Sunday Times Teaser 2594 – Just Enough

by Andrew Skidmore

Published: 10 June 2012 (link)

I have two identical bags of balls, each containing two reds, two yellows and two blues. Blindfolded throughout, I remove just enough balls from the first to be sure that the removed balls include at least two colours. I place these balls in the second bag. Then I remove some balls from the second bag and put them in the first: I move just enough to be sure that the first bag will now contain at least one ball of each colour.

What (as a percentage) is the probability that all the balls left in the second bag are the same colour?

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Brian Gladman permalink


from itertools import permutations

# the three ball colours
yellow, blue, red = 1, 2, 3
# the initial bag contents
bag = ( yellow, blue, red, yellow, blue, red )

count_all = count_same = 0
# we need three balls from the first bag to
# ensure that we have at least two colours
for p in permutations(bag, 3):
  # the possibilities for bag one and two are now:
  #
  #  1       2
  # ybr  yyybbbrrr 
  # yyr  yybbbbrrr
  # yyb  yybbbrrrr   
  # rry  yyybbbbrr   
  # rrb  yyyybbbrr
  # bby  yyybbrrrr
  # bbr  yyyybbrrr
  #
  # to be sure that bag 1 has at least one ball
  # of each colour we have to transfer at least
  # six balls back from bag two to bag one, which
  # leaves three balls in bag two
  for q in permutations(bag + p, 3):
    # count the permutations that leave all balls 
    # in bag two of the same colour
    if q[0] == q[1] == q[2]:
      count_same += 1
    # count all permutations
    count_all += 1
fs = 'There is a {} percent probability.'
print(fs.format(100 * count_same / count_all))

from itertools import permutations

# the three ball colours

yellow, blue, red = 1, 2, 3

# the initial bag contents

bag = ( yellow, blue, red, yellow, blue, red )

count_all = count_same = 0

# we need three balls from the first bag to

# ensure that we have at least two colours

for p in permutations(bag, 3):

# the possibilities for bag one and two are now:

# 1 2

# ybr yyybbbrrr

# yyr yybbbbrrr

# yyb yybbbrrrr

# rry yyybbbbrr

# rrb yyyybbbrr

# bby yyybbrrrr

# bbr yyyybbrrr

# to be sure that bag 1 has at least one ball

# of each colour we have to transfer at least

# six balls back from bag two to bag one, which

# leaves three balls in bag two

for q in permutations(bag + p, 3):

# count the permutations that leave all balls

# in bag two of the same colour

if q[0] == q[1] == q[2]:

count_same += 1

# count all permutations

count_all += 1

fs = 'There is a {} percent probability.'

print(fs.format(100 * count_same / count_all))

Frits permalink

Using itertools.combinations().


from itertools import combinations

# the initial bag contents
bag = "yrbyrb"

# using: number perm occurences = 6 * number comb ouccurences

occ333 = 6 * sum(len(set(q)) == 1 for q in combinations("aaabbbccc", 3))
occ234 = 6 * sum(len(set(q)) == 1 for q in combinations("aaaabbccc", 3))

# we need 6 balls from the second bag with 9 balls
count_all, count_same = 9 * 8 * 7 * 6 * 5 * 4, 0
# we need three balls from the first bag 
for p in combinations(bag, 3):
  count_same += occ333 if len(set(p)) == 3 else occ234
print(f"There is a {(600 * count_same / count_all):.2} percent probability")

from itertools import combinations

# the initial bag contents

bag = "yrbyrb"

# using: number perm occurences = 6 * number comb ouccurences

occ333 = 6 * sum(len(set(q)) == 1 for q in combinations("aaabbbccc", 3))

occ234 = 6 * sum(len(set(q)) == 1 for q in combinations("aaaabbccc", 3))

# we need 6 balls from the second bag with 9 balls

count_all, count_same = 9 * 8 * 7 * 6 * 5 * 4, 0

# we need three balls from the first bag

for p in combinations(bag, 3):

count_same += occ333 if len(set(p)) == 3 else occ234

print(f"There is a {(600 * count_same / count_all):.2} percent probability")

Sunday Times Teaser 2594 – Just Enough

by Andrew Skidmore

Published: 10 June 2012 (link)

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