One ace per hand

Fifty-two cards are shuffled and dealt to 4 players, 13 cards each. What's the probability that each player gets exactly one ace? Watch the cards being dealt one at a time and pay attention to where the aces land:

Deal repeatedly and watch the success rate — after enough deals, it settles at about 10.5%:

The conditional chain

Instead of counting arrangements (which involves enormous factorials), use conditional probability. Track the aces one at a time.

The counting approach

We can verify using combinatorics. The total number of ways to distribute 52 cards into four hands of 13:

$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13} = \frac{52!}{(13!)^4}$

For the favorable outcomes: first distribute the 4 aces (one per player) in $4! = 24$ ways, then distribute the remaining 48 cards into four hands of 12:

$4! \times \frac{48!}{(12!)^4}$

Why this probability is what it is

The value 10.55% might seem low, but consider: with 4 aces and 4 players, there are $4^4 = 256$ unrestricted ways to assign aces to players (if each ace independently chose a random player). Only $4! = 24$ of those assignments put each ace in a different hand. If the assignments were independent, $P = 24/256 = 9.375\%$.

The actual cards aren't independent (dealing one card affects where the next goes), which slightly increases the probability to 10.55%. The dealing process is "sampling without replacement," which makes even distribution slightly more likely than the independent case.

Competition appearances

This problem appears in quant interviews as a warm-up at Goldman Sachs and Morgan Stanley, in bridge as the probability of a "flat" ace distribution (important for bidding strategy), in poker and card game analysis, and in AMC/AIME problems about distributing objects into groups with constraints.

This technique works for any problem about distributing special items among groups. Explore how the probability changes with different numbers of special cards and players:

The takeaway

Instead of counting a complex combinatorial event all at once, break it into a chain of simple conditional steps. Each step asks "does the next special item go where we need it?" and the chain rule multiplies the answers. This is less error-prone and makes the structure easier to see than factorials do.