Solution
One Ace Apiece
Show the problem again
A standard 52-card deck is shuffled uniformly at random and dealt into four hands of 13 cards each. What is the probability that each hand contains exactly one ace? Give the exact reduced fraction, or a decimal rounded to four places.
Worked solution
The answer is 2197/20825 ≈ 0.1055. Only the positions of the four aces matter: they occupy a uniformly random set of 4 of the 52 deal slots. Place them one at a time. The first ace can go anywhere. The second must avoid the first ace's hand, which has 39 of the remaining 51 slots. The third has 26 of 50, and the fourth has 13 of 49: (39/51)(26/50)(13/49) = 13182/124950 = 2197/20825. Equivalently, 13^4 · 4!/(52·51·50·49) after ordering the aces. An even split happens only about 10.5% of the time, against roughly 58% for a 2-1-1-0 split.
Source: Classic card-dealing exercise found in many probability texts (for example Feller, An Introduction to Probability Theory). Statement written for AxiomIQ.