Sunday Times Teaser 3275 – Against the Odds

by BRG on June 29, 2025

by Andrew Skidmore

Published Sunday June 29 2025 (link)

I have taken a number of playing cards (more red than black) from a standard pack of 52 cards and distributed them in a certain way in two piles, each containing both red and black cards.

A card is taken at random from the smaller pile and placed in the larger pile. The larger pile is shuffled; a card is taken at random from it and placed in the smaller pile.

There is a one in eight chance that the number of red and black cards in each pile will be the same as they were originally.

How many black cards in total were used?

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BRG permalink


# pile   red black        b1.(b2 + 1) + r1.(r2 + 1)     1
# one     r1    b1   p = --------------------------  =  -
# two     r2    b2        (b1 + r1).(b2 + r2 + 1)       8

# the maximum number of cards of one colour
max_c = 26
  
# the total number of red cards
for nr in range(3, max_c + 1):
  # the fewer number of black cards
  for nb in range(2, nr):
    
    # the number of red cards in pile one 
    for r1 in range(1, nr):
      r2 = nr - r1
      # the number of black cards in pile one
      for b1 in range(1, nb):
        b2 = nb - b1
        # analysis shows that (b1.r2 + r1.b2) mod 7 == 0        
        if not (b1 * r2 + b2 * r1) % 7:
          
          # find the probability that the piles are not changed by the two moves
          t = b1 * (b2 + 1) + r1 * (r2 + 1)
          b = (b1 + r1) * (b2 + r2 + 1)
          if b == 8 * t:
            print(f"{nb} black cards [{r1 = }, {r2 = }, {b1 = }, {b2 = }].")

# pile red black b1.(b2 + 1) + r1.(r2 + 1) 1

# one r1 b1 p = -------------------------- = -

# two r2 b2 (b1 + r1).(b2 + r2 + 1) 8

# the maximum number of cards of one colour

max_c = 26

# the total number of red cards

for nr in range(3, max_c + 1):

# the fewer number of black cards

for nb in range(2, nr):

# the number of red cards in pile one

for r1 in range(1, nr):

r2 = nr - r1

# the number of black cards in pile one

for b1 in range(1, nb):

b2 = nb - b1

# analysis shows that (b1.r2 + r1.b2) mod 7 == 0

if not (b1 * r2 + b2 * r1) % 7:

# find the probability that the piles are not changed by the two moves

t = b1 * (b2 + 1) + r1 * (r2 + 1)

b = (b1 + r1) * (b2 + r2 + 1)

if b == 8 * t:

print(f"{nb} black cards [{r1 = }, {r2 = }, {b1 = }, {b2 = }].")

Frits permalink


from itertools import combinations

# we have more red cards than black 
for b, r in combinations(range(2, 27), 2):
  # redistribute them in a certain way in two piles
  for b1 in range(1, b):
    b2 = b - b1
    for r1 in range(1, r):
      r2 = r - r1 
      # assume pile 1 is the smaller pile
      if b2 + r2 <= b1 + r1: continue 
      # p = chance moving a black card and moving a black card back plus
      #     chance moving a red card and moving a red card back 
      # p = b1 / (b1 + r1) * (b2 + 1) / (b2 + r2 + 1) + 
      #     r1 / (b1 + r1) * (r2 + 1) / (b2 + r2 + 1) = 1/8
      
      if (b2 + r2 + 1) * (b1 + r1) == 8 * (b1 * (b2 + 1) + r1 * (r2 + 1)):
        print("answer:", b)

from itertools import combinations

# we have more red cards than black

for b, r in combinations(range(2, 27), 2):

# redistribute them in a certain way in two piles

for b1 in range(1, b):

b2 = b - b1

for r1 in range(1, r):

r2 = r - r1

# assume pile 1 is the smaller pile

if b2 + r2 <= b1 + r1: continue

# p = chance moving a black card and moving a black card back plus

# chance moving a red card and moving a red card back

# p = b1 / (b1 + r1) * (b2 + 1) / (b2 + r2 + 1) +

# r1 / (b1 + r1) * (r2 + 1) / (b2 + r2 + 1) = 1/8

if (b2 + r2 + 1) * (b1 + r1) == 8 * (b1 * (b2 + 1) + r1 * (r2 + 1)):

print("answer:", b)

Frits permalink

Only six r1 iterations.


