You’re in a casino offered a curious card game involving a standard 52-card deck. The rules seem simple—and tempting:

Game Mechanics

  1. A full 52-card deck is well shuffled.
  2. Cards are turned over two at a time until the deck is exhausted.
  3. For each pair:
    • If both are red, they go into your pile.
    • If both are black, they go into the dealer’s pile.
    • If one red and one black, both are discarded.
  4. At the end:
    • If your pile has more cards, you win $100.
    • If not (tie or dealer has more), you win nothing.
  5. You get to choose the entry fee before playing.

What Should You Pay?

Let’s analyze the symmetry and probability:

  • The deck has 26 red and 26 black cards.
  • Pairs are drawn randomly.
  • Every red card that goes to you is matched by a black card that can go to the dealer.
  • Whenever two red cards go to you, a black-black pair is equally likely.

Key Insight

Over many games, you’ll find that the number of red-red pairs (your gain) and black-black pairs (dealer’s gain) will be exactly equal in expectation. In fact, the red and black pairs are symmetric due to identical counts and uniform shuffling.

Mathematical Outcome

The expected number of red pairs equals the expected number of black pairs. So, the expected win is $0.

Moreover, the dealer wins ties, so your chances of actually winning $100 are 0%.

Hence, the expected value of the game is zero.

Final Answer

The maximum fee you should pay is $0.

Playing this game has zero expected value—it’s a cleverly disguised losing proposition.

Reference