Color Elimination Card Game

Expectation · Medium · Free problem

Jane proposes the following game. Start with a standard 52-card deck (26 red, 26 black). You take turns drawing two cards at a time. Jane goes first.

  • If both cards are black, Jane keeps them.
  • If both cards are red, you keep them.
  • If one is red and one is black, both are discarded.

When the deck is exhausted, you and Jane combine your kept cards, shuffle them back together, and repeat the process. The game continues until one color is completely eliminated.

If there are red cards remaining at the end, you win $\

$. What is a fair price to pay for this game?

Hints

  1. Think about what the discard rule does to the relative counts of red and black cards. Is there a quantity that is preserved across rounds?
  2. In each round, mixed pairs remove one red and one black card together. Count the red pairs kept ($r$) and black pairs kept ($b$) using the total card accounting equations -- what relationship must $r$ and $b$ satisfy?
  3. After every round the reshuffled deck has exactly r$ red and b$ black cards with $r = b$. If the counts are always equal, the only way to eliminate one color is to eliminate both at once -- so red cards never survive alone.

Worked Solution

Setup: Start with a standard 52-card deck (26 red, 26 black). Pairs are drawn and sorted: black pairs to Jane, red pairs to you, mixed pairs discarded. Kept cards are reshuffled and the process repeats until one color is eliminated. You win $\ $ if red cards remain. We need the fair price.

Key Observation -- The Discard Rule Preserves Balance: Every mixed pair that is discarded removes exactly one red and one black card together. Same-color pairs remove cards of only one color. The critical question is whether the reshuffled deck can ever become imbalanced.

Counting One Full Round: Suppose in a given round the deck has $R$ red and $B$ black cards with $R = B$ (true initially since $R = B = 26$). After drawing all pairs: - Let $r$ = number of red pairs you keep, $b$ = number of black pairs Jane keeps, $d$ = number of mixed pairs discarded. - Total cards drawn: r + 2b + 2d = R + B$, so $r + b + d = \frac{R + B}{2}$. - Red card accounting: r + d = R$. - Black card accounting: b + d = B$. - Subtracting: r - 2b = R - B = 0$, so $r = b$.

After the round, the reshuffled pile contains r$ red cards and b = 2r$ black cards. The red and black counts are equal again.

Invariant: If the deck starts with equal numbers of red and black cards, it will have equal numbers after every round. This holds at every stage of the game.

Conclusion: The game ends only when one color is fully eliminated. But since red and black counts are always equal, one color cannot vanish without the other vanishing simultaneously. The game always terminates with zero cards of both colors -- no red cards ever survive.

You can never win. The expected payout is $\$0$.

$\boxed{\text{Fair price} = \$0}$

Intuition

This is a beautiful invariant problem disguised as a game theory question. The trick is to ignore the sequential mechanics of drawing and focus on what is conserved. The discard rule is the key: mixed pairs always remove one red and one black, maintaining perfect balance. Same-color pairs kept by each player also maintain balance because they come back in the reshuffle. So the red-black parity is locked at equality forever.

This type of invariant argument shows up in many interview problems. The general strategy is: instead of simulating a complex process, find a quantity that never changes (or changes in a predictable way) and use that to determine the outcome without working through the details. Here, the invariant (equal counts of red and black) immediately tells you the outcome: both colors die together, so you can never win. Whenever a problem feels like it might have a "trick" answer, look for conservation laws.

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