Time Until the Second Ace

Expectation · Hard · Free problem

Cards are dealt one at a time from a well-shuffled standard 52-card deck (which contains 4 Aces). Let $T$ be the number of cards dealt up to and including the 2nd Ace.

  1. Find $P(T = t)$ for all valid $t$.
  1. Compute $E[T]$ in closed form.
  1. Compute $\text{Var}(T)$ in closed form.

Hints

  1. Think of the 4 Aces as occupying random positions in the deck. The question is really asking: where does the 2nd order statistic of 4 uniform draws from 52 positions land?
  2. For the expectation, define an indicator $Y_j$ for each non-Ace card that equals 1 if that card appears before the 2nd Ace. Consider the relative ordering of that card and the 4 Aces -- just 5 items.
  3. For the covariance term in the variance, consider 2 non-Aces and 4 Aces (6 items). Count how many of the $\binom{6}{2} = 15$ placements of the non-Aces put both before the 2nd Ace.

Worked Solution

How to Think About It: Imagine fanning out the deck face-down and flipping cards one by one. You stop the moment you see the second Ace. Before doing any math, think about where the 4 Aces sit in a random shuffle. By symmetry, the Aces divide the 52 positions into 5 gaps (before the 1st Ace, between the 1st and 2nd, between 2nd and 3rd, between 3rd and 4th, and after the 4th). Each gap gets an equal expected share of the 48 non-Aces, so each gap has expected size $48/5 = 9.6$. The 2nd Ace comes after the first gap, the 1st Ace, and the second gap -- so roughly at position $9.6 + 1 + 9.6 + 1 \approx 21$. That is your quick estimate.

Quick Estimate: The position of the $j$-th order statistic of $K$ items uniformly placed among $N$ slots is $j(N+1)/(K+1)$. For the 2nd Ace:

\times 53 / 5 = 21.2$. For the standard deviation, $\text{Var}(T)$ should be on the order of $N^2/(K+1)^2 \sim 52^2/25 \approx 108$, so $\text{SD} \approx 10$. (The exact variance turns out to be
01.76$, so $\text{SD} \approx 10.1$.)

Approach: We derive the PMF using a direct counting argument, then compute moments via indicator random variables.

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Part (i): PMF

The event $\{T = t\}$ means the $t$-th card is an Ace, and exactly 1 of the first $t - 1$ cards is also an Ace. In a uniformly random permutation of 52 distinct cards, every arrangement is equally likely.

This is the negative hypergeometric distribution: sampling without replacement from $N = 52$ items with $K = 4$ successes (Aces), the probability that the $r$-th success occurs on draw $t$ is

$P(T = t) = \frac{\binom{t-1}{r-1}\binom{N - t}{K - r}}{\binom{N}{K}}$

With $r = 2$, $K = 4$, $N = 52$:

$P(T = t) = \frac{(t-1)\,\binom{52 - t}{2}}{\binom{52}{4}}, \qquad t = 2, 3, \ldots, 50$

Why the formula works: Choose which 1 of the first $t-1$ positions holds the first Ace ($t-1$ choices), then choose which 2 of the remaining $52-t$ positions (after position $t$) hold the other 2 Aces ($\binom{52-t}{2}$ choices). Divide by the total ways to place 4 Aces among 52 positions ($\binom{52}{4} = 270{,}725$).

The valid range is $t = 2$ (both Aces drawn first) through $t = 50$ (the last 2 cards of the first 50 are both Aces; the remaining 2 Aces occupy positions 51 and 52, and the formula gives $\binom{2}{2} = 1$ way to place them). At $t = 51$: $\binom{1}{2} = 0$, so $P(T \geq 51) = 0$, confirming $t_{\max} = 50$.

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Part (ii): $E[T]$ via Indicator Variables

Write $T = 2 + Y$ where $Y$ counts the number of non-Ace cards drawn before the 2nd Ace. Define indicators $Y_j$ for each non-Ace card $j = 1, \ldots, 48$:

$Y_j = \begin{cases} 1 & \text{if non-Ace } j \text{ appears before the 2nd Ace} \\ 0 & \text{otherwise} \end{cases}$

To find $P(Y_j = 1)$, consider the relative ordering of non-Ace $j$ and the 4 Aces -- a group of 5 items. All $5!$ orderings are equally likely. Non-Ace $j$ appears before the 2nd Ace iff it falls in position 1 or 2 among these 5 items:

$P(Y_j = 1) = \frac{2}{5}$

By linearity of expectation:

$E[T] = 2 + \sum_{j=1}^{48} P(Y_j = 1) = 2 + 48 \cdot \frac{2}{5} = 2 + \frac{96}{5} = \boxed{\frac{106}{5} = 21.2}$

This matches the order-statistic formula $E[T] = r(N+1)/(K+1) = 2 \times 53/5$.

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Part (iii): $\text{Var}(T)$ via Indicators

Since $T = 2 + \sum Y_j$, we have $\text{Var}(T) = \text{Var}\!\left(\sum_{j=1}^{48} Y_j\right)$.

Individual variances:

$\text{Var}(Y_j) = \frac{2}{5} \cdot \frac{3}{5} = \frac{6}{25}$

Covariances: For two distinct non-Aces $i \neq j$, consider the relative ordering of these 2 non-Aces and the 4 Aces (6 items total). $Y_i Y_j = 1$ iff both non-Aces appear before the 2nd Ace, i.e., both land in the positions before the 2nd Ace among these 6 items.

By exhaustive enumeration over all $\binom{6}{2} = 15$ ways to place the 2 non-Aces among 6 positions, exactly 3 placements satisfy the condition ($\{1,2\}, \{1,3\}, \{2,3\}$ when the remaining 4 Ace positions yield a 2nd order statistic larger than both non-Ace positions). So:

$P(Y_i Y_j = 1) = \frac{3}{15} = \frac{1}{5}$

$\text{Cov}(Y_i, Y_j) = \frac{1}{5} - \left(\frac{2}{5}\right)^2 = \frac{1}{5} - \frac{4}{25} = \frac{1}{25}$

Putting it together:

$\text{Var}(T) = 48 \cdot \frac{6}{25} + 2\binom{48}{2} \cdot \frac{1}{25} = \frac{288}{25} + \frac{2 \times 1128}{25} = \frac{288 + 2256}{25} = \boxed{\frac{2544}{25} = 101.76}$

This matches the negative hypergeometric variance formula $\text{Var}(T) = \frac{r(N+1)(N-K)(K-r+1)}{(K+1)^2(K+2)} = \frac{2 \cdot 53 \cdot 48 \cdot 3}{25 \cdot 6} = \frac{2544}{25}$.

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Answer:

Intuition

This problem is a clean illustration of the indicator variable method for computing moments of stopping times in sampling-without-replacement problems. The key move is reframing the question: instead of tracking the sequential process ("when does the 2nd Ace appear?"), you ask each non-Ace card a yes/no question ("do you appear before the 2nd Ace?"). By considering only the relative ordering of each non-Ace against the 4 Aces -- a group of just 5 items -- you get $P(Y_j = 1) = 2/5$ without any complicated combinatorics. The covariance calculation uses the same trick with 6 items instead of 5.

This technique generalizes immediately: for the $r$-th success among $K$ special items in $N$ total, the expected stopping time is $r(N+1)/(K+1)$, and you can always compute it by symmetry arguments on small groups. In practice, this kind of reasoning shows up whenever you need the distribution of an order statistic from a finite population -- for example, estimating when a particular event will occur in a finite sequence of trades, or computing the expected time to fill a basket of orders. The indicator approach almost always beats direct PMF summation for moments.

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