[Math] probability of having an ace on each hand when dealing 52 cards to 4 hands

probabilityprobability theory

The question from DeGroot's book: "probability and statistics"

"Suppose that a deck of 52 cards containing four aces is shuffled thoroughly
and the cards are then distributed among four players so that each player
receives 13 cards. We shall determine the probability that each player will receive one ace."

I have so far 3 methods to solve this problem, all of them yield the number provided in the book. Below is my 4th method. Alas, it doesn't give the same answer:

since the question doesn't concern which suit of ace goes to which hand, and the differences between remaining cards are irrelevant, I simplify the question as:

"distribute 52 cards containing 4 A's and 48 B's to 4 piles so that each pile
receives 13 cards. determine the probability that each pile will receive one A."

to solve this simplified question, I focus on distribution of the 4 A's, then distribute the 48 B's to 4 piles to make each 13-card pile:

  1. there is only 1 way to distribute 4 A's to 4 piles.
  2. I summed up below sets of outcomes to compute the sample space:

number of combinations that one hand get all 4 A's: 4
number of combinations that one hand get 3 A's: 4×3
number of combinations that 2 hands each get 2 A's, the other 2 hands get 0 A: 6
number of combinations that one hand get 2 A's, two other hands each get 1 A: 12
number of combinations that each hand get 1 A: 1

so, here is the number:
1/35=0.0286
which is about 3.7 time smaller chance than the "correct answer" (0.1055).

Why does this way of counting not work ?

Thanks!

P.S.

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I think the first method given by the author partially applied this simplification: it doesn't distinguish different suit of aces.

the method is:
The number of possible different combinations of the four positions in the deck
occupied by the four aces is C(52,4). and there are exactly 13^4 of combinations for each ace in each player's hand. the result is:

p=13^4/C(52,4)= 0.1055

Best Answer

Your simplification doesn't work. Without the simplification, there are $24$ ways to give each player an ace while there are four ways to give all aces to one person. Your simplification skews this relation, and thus the probability.

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