Sunday Times Teaser 3102 – Jokers

by BRG on March 4, 2022

by Victor Bryant

Published Sunday March 06 2022 (link)

A group of us wanted to play a card game. We had a standard 52-card pack but we needed to deal out a whole pack with each person getting the same number of cards, so I added just enough “jokers” to make this possible. Then I randomly shuffled the enlarged pack (in this shuffled pack there was more than a 50:50 chance that there would be at least two jokers together). Then I dealt the cards out, each of us getting the same number (more than three). There was more than a 50:50 chance that my cards would not include a joker.

How many of us were there?

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Brian Gladman permalink


from itertools import count
from math import prod

# To put J jokers into a pack of 52 cards with no adjacent
# jokers, we can put them before the start, after the end 
# or in the 51 gaps between the cards. The number of ways
# of doing this is C(53, J). The total number of ways to 
# put J jokers in a pack of 52 cards is C(52 + J, J). THe 
# probability of a random shuffle with no adjacent jokers 
# is hence C(53, J) / C(52 + J, J).

def adj_joker_prob(nj):
  n, d, t, b = 1, 1, 53, 53 
  for _ in range(nj):
    n *= t
    d *= b
    t -= 1
    b += 1
  return d - n, d

# consider the number of players
for players in count(2):
  # find the minimum number of jokers to be added
  jokers = -52 % players
  # and the cards per player
  cpp = (52 + jokers) // players
  
  # there are more than three cards per player and the
  # adjacent joker probability is greater than 50:50
  n, d = adj_joker_prob(jokers)
  if cpp <= 3 or 2 * n <= d:
    continue

  # calculate the probability that <c> cards have no jokers
  n = prod(52 - i for i in range(cpp))
  d = prod(52 + jokers - i for i in range(cpp))
  # which must be greater than 50:50
  if 2 * n <= d:  
    continue
  print(f"{players} players ({jokers} jokers).")
  break

from itertools import count

from math import prod

# To put J jokers into a pack of 52 cards with no adjacent

# jokers, we can put them before the start, after the end

# or in the 51 gaps between the cards. The number of ways

# of doing this is C(53, J). The total number of ways to

# put J jokers in a pack of 52 cards is C(52 + J, J). THe

# probability of a random shuffle with no adjacent jokers

# is hence C(53, J) / C(52 + J, J).

def adj_joker_prob(nj):

n, d, t, b = 1, 1, 53, 53

for _ in range(nj):

n *= t

d *= b

t -= 1

b += 1

return d - n, d

# consider the number of players

for players in count(2):

# find the minimum number of jokers to be added

jokers = -52 % players

# and the cards per player

cpp = (52 + jokers) // players

# there are more than three cards per player and the

# adjacent joker probability is greater than 50:50

n, d = adj_joker_prob(jokers)

if cpp <= 3 or 2 * n <= d:

continue

# calculate the probability that <c> cards have no jokers

n = prod(52 - i for i in range(cpp))

d = prod(52 + jokers - i for i in range(cpp))

# which must be greater than 50:50

if 2 * n <= d:

continue

print(f"{players} players ({jokers} jokers).")

break

An alternative that considers jokers before players.


from itertools import count
from math import prod

# To put J jokers into a pack of 52 cards with no adjacent
# jokers, we can put them before the start, after the end 
# or in the 51 gaps between the cards. The number of ways
# of doing this is C(53, J). The total number of ways to 
# put J jokers in a pack of 52 cards is C(52 + J, J). THe 
# probability of a random shuffle with no adjacent jokers 
# is hence C(53, J) / C(52 + J, J).

n, d, t, b = 1, 1, 53, 53
for j_min in count(1):
  n *= t
  d *= b
  if 2 * n < d:
    break
  t -= 1
  b += 1

# consider the number of jokers added
jokers, found = j_min - 1, False
while not found:
  jokers += 1
  # consider the number of players
  for np in count(2):
    # the number of players divides the number of cards
    c, r = divmod(52 + jokers, np)
    if r:
      continue
    # each player has more than three cards
    if c <= 3:
      break
    # calculate the probability of no jokers in <c> cards
    n = prod(52 - i for i in range(c))
    d = prod(52 + jokers - i for i in range(c))
    # which must be greater than 50:50
    if 2 * n > d:  
      print(f"{np} players ({jokers} jokers).")
      found = True

from itertools import count

from math import prod

# To put J jokers into a pack of 52 cards with no adjacent

# jokers, we can put them before the start, after the end

# or in the 51 gaps between the cards. The number of ways

# of doing this is C(53, J). The total number of ways to

# put J jokers in a pack of 52 cards is C(52 + J, J). THe

# probability of a random shuffle with no adjacent jokers

# is hence C(53, J) / C(52 + J, J).

n, d, t, b = 1, 1, 53, 53

for j_min in count(1):

n *= t

d *= b

if 2 * n < d:

break

t -= 1

b += 1

# consider the number of jokers added

jokers, found = j_min - 1, False

while not found:

jokers += 1

# consider the number of players

for np in count(2):

# the number of players divides the number of cards

c, r = divmod(52 + jokers, np)

if r:

continue

# each player has more than three cards

if c <= 3:

break

# calculate the probability of no jokers in <c> cards

n = prod(52 - i for i in range(c))

d = prod(52 + jokers - i for i in range(c))

# which must be greater than 50:50

if 2 * n > d:

print(f"{np} players ({jokers} jokers).")

found = True

Erling Torkildsen permalink


# Consider 3 - 17 players
for p in range(3, 18):
  if 52 % p != 0:
    
    # Jokers to be added
    j = p * (52 // p + 1) - 52

    # If we ignore anomalies with top and bottom cards
    # P(not(JJ)) are close to h
    h = 1
    for n in range(j):
      h *= (j + 52 - 3 * n) / (j + 52 - n)

    # Check when P(JJ) = 1 - P(not(JJ)) > 0.5
    if 1 - h > 0.5:

      # P(no J) is built by k
      k = 1
      for m in range((52 + j) // p):
        k *= (52 - m) / (j + 52 - m)
      if k > 0.5:
        print(f"{p} players ({j} jokers added)")

# Consider 3 - 17 players

for p in range(3, 18):

if 52 % p != 0:

# Jokers to be added

j = p * (52 // p + 1) - 52

# If we ignore anomalies with top and bottom cards

# P(not(JJ)) are close to h

h = 1

for n in range(j):

h *= (j + 52 - 3 * n) / (j + 52 - n)

# Check when P(JJ) = 1 - P(not(JJ)) > 0.5

if 1 - h > 0.5:

# P(no J) is built by k

k = 1

for m in range((52 + j) // p):

k *= (52 - m) / (j + 52 - m)

if k > 0.5:

print(f"{p} players ({j} jokers added)")

Sunday Times Teaser 3102 – Jokers

by Victor Bryant

Published Sunday March 06 2022 (link)

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