Two Subdecks: Race to the First Ace
A standard 52-card deck (4 aces, 48 non-aces) is split into two subdecks:
- Deck A: 39 cards containing 3 aces and 36 non-aces.
- Deck B: 13 cards containing 1 ace and 12 non-aces.
Two players each independently draw cards one at a time from the top of their respective deck (after a uniformly random shuffle). Each player stops when they draw their first ace. Let $N_A$ and $N_B$ be the number of cards drawn (including the ace) from Deck A and Deck B respectively.
The player with the smaller draw count wins (ties are ignored).
- Compute $P(N_B < N_A)$ -- the probability that the 13-card deck wins.
- Compute $P(N_A < N_B)$ -- the probability that the 39-card deck wins.
- Which deck should you choose?
Hints
- The ace in Deck B is equally likely to be in any of the 13 positions. For Deck A, think about the probability that the first $k$ cards are all non-aces.
- Use $P(N_A > k) = \binom{36}{k}/\binom{39}{k}$ -- the probability that $k$ cards drawn from Deck A are all non-aces. Then $P(N_B < N_A) = \frac{1}{13}\sum_{b=1}^{12} P(N_A > b)$.
- The expected first-ace positions are $E[N_B] = 7$ and $E[N_A] = 10$. This already suggests Deck B is better, and the exact calculation confirms it.
Worked Solution
How to Think About It: Each deck is randomly shuffled, so the position of the first ace is uniformly distributed among the "possible" positions. For Deck B (13 cards, 1 ace), the ace is equally likely to be in any of the 13 positions, so $N_B$ is uniform on $\{1, 2, \ldots, 13\}$. For Deck A (39 cards, 3 aces), the position of the first ace is determined by the arrangement of 3 aces among 39 cards. The key quantity is $P(N_A = k)$ -- the probability the first ace in Deck A appears at position $k$.
Quick Estimate: Deck B has expected first-ace position $E[N_B] = (13+1)/2 = 7$. For Deck A, the expected position of the first of 3 aces among 39 cards is $(39+1)/(3+1) = 10$. So on average, Deck B draws fewer cards. This suggests Deck B is the better choice. But we need exact probabilities.
Approach: Compute the PMFs of $N_A$ and $N_B$, then sum $P(N_B < N_A) = \sum_{b=1}^{12} \sum_{a=b+1}^{37} P(N_B = b) P(N_A = a)$.
Formal Solution:
PMF of $N_B$: With 1 ace in 13 cards, the ace is equally likely in any position: $P(N_B = b) = \frac{1}{13}, \quad b = 1, 2, \ldots, 13$
PMF of $N_A$: The first ace appears at position $k$ means positions
More directly: the first ace is at position $k$ when all $k-1$ cards before it are non-aces and the $k$-th card is an ace: $P(N_A = k) = \frac{\binom{36}{k-1}}{\binom{39}{k}} \cdot \frac{3}{39 - (k-1)} \cdot \frac{\binom{39-k}{2}}{\ldots}$
The cleanest derivation uses the fact that the first ace among 3 aces in 39 cards has the negative hypergeometric distribution. Equivalently, the positions of the 3 aces are a random size-3 subset of $\{1, \ldots, 39\}$, and we want the minimum. The CDF is: $P(N_A > k) = \frac{\binom{36}{k}}{\binom{39}{k}} = \frac{36! \cdot (39-k)!}{(36-k)! \cdot 39!}$
This is the probability that all $k$ of the first positions are non-aces, i.e., the $k$ drawn cards come from the 36 non-aces.
So: $P(N_A = k) = P(N_A > k-1) - P(N_A > k) = \frac{\binom{36}{k-1}}{\binom{39}{k-1}} - \frac{\binom{36}{k}}{\binom{39}{k}}$
for $k = 1, 2, \ldots, 37$ (the latest the first ace can appear is position 37, since there are 3 aces among 39 cards).
Computing the probabilities: Since $N_A$ and $N_B$ are independent:
$P(N_B < N_A) = \sum_{b=1}^{12} P(N_B = b) \cdot P(N_A > b) = \frac{1}{13} \sum_{b=1}^{12} \frac{\binom{36}{b}}{\binom{39}{b}}$
$P(N_A < N_B) = \sum_{a=1}^{37} P(N_A = a) \cdot P(N_B > a) = \sum_{a=1}^{12} P(N_A = a) \cdot \frac{13 - a}{13}$
(For $a \geq 13$, $P(N_B > a) = 0$ since $N_B \leq 13$.)
Computing $P(N_A > b) = \binom{36}{b}/\binom{39}{b}$ for each $b$:
| $b$ | $P(N_A > b)$ | |-----|---------------| | 1 | $36/39 = 12/13 \approx 0.9231$ | | 2 | $(36 \cdot 35)/(39 \cdot 38) = 1260/1482 \approx 0.8502$ | | 3 | $(36 \cdot 35 \cdot 34)/(39 \cdot 38 \cdot 37) = 42840/54834 \approx 0.7813$ | | 4 | $\approx 0.7157$ | | 5 | $\approx 0.6531$ | | 6 | $\approx 0.5934$ | | 7 | $\approx 0.5363$ | | 8 | $\approx 0.4818$ | | 9 | $\approx 0.4296$ | | 10 | $\approx 0.3796$ | | 11 | $\approx 0.3318$ | | 12 | $\approx 0.2860$ |
Sum $= \sum_{b=1}^{12} P(N_A > b) \approx 7.3619$
$P(N_B < N_A) = \frac{7.3619}{13} \approx 0.5663$
For $P(\text{tie})$: $P(N_A = N_B) = \frac{1}{13}\sum_{k=1}^{13} P(N_A = k) = \frac{1}{13}\sum_{k=1}^{13}[P(N_A > k-1) - P(N_A > k)]$.
Using $P(N_A > 0) = 1$ and $P(N_A > 13) \approx 0.1498$: $P(N_A = N_B) = \frac{1 - 0.1498}{13} \approx 0.0654$
$P(N_A < N_B) = 1 - P(N_B < N_A) - P(N_A = N_B) \approx 1 - 0.5663 - 0.0654 \approx 0.3683$
Answer: $P(N_B < N_A) \approx 0.566$ and $P(N_A < N_B) \approx 0.368$. Choose Deck B (the 13-card deck). It wins about 57% of the time (ignoring ties), because concentrating 1 ace among 13 cards gives a lower expected first-ace position than spreading 3 aces among 39 cards.
Intuition
The fundamental insight is that the expected position of the first success in sampling without replacement from a population of $N$ items with $K$ successes is $(N+1)/(K+1)$. For Deck A: $(39+1)/(3+1) = 10$. For Deck B: $(13+1)/(1+1) = 7$. So Deck B finds its ace about 3 draws sooner on average.
But this problem asks for the win probability, not just the expected value. The full distribution matters, and Deck B wins because its distribution is more tightly concentrated around a lower mean. Even though Deck A occasionally finds an ace early (it has 3 chances), Deck B's single ace being in a much smaller deck gives it a stochastic advantage. This is a nice example of how concentrating resources (one ace in a small deck) beats spreading them out (three aces in a large deck) when you care about the minimum draw count.