Answer: The Collector's Last Coupon

Key Idea / Intuition

The game ends the moment the second color appears for the first time — equivalently, it ends at the flip that breaks the initial "all one color" run. The first card is always some color. We then wait until the first card of the opposite color appears. By symmetry, what matters is only the position of the first card of the minority color among a deck of $2n$ cards. This turns into a clean question about the position of a specific card type in a random ordering.

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Formal Proof / Solution

Setup. After flipping the first card (say it's red, by symmetry the argument is identical for black), we need the first black card among the remaining $2n - 1$ cards, of which exactly $n$ are black.

Key observation. Consider the positions of the $n$ black cards among the remaining $2n-1$ slots. By symmetry, the first black card is equally likely to be in any of the $2n-1$ positions — but that's not quite right. We need the minimum order statistic of $n$ uniform draws without replacement from $\{1, 2, \ldots, 2n-1\}$.

Expected position of the first black card. If we pick $n$ items uniformly at random (without replacement) from $\{1, 2, \ldots, 2n-1\}$, the expected value of the minimum is:

$\mathbb{E}[\min] = \frac{2n - 1 + 1}{n + 1} = \frac{2n}{n+1}.$

This uses the standard result: for the minimum of $n$ items chosen without replacement from $\{1, \ldots, N\}$,

$\mathbb{E}[\text{min}] = \frac{N+1}{n+1}.$

Here $N = 2n-1$, so $\mathbb{E}[\text{min}] = \frac{2n}{n+1}$.

Total expected flips. We flip the first card (1 flip always), then wait $\mathbb{E}[\text{min}]$ more flips to see the first opposite-color card:

$\mathbb{E}[\text{total}] = 1 + \frac{2n}{n+1} = \frac{n+1+2n}{n+1} = \frac{3n+1}{n+1}.$

Sanity checks:

• $n=1$: deck has 1 red, 1 black. You always flip exactly 2 cards. Formula gives $\frac{4}{2} = 2$. ✓

• $n=2$: deck has 2 red, 2 black. Formula gives $\frac{7}{3} \approx 2.33$.

  Manual check: P(stop at flip 2) = P(second card differs from first) = $\frac{2}{3}$; P(stop at flip 3) = $\frac{1}{3}$ (second card same, third must differ). $\mathbb{E} = 2 \cdot \frac{2}{3} + 3 \cdot \frac{1}{3} = \frac{7}{3}$. ✓

The elegant formula:

$\boxed{\mathbb{E}[\text{cards flipped}] = \frac{3n+1}{n+1}}$

As $n \to \infty$, this approaches $3$ — a beautiful limit saying that in a very large balanced deck, you expect to flip only about 3 cards before seeing both colors.

Why the limit is 3: In a huge deck, the probability the second card already differs from the first is $\approx 1/2$. If not, the third card differs with probability $\approx 1/2$, and so on. This geometric-like structure gives $\mathbb{E} \approx 1 + 2 = 3$ via a geometric series argument (expectation of a geometric with $p=1/2$, shifted).