Question

Question:What is the probability that each player has a hand containing an ace when the 52 cards of a standard deck are dealt to four players?

Short answer

Answer

\(\frac{{2197}}{{20,825}} \approx 0.1055\)

Step-by-step solution

Probability for Deck of Cards and Combination formula

A standard deck of cards contains\(52\)cards, of which\(26\)are red and\(26\)are black,\(13\)are of each suit (hearts, diamonds, spades, clubs) and of which\(4\)are of each denomination (\(A,\;2\)to\(10,\;\;{\rm{J}},\;{\rm{Q}},\;{\rm{K}}\)). The face cards are the jacks\(J\), queens\(Q\)and kings\({\rm{K}}\).

Product rule If one event can occur in\(m\)ways AND a second event can occur in\(n\)ways, then the number of ways that the two events can occur in sequence is then\(m \cdot n\).

Definition permutation (order is important):

No repetition allowed:\(P(n,r) = \frac{{n!}}{{(n - r)!}}\)

Repetition allowed:\({{\bf{n}}^r}\)

Definition combination (order is not important):

No repetition allowed: \(C(n,r) = \left( {\begin{aligned}{*{20}{l}}n\\r\end{aligned}} \right) = \frac{{n!}}{{r!(n - r)!}}\)

Repetition allowed:\(C(n + r - 1,r) = \frac{{(n + r - 1)!}}{{r!(n - 1)!}}\)

with \(n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1\).

Use Probability and Combination for Deck of Cards 

The order of the cards does not matter (since a different order will result in the same hand), thus we need to use the definition of a combination.

Total hands when the selected \(13\) cards from the \(52\)cards in the deck for the first person. Repetition is not allowed (since one card cannot be drawn more than once).

\(\begin{aligned}{}C(52,13) &= \frac{{52!}}{{13!(52 - 13)!}}\\C(52,13) &= \frac{{52!}}{{13!39!}}\\C(52,13) &= 635,013,559,600C(39,13)\end{aligned}\)

Then \(13\) cards have to be selected from the remaining \(39\)cards for the second person.

\(\begin{aligned}{}C(39,13) &= \frac{{39!}}{{13!(39 - 13)!}}\\C(39,13) &= \frac{{39!}}{{13!26!}}\\C(39,13) &= 8,122,425,444\end{aligned}\)

Then \(13\) cards have to be selected from the remaining \(26\) cards for the third person.

\(\begin{aligned}{}C(26,13) &= \frac{{26!}}{{13!(26 - 13)!}}\\C(26,13) &= \frac{{26!}}{{13!13!}}\\C(26,13) &= 10,400,600\end{aligned}\)

Then \(13\) cards have to be selected from the remaining \(13\) cards for the fourth person.

\(\begin{aligned}{}C(26,13) &= \frac{{13!}}{{13!(13 - 13)!}}\\C(26,13) &= \frac{{13!}}{{13!0!}} &= 1{\rm{ }}\\{\rm{Use the product rule }} &= 635,013,559,600 \cdot 8,122,425,444 \cdot 10,400,600 \cdot 1\\{\rm{Use the product rule }} &= 53,644,737,765,488,792,839,237,440,000\end{aligned}\)

Use Probability for Deck of Cards and Combination for the case of Ace in each hand

Each hand contains ace.

When first deal the \(4\) aces to the \(4\) people. The order will matter (because a different order will lead to different people getting different aces) and thus it has to use a permutation in this case.

\(\begin{aligned}{}P(4,4) &= \frac{4}{{(4 - 4)!}}\\P(4,4) &= \frac{{4!}}{{0!}}\\P(4,4) &= 4!\\P(4,4) &= 24\end{aligned}\)

Now select \(12\) cards from the \(48\) cards in the deck for the first person. Repetition is not allowed (since one card cannot be drawn more than once).

\(\begin{aligned}{}C(48,12) &= \frac{{48!}}{{12!(48 - 12)!}}\\C(48,12) &= \frac{{48!}}{{12!36!}}\\C(48,12) &= 69,668,534,468\end{aligned}\)

Then \(12\) cards have to be selected from the remaining 36 cards for the second person.

\(\begin{aligned}{}C(36,12) &= \frac{{36!}}{{12!(36 - 12)!}}\\P(36,12) &= \frac{{36!}}{{12!24!}}\\C(36,12) &= 1,251,677,700\end{aligned}\)

Then \(12\) cards have to be selected from the remaining \(24\) cards for the third person.

\(\begin{aligned}{}C(24,12) &= \frac{{24!}}{{12!(24 - 12)!}}\\C(24,12) &= \frac{{24!}}{{12!12!}}\\C(24,12) &= 2,704,156\end{aligned}\)

Then \(13\) cards have to be selected from the remaining \(12\) cards for the fourth person.

\(\begin{aligned}{}C(12,12) &= \frac{{12!}}{{12!(12 - 12)!}}\\C(12,12) &= \frac{{12!}}{{12!0!}}\\C(12,12) &= 1{\rm{ }}\\{\rm{Use the product rule }} &= 69,668,534,468 \cdot 1,251,677,700 \cdot 2,704,156 \cdot 1\\{\rm{Use the product rule }} &= 235,809,301,462,142,612,780,721,600\end{aligned}\)

Use Probability for Favourable outcomes 

The probability is the number of favourable outcomes divided by the number of possible outcomes:

\(\begin{aligned}{}P({\rm{ Each is dealt an ace }}) &= \frac{{{\rm{ of favorable outcomes }}}}{{{\rm{ of possible outcomes }}}}\\P({\rm{ Each is dealt an ace }}) &= \frac{{P(4,4) \cdot C(48,12) \cdot C(36,12) \cdot C(24,12) \cdot C(12,12)}}{{C(52,13) \cdot C(39,13) \cdot C(26,13) \cdot C(13,13)}}\\P({\rm{ Each is dealt an ace }}) &= \frac{{235,809,301,462,142,612,780,721,600}}{{53,644,737,765,488,792,839,237,440,000}}\\P({\rm{ Each is dealt an ace }}) &= \frac{{2197}}{{20,825}}\\P({\rm{ Each is dealt an ace }}) \approx 0.1055\end{aligned}\)