One Ace Each: Dealing 52 Cards to Four Players
A standard 52-card deck is shuffled and dealt out completely to four players, 13 cards each. What is the probability that every player receives exactly one ace?
Hints
- Focus only on where the four aces land; the other 48 cards are irrelevant. Place the aces one at a time.
- The first ace can go anywhere. The second ace must avoid the 12 remaining slots in the first ace's hand: 39 good slots out of 51.
- Multiply: $1 \cdot \frac{39}{51} \cdot \frac{26}{50} \cdot \frac{13}{49}$.
Worked Solution
How to Think About It: A completed deal is a uniformly random partition of the deck into four labeled hands of 13. The only thing that matters is which hand each ace lands in, so you can ignore the 48 non-aces and just track the four aces as they are placed into the 52 slots one at a time.
Quick Estimate: Each ace after the first needs to avoid the hands already holding an ace. The three factors are roughly $0.76$, $0.52$ and $0.27$; their product is about $0.105$.
Formal Solution:
*Step 1 -- Sequential placement of aces.* Think of the 52 card positions as 4 hands of 13 slots. Drop the aces in one by one into uniformly random unused slots.
- Ace 1 lands in some hand (probability 1).
- Ace 2 must avoid the 12 remaining slots of that hand: $39/51$.
- Ace 3 must avoid the two hands already holding an ace ($2 \times 12 = 24$ slots): $26/50$.
- Ace 4 must land in the last empty-of-aces hand: $13/49$.
$$P = \frac{39}{51} \cdot \frac{26}{50} \cdot \frac{13}{49} = \frac{13{,}182}{124{,}950} = \frac{2197}{20825} \approx 0.1055.$$
*Step 2 -- Combinatorial check.* Count deals directly: assign the aces to the four hands ($4!$ ways), then fill each hand with non-aces: $\binom{48}{12}\binom{36}{12}\binom{24}{12}$ ways, out of $\binom{52}{13}\binom{39}{13}\binom{26}{13}$ total deals. This ratio simplifies to the same $2197/20825$.
*Step 3 -- Interpretation.* $2197 = 13^3$ and $20825 = 51 \cdot 50 \cdot 49 / 6$; the answer is $13^3 / \binom{51}{3}$: choose the three slots of the other three aces relative to the first, and they must fall one per hand.
Answer: $P = \dfrac{39 \cdot 26 \cdot 13}{51 \cdot 50 \cdot 49} = \dfrac{2197}{20825} \approx 0.1055$.
Intuition
Only the positions of the four aces matter, and each successive ace must land in a hand that does not yet hold one: $39/51$, then $26/50$, then $13/49$. The answer, about $10.5\%$, is much smaller than the naive "one in four" many people guess, because "aces spread evenly" is one specific pattern among the many ways four aces can be distributed. Sequential placement of the special items is the cleanest way to handle "each group gets exactly one" questions, whether the items are aces, defaults in a portfolio, or fills across venues.