Three-Card Straight From Combined Decks
You take $k \geq 1$ standard 52-card decks and shuffle them together into one big deck. You draw 3 cards uniformly at random without replacement.
A "three-card straight" means the three cards have three consecutive ranks (e.g., 5-6-7 or Q-K-A). An Ace can serve as either the low card (A-2-3) or the high card (Q-K-A). Suits don't matter -- straight flushes count too.
Let $p_k$ be the probability the three drawn cards form a three-card straight. Find $\displaystyle\lim_{k \to \infty} p_k$.
Hints
- What happens to sampling without replacement when the number of copies of each card grows without bound?
- In the limit, the three draws are independent and uniform over the 52 card types. Count favorable outcomes by choosing the lowest rank, then the ordering, then the suits.
- There are 12 valid starting ranks (A through Q), $3! = 6$ orderings of the three ranks, and $4^3 = 64$ suit combinations. The total sample space is $52^3$.
Worked Solution
How to Think About It: When you combine $k$ decks and only draw 3 cards, the key question is: what happens to "without replacement" as $k$ grows? With $52k$ cards in the deck, removing 3 is negligible -- the conditional probabilities barely shift from the unconditional ones. In the limit, drawing without replacement from $k$ decks behaves exactly like drawing with replacement from a single deck. So the problem reduces to: pick 3 cards independently and uniformly from a standard 52-card deck. What is the probability they form a consecutive-rank triple?
Quick Estimate: There are 13 ranks. The valid lowest ranks for a 3-card straight are A through Q, giving 12 starting points. For each starting point, the 3 ranks are fixed. Each rank has 4 suits, so $4^3 = 64$ suit combinations per straight. The 3 cards can arrive in any of $3! = 6$ orderings. Total favorable outcomes: $12 \times 6 \times 64 = 4{,}608$. Total outcomes (with replacement): $52^3 = 140{,}608$. Probability: $4{,}608 / 140{,}608 = 72/2{,}197 \approx 0.03277$. So roughly a 3.3% chance.
Approach: Formalize the with-replacement argument via the $k \to \infty$ limit.
Formal Solution:
When $k \to \infty$, the deck has $52k$ cards. The probability of drawing any particular card on the first draw is $k/(52k) = 1/52$. After removing one card, the probability of drawing a particular card on the second draw is $k/(52k - 1)$ (or $(k-1)/(52k-1)$ if it shares a rank-suit with the first card). Either way, as $k \to \infty$ this approaches $1/52$. The same holds for the third draw. So in the limit, the three draws are independent, each uniform over the 52 card types.
Now count three-card straights under with-replacement sampling:
- Number of valid rank triples: The lowest rank in a straight can be A, 2, 3, ..., Q -- that is 12 choices. Each choice fixes a unique triple of consecutive ranks.
- Orderings: The three ranks can appear in any of $3! = 6$ orders among the three draws.
- Suit choices: Each of the 3 cards can be any of 4 suits independently, giving $4^3 = 64$ combinations.
- Total favorable: $12 \times 6 \times 64 = 4{,}608$.
- Total outcomes: $52^3 = 140{,}608$.
Therefore:
$$\lim_{k \to \infty} p_k = \frac{12 \times 3! \times 4^3}{52^3} = \frac{4{,}608}{140{,}608} = \frac{72}{2{,}197}$$
Answer: $\displaystyle\lim_{k \to \infty} p_k = \frac{72}{2{,}197} \approx 0.03277$.
Intuition
The core idea is that sampling without replacement from a huge population converges to sampling with replacement. This is the same principle behind why polling 1,000 people gives nearly the same margin of error whether the country has 10 million or 300 million residents -- once the sample is a tiny fraction of the population, the dependence between draws vanishes. In quant work, this shows up whenever you approximate hypergeometric draws with binomial ones, which is valid as long as your sample is small relative to the pool.
The counting itself is a clean exercise in multiplicative counting: fix the rank pattern (12 choices for the base), account for the order the cards appear (3! permutations), and independently assign suits ($4^3$). A common mistake is to forget that the three cards can be drawn in any order, or to accidentally exclude A-2-3 or Q-K-A by miscounting the number of valid starting ranks.