Sunday Times Teaser 3300 – A Faulty Bet

by BRG on December 21, 2025

by Peter Rowlett

Published Sunday December 21 2025 (link)

I offer my friends a bet: they pay £1 to play, then draw a card at random from a standard 52-card deck. If they draw the Queen of Hearts, I pay them a £30 prize, otherwise I keep their £1.

Although I have many friends, I have a reasonable chance of avoiding paying a prize. But I make a mistake! As I go around the group offering the bet to each friend, I forget to take their chosen card back and shuffle it into the deck, meaning each new bet is made with fewer cards. As a result, my chances of having to pay out are nearly 50 per cent greater. In fact if I had just one more friend, then my chances of having to pay out would be more than 50 per cent greater than if I hadn’t made the mistake.

How many friends did I have in the group?

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4 Comments Leave one →

BRG permalink


try:
  from gmpy2 import mpq as RF
except ImportError:
  from fractions import Fraction as RF
 
pack, cpf, cpr = 53, 1, 1
# consider the number of friends
for friends in range(1, 53):
  pack -= 1
  # the cumulative probability of not paying out with the full pack
  cpf *= 1 - RF(1, 52)
  # the cumulative probability of not paying out with the reducing pack
  cpr *= 1 - RF(1, pack)
  # the ratio of the two probabilities of paying out
  if RF(1 - cpr, 1 - cpf) > 3 / 2:
    print(f"Peter had {friends - 1} friends.")
    break

try:

from gmpy2 import mpq as RF

except ImportError:

from fractions import Fraction as RF

pack, cpf, cpr = 53, 1, 1

# consider the number of friends

for friends in range(1, 53):

pack -= 1

# the cumulative probability of not paying out with the full pack

cpf *= 1 - RF(1, 52)

# the cumulative probability of not paying out with the reducing pack

cpr *= 1 - RF(1, pack)

# the ratio of the two probabilities of paying out

if RF(1 - cpr, 1 - cpf) > 3 / 2:

print(f"Peter had {friends - 1} friends.")

break

Frits permalink


deck = 52 
# consider the number of cards in the pack
for n in range(deck, 0, -1):
  m = deck + 1 - n
  if 2 * m * deck ** (m - 1) > 3 * (deck**m - (deck - 1)**m):
    print(f"answer: {deck - n} friends")
    break

deck = 52

# consider the number of cards in the pack

for n in range(deck, 0, -1):

m = deck + 1 - n

if 2 * m * deck ** (m - 1) > 3 * (deck**m - (deck - 1)**m):

print(f"answer: {deck - n} friends")

break

Frits permalink

I didn’t code the “nearly 50 per cent greater” requirement. It is not likely that it is exactly 50 per cent greater for the reported answer.

Reply

BRG permalink

Floating point is accurate enough to give the answer:


# three way compare
cmp3w = lambda x, y: (x > y) - (x < y)

# The cumulative probability of NOT paying out with N friends and the use of:
# 
# (a) a full pack (cpfp):     = (51/52)^N
# 
# (b) a reducing pack (cprp): = (1 - 1/52).(1 - 1/51) ... (1 - 1 / (53 - N))
#                             = (51/52).(50/51).(49/50) ... (52 - N) / (53 - N)
#                             = (52 - N) / 52

cpfp = (51 / 52) ** 53
for friends in range(52, 0, -1):
  cpfp *= (52 / 51)
  cprp = 1 - friends / 52
  if (cmp := cmp3w((1 - cprp) / (1 - cpfp), 3 / 2)) == -1:
    print(f"Peter had {friends} friends.")
    break
  elif not cmp:
    print('bad result')

# three way compare

cmp3w = lambda x, y: (x > y) - (x < y)

# The cumulative probability of NOT paying out with N friends and the use of:

# (a) a full pack (cpfp): = (51/52)^N

# (b) a reducing pack (cprp): = (1 - 1/52).(1 - 1/51) ... (1 - 1 / (53 - N))

# = (51/52).(50/51).(49/50) ... (52 - N) / (53 - N)

# = (52 - N) / 52

cpfp = (51 / 52) ** 53

for friends in range(52, 0, -1):

cpfp *= (52 / 51)

cprp = 1 - friends / 52

if (cmp := cmp3w((1 - cprp) / (1 - cpfp), 3 / 2)) == -1:

print(f"Peter had {friends} friends.")

break

elif not cmp:

print('bad result')

Sunday Times Teaser 3300 – A Faulty Bet

by Peter Rowlett

Published Sunday December 21 2025 (link)

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