S2T2

Jim Randell 4:55 pm on 23 April 2021 Permalink Reply
Tags: by: Danny Roth ( 61 )   

• 

  Jim Randell 5:30 pm on 23 April 2021 Permalink | Reply

  The equations can be solved manually without too much trouble to give a direct solution, but here is a Python program that solves the puzzle. It runs in 45ms.

  Run: [ @replit ]

  from enigma import P, irange, peek, printf
  
  # the cards in order (highest to lowest)
  cards = list(v + s for v in "AKQJX98765432" for s in "SHDC")
  
  # index of certain cards we are interested in
  (i6H, i7H) = (cards.index(c) for c in ('6H', '7H'))
  
  # count number of pairs where X+Y play together
  # given G chose card index g and M chose card index m
  N = lambda g, m: P(g, 2) + P(51 - m, 2)
  
  # choose a card for G
  for g in irange(0, i7H - 2):
  
    # count pairs if M had chosen 6H or 7H
    ns = { 3 * N(g, i6H), 2 * N(g, i7H) }
    if len(ns) > 1: continue
  
    # choose a card for M
    for m in irange(g + 1, i7H - 1):
  
      # that gives the correct probability
      if N(g, m) in ns:
        # output solution
        printf("G={G} [{g}]; n={n}; M={M} [{m}]", G=cards[g], M=cards[m], n=peek(ns))
  

  Solution: George’s card is A♣. Martha’s card is 9♠.

  ---

  Manually:

  There are 52 cards to start with, and after George has chosen one (at index g in the list of cards) there are 51 remaining.

  Martha chooses a lower card than George (at index m in the list of cards), and now there are 50 cards remaining.

  The other couple (X and Y) then choose cards, and if they are both lower valued than Martha’s card, or higher valued than George’s card then they will play together.

  The number of possible (unordered) pairs of cards for X+Y is: P(50, 2) = 2450, but that is the same in all scenarios, so we can just compare the number of pairs which result in X+Y playing together (n = 2450p).

  The number of pairs which are better than George’s card is: P(g, 2)

  And the number of pairs which are worse than Martha’s card is: P(51 − m, 2)

  If Martha had chosen 6♥ (the card at index 33) the probability is p/3, so:

  P(g, 2) + P(51 − 33, 2) = n / 3

  If Martha had chosen 7♥ (the card at index 29) the probability is p/2, so:

  P(g, 2) + P(51 − 29, 2) = n / 2

  Together these solve to give:

  P(22, 2) − P(18, 2) = n / 6
  ⇒ n = 936
  g² − g − 6 = 0
  (g + 2)(g − 3) = 0
  ⇒ g = 3

  So G chose the card at index 3 = A♣.

  (And the probability is p = 936/2450 ≈ 0.382).

  To find Martha’s card:

  P(g, 2) + P(51 − m, 2) = 936
  m² − 101m + 2556 = 936
  m² − 101m + 1620 = 0
  (m − 20)(m − 81) = 0
  ⇒ m = 20

  So M chose the card at index 20 = 9♠.

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• 

  Frits 11:45 pm on 23 April 2021 Permalink | Reply

  from enigma import div
  
  # chance is numerator / denominator
  # P = lambda g, m: ((m - 1) * (m - 2)  + (52 - g) * (51 - g)) / (50 * 49)
  
  # range cards: 1 - 52
  # H7 is card 23, H6 = 19
  
  suits = ["clubs", "diamonds", "hearts", "spades"]
  nums = ["2", "3", "4", "5", "6", "7", "8", "9", "10", 
          "jack", "queen", "king", "ace"]
    
  # check values for George
  for g in range(25, 53):
    G = (52 - g) * (51 - g)
    
    h7_numerator = (23 - 1) * (23 - 2) + G
    if h7_numerator % 3: continue
    
    h6_numerator = 2 * h7_numerator // 3
    if h6_numerator != (19 - 1) * (19 - 2) + G: continue
    
    m_numerator = 2 * h7_numerator
    
    # (m - 1) * (m - 2) + G = m_numerator, m^2 -3m + (2 + G - m_numerator) = 0
    m = div(3 + (1 + 4 * (m_numerator - G)) ** 0.5, 2)
    if m is None: continue
    m = int(m)
    
    print(f"George and Martha drew {nums[(g - 1) // 4]} of {suits[(g - 1) % 4]}"
          f" and {nums[(m - 1) // 4]} of {suits[(m - 1) % 4]}")
  

  With more analysis:

  from enigma import div
  
  # range cards: 1 - 52
  
  suits = ["clubs", "diamonds", "hearts", "spades"]
  nums = ["2", "3", "4", "5", "6", "7", "8", "9", "10", 
          "jack", "queen", "king", "ace"]
  
  # numerator of P = (m - 1) * (m - 2) + (52 - g) * (51 - g) 
  # P is twice the chance for H7 (card 23)
  
  # (m - 1) * (m - 2) + (52 - g) * (51 - g) = 
  # 2 * ((23 - 1) * (23 - 2) + (52 - g) * (51 - g))
  
  # g = 103 / 2 - ((4 * m^2 - 12*m - 3687) ^ 0.5) / 2
  
  # P is three times the chance for H6 (card 19)
  
  # (m - 1) * (m - 2) + (52 - g) * (51 - g) = 
  # 3 * ((19 - 1) * (19 - 2) + (52 - g) * (51 - g))
  
  # g = 103 / 2 - ((2 * m^2 - 6*m - 1831) ^ 0.5) / 2
  
  # (4 * m^2 - 12*m - 3687) - (2 * m^2 - 6*m - 1831) = 0
  # 2 * m^2 - 6 * m - 1856 = 0
  m = div(6 + (36 + 4 * 2 * 1856) ** 0.5, 4)
  if m is None: exit() 
  m = int(m)
  
  g = div(103 - (4 * m ** 2 - 12 * m - 3687) ** 0.5, 2)
  if g is None: exit() 
  g = int(g)
  
  print(f"George and Martha drew {nums[(g - 1) // 4]} of {suits[(g - 1) % 4]}"
        f" and {nums[(m - 1) // 4]} of {suits[(m - 1) % 4]}")
  

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