Matching Royals: Each Player Gets One of Each Rank
A reduced deck of 16 cards consists of 4 ranks (Jack, Queen, King, Ace), each appearing in all 4 suits. The deck is dealt evenly to 4 players, so each player receives exactly 4 cards.
What is the probability that every player receives exactly one card of each rank? Round your answer to the nearest ten-thousandth.
Hints
- Think about the problem rank by rank: for each of the 4 ranks, how many ways can you assign the 4 suited cards to the 4 players?
- The total number of ways to deal 16 cards into 4 hands of 4 is the multinomial $\binom{16}{4,4,4,4} = 16!/(4!)^4$.
- Each rank contributes $4!$ favorable permutations independently, giving $(4!)^4$ favorable outcomes total.
Worked Solution
How to Think About It: We are dealing 16 distinct cards into 4 hands of 4 cards each, and we want each hand to be a "rainbow" -- one card from each rank. Think of it rank by rank: the 4 Jacks go to 4 players, the 4 Queens go to 4 players, and so on. Each rank's 4 cards must be distributed one-per-player. The total number of deals is the multinomial, and the favorable outcomes are the product of permutations for each rank.
Quick Estimate: There are 4 ranks, each with 4 cards to assign to 4 players. For the first rank, all $4! = 24$ assignments are equally likely (one card per player). For the second rank, given the first is "used up" in terms of rank slots but not in terms of card slots, we again have $4!$ ways. But the constraint interacts with the dealing -- let me just compute directly.
Approach: Count favorable outcomes (one card of each rank per player) over total outcomes (arbitrary deals of 4 cards each).
Formal Solution:
Total deals: We distribute 16 distinct cards into 4 groups of 4. The number of ways is the multinomial coefficient:
$$\binom{16}{4, 4, 4, 4} = \frac{16!}{(4!)^4}$$
Favorable deals: For each rank (4 cards, one per suit), we assign one card to each of the 4 players. That is a permutation of 4 items: $4!$ ways per rank. Since there are 4 ranks and the assignments are independent:
$$\text{Favorable} = (4!)^4$$
Probability:
$$P = \frac{(4!)^4}{\binom{16}{4,4,4,4}} = \frac{(4!)^4}{\frac{16!}{(4!)^4}} = \frac{(4!)^8}{16!}$$
Now compute numerically:
- $4! = 24$
- $(4!)^8 = 24^8 = 110{,}075{,}314{,}176$
- $16! = 20{,}922{,}789{,}888{,}000$
$$P = \frac{110{,}075{,}314{,}176}{20{,}922{,}789{,}888{,}000} \approx 0.005260$$
Rounding to the nearest ten-thousandth: $P \approx 0.0053$.
Let me verify: $24^8 = 24^2 \times 24^2 \times 24^2 \times 24^2 = 576^4$. $576^2 = 331{,}776$. $576^4 = 331{,}776^2 = 110{,}075{,}314{,}176$. Check. $16! = 20{,}922{,}789{,}888{,}000$. The ratio is $110{,}075{,}314{,}176 / 20{,}922{,}789{,}888{,}000 \approx 0.005260$.
Answer: The probability that each player receives one card of each rank is:
$$P = \frac{(4!)^8}{16!} \approx 0.0053$$
Intuition
This is a counting problem with a clean combinatorial structure. The key insight is to decompose by rank: within each rank, the 4 cards (one per suit) must be distributed one to each player. That is simply a permutation -- $4!$ ways per rank -- and the ranks are independent once you frame it this way. The total is a multinomial coefficient because you are partitioning 16 distinct objects into 4 groups.
The probability works out to about half a percent, which makes intuitive sense: there are many ways to deal 16 cards, and requiring each hand to be a perfect rainbow is quite restrictive. Problems like this -- counting favorable configurations in a random partition -- are common in combinatorial probability. The technique of decomposing by "type" (here, rank) and multiplying independent permutations is broadly applicable to card-dealing, ball-in-urn, and assignment problems.