On: Coin Puzzle: Predict the Other's Coin

[pliny]

Your analysis is incorrect because A and B are allowed to decide on a strategy ahead of time - and there are two simple and similar strategies that allow A and B to guess correctly, together, half the time.

The first is described by: 'If I get heads, I will guess you got heads, otherwise I will guess you got tails' The second: 'If I get heads, I will guess you got tails, otherwise I will guess you got heads'

These are functionally identical, in that in half of the cases they get the same result (first strategy wins these) and in the other half they get opposite results (second strategy wins these) - they get 2$ for winning and give 1$ when they lose, since they win half the time under either strategy C should not play this game.

[kator → pliny]

Play with different strategies:

  import random
 
  # Modify strategy here
  def turn():
      toss = random.randint(0, 1)
      guess = random.randint(0, 1)
      # Uncomment to make A/B win
      # guess = toss
      return toss, guess

  random.seed()

  score_ab, score_c = 0, 0

  for i in range(1, 1001):
      score_ab -= 1
      score_c  += 1
      toss_a, guess_a = turn()
      toss_b, guess_b = turn()

      if toss_a == guess_b and toss_b == guess_a:
        score_ab += 3
        score_c  -= 3

      print i, 'ab:', score_ab, 'c:', score_c

  if score_ab > score_c:
      print "Players A, B win"
  else:
      print "Player C wins"