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

You don’t even need to rely on probability or expected values! Let’s look at the deterministic math of every single deck:

  • Every mixed pair (discarded) consumes exactly 1 red card and 1 black card.
  • The remaining cards form the mono-color pairs (your pile and the dealer’s pile).

Mathematical Outcome

Let \( M \) be the number of mixed pairs. These \( M \) pairs consume exactly \( M \) red cards and \( M \) black cards from the deck.

This leaves exactly \( 26 - M \) red cards for your pile, and \( 26 - M \) black cards for the dealer’s pile.

Because your pile and the dealer’s pile will always have the exact same number of cards, the game always ends in a tie, regardless of how the deck is shuffled.

Since the rules state you only win if you have more cards, your chances of winning are exactly 0%.

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