 2 * nir, (p1, p2, p3, p4)))

# find solutions that are unique given the total number of pots

unique = partition_unique(sols, lambda x: x[2], lambda x: x[3])[0]

for tpa, nir, tpots, p4 in unique:

print(f"{tpots} pots in {tpa} attempts (2 runs of {nir}) -> pot successes {p4}")

Frits permalink

Hello Brian,

… “I then potted several balls” … so nir > 1?

I also have seen definitions of “several” as “more than two but fewer than many”.

Reply
Brian Gladman permalink

Yes, I agree that 2 is better than 1.

Reply

Sunday Times Teaser 3030 – Pot Success

by John Owen

Sunday October 18 2020 (link)

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