#   b1     ( 7.(r2 + 1) - b2 )
#   -- = - ( --------------- )   credit: B. Gladman
#   r1     ( 7.(b2 + 1) - r2 )

# b1 + b2 <= 25
for b2 in range(1, 25):
  x = 7 * (b2 + 1)
  for r2 in range(1, 26):
    num = 7 * (r2 + 1) - b2
    # numerator is increasing and denominator is decreasing by 1
    if num > 0 and x >= 25:  break
    if (den := x - r2) == 0 or not num: continue
    inc_b1, _r = divmod(-num, den)
    if not (0 <= inc_b1 < 25): continue
    
    # b1 <= 25 - b2, r1.-num / den <= (25 - b2), r1 <= (b2 - 25).den / num
    max_r1 = min(25 - r2, b2 + r2 - 2, (b2 - 25) * den // num)
    
    # b1 must be at least 1 
    if max_r1 < -den / num: continue
    
    # reverse looping in order to be able to break
    for r1 in range(max_r1, 0, -1):
      b1, _r = divmod(-r1 * num, den)
      # break for low b1 or if b1 increase < 1 and too many black cards
      if b1 < 1 or (not inc_b1 and b1 + b2 > r1 + r2): break
      if not _r and b1 + b2 < r1 + r2:
        print(f"{b1 + b2} black cards [{r1 = }, {r2 = }, {b1 = }, {b2 = }]")

# b1 ( 7.(r2 + 1) - b2 )

# -- = - ( --------------- ) credit: B. Gladman

# r1 ( 7.(b2 + 1) - r2 )

# b1 + b2 <= 25

for b2 in range(1, 25):

x = 7 * (b2 + 1)

for r2 in range(1, 26):

num = 7 * (r2 + 1) - b2

# numerator is increasing and denominator is decreasing by 1

if num > 0 and x >= 25: break

if (den := x - r2) == 0 or not num: continue

inc_b1, _r = divmod(-num, den)

if not (0 <= inc_b1 < 25): continue

# b1 <= 25 - b2, r1.-num / den <= (25 - b2), r1 <= (b2 - 25).den / num

max_r1 = min(25 - r2, b2 + r2 - 2, (b2 - 25) * den // num)

# b1 must be at least 1

if max_r1 < -den / num: continue

# reverse looping in order to be able to break

for r1 in range(max_r1, 0, -1):

b1, _r = divmod(-r1 * num, den)

# break for low b1 or if b1 increase < 1 and too many black cards

if b1 < 1 or (not inc_b1 and b1 + b2 > r1 + r2): break

if not _r and b1 + b2 < r1 + r2:

print(f"{b1 + b2} black cards [{r1 = }, {r2 = }, {b1 = }, {b2 = }]")

BRG permalink

Here is my analytically based solution:


#   b1     ( 7.(r2 + 1) - b2 )
#   -- = - ( --------------- )
#   r1     ( 7.(b2 + 1) - r2 )
#
# since the left hand side is positive the numerator and denominator
# on the right must have opposite signs and both are positive if b2
# and r2 are 3 or more; hence one of (b2, r2) is less than 3.

# the r2/b2 value that is in (1, 2)
for p in (1, 2):
  # the other r2/b2 value
  for q in range(1, 26):
    
    for r2, b2 in ((p, q), (q, p)):
      # the numerator and denominator values
      num = 7 * (r2 + 1) - b2      
      den = 7 * (b2 + 1) - r2
      if num * den >= 0:
        continue
      # consider possible values for r1
      for r1 in range(max(1, b2 - r2 + 1), 27 - r2):
        # check for an integer b1 value
        b1, _r = divmod(-r1 * num, den)
        if b1 > 25:
          break
        # there are fewer total black cards than red ones
        if not _r and b2 < b1 + b2 < r1 + r2:
          print(f"{b1 + b2} black cards [{r1 = }, {r2 = }, {b1 = }, {b2 = }].")

# b1 ( 7.(r2 + 1) - b2 )

# -- = - ( --------------- )

# r1 ( 7.(b2 + 1) - r2 )

# since the left hand side is positive the numerator and denominator

# on the right must have opposite signs and both are positive if b2

# and r2 are 3 or more; hence one of (b2, r2) is less than 3.

# the r2/b2 value that is in (1, 2)

for p in (1, 2):

# the other r2/b2 value

for q in range(1, 26):

for r2, b2 in ((p, q), (q, p)):

# the numerator and denominator values

num = 7 * (r2 + 1) - b2

den = 7 * (b2 + 1) - r2

if num * den >= 0:

continue

# consider possible values for r1

for r1 in range(max(1, b2 - r2 + 1), 27 - r2):

# check for an integer b1 value

b1, _r = divmod(-r1 * num, den)

if b1 > 25:

break

# there are fewer total black cards than red ones

if not _r and b2 < b1 + b2 < r1 + r2:

print(f"{b1 + b2} black cards [{r1 = }, {r2 = }, {b1 = }, {b2 = }].")

Sunday Times Teaser 3275 – Against the Odds

by Andrew Skidmore

Published Sunday June 29 2025 (link)

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