S2T2

Jim Randell 8:40 am on 25 March 2019 Permalink Reply
Tags: by: Victor Bryant ( 140 )   

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  Jim Randell 8:41 am on 25 March 2019 Permalink | Reply

  This Python 3 program works by finding possible (prime, X, Y) tuples for choosing a pair of X socks from X+Y, and then links together the appropriate number of tuples to solve the puzzle.

  It runs in 75ms.

  Run: [ @replit ]

  from enigma import (irange, is_prime, fraction, printf)
  
  # min and max
  (m, M) = (2, 49)
  
  # find primes of the form (X + Y)(X + Y - 1) / X(X - 1)
  # where X + Y =< M and X >= m, Y >= m
  rs = list()
  for X in irange(max(m, 2), M - m):
    for Y in irange(m, M - X):
      (p, r) = divmod((X + Y) * (X + Y - 1), X * (X - 1))
      if not (r == 0 and is_prime(p)): continue
      rs.append((p, X, Y))
  
  # solve for k steps
  def solve(k, ns=None, ps=None):
    # are we done?
    if k == 0:
      yield (ns, ps)
    else:
      if not ns:
        # first step, use both X and Y
        for (p, X, Y) in rs:
          yield from solve(k - 1, [X, Y], [p])
      else:
        # subsequent steps match sum on X
        T = sum(ns)
        for (p, X, Y) in rs:
          if X == T and p not in ps:
            yield from solve(k - 1,  ns + [Y], ps + [p])
  
  # find a sequence of 4 primes
  for ((B, W, R, G, Y), (p1, p2, p3, p4)) in solve(4):
    # calculate the probability of a yellow pair
    T = B + W + R + G + Y
    (a, b) = fraction(Y * (Y - 1), T * (T - 1))
    # output solution
    printf("P(yellow) = {a}/{b} [B={B} W={W}, p1={p1}; R={R} p2={p2}; G={G} p3={p3}; Y={Y} p4={p4}; T={T}]")
  

  Solution: There is a 1 in 6 chance of getting a yellow pair.

  There are two scenarios which give the answer:

  From 3 black and 3 white, the probability a black pair is 1/5.
  From 6 grey and 9 red, the probability of a grey pair is 1/7.
  From 15 pink and 6 green, the probability of a pink pair is 1/2.
  From 21 brown and 15 yellow, the probability of a brown pair is 1/3.

  From 3 black and 4 white, the probability a black pair is 1/7.
  From 7 grey and 8 red, the probability of a grey pair is 1/5.
  From 15 pink and 6 green, the probability of a pink pair is 1/2.
  From 21 brown and 15 yellow, the probability of a brown pair is 1/3.

  But in both cases the probability of a yellow pair (from 21 brown and 15 yellow) is 1/6.

  In the puzzle I interpreted “some” to mean “at least 2”, and this gives the solution above. However, if we interpret it as “at least 1” then you get 2 further solutions, which give an answer of 1/14. (You can change the value of [[ m ]] to [[ 1 ]] in the program to see this).

  Interestingly there are no solutions (even with the relaxation of “some”) where all the numbers of different coloured socks are all even. Which certainly is an “odd” way to acquire socks.

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