Question

Question: What is the probability that a \(13\) -card bridge hand contains

a) all \(13\) hearts?

b) \(13\) cards of the same suit?

c) seven spades and six clubs?

d) seven cards of one suit and six cards of a second suit?

e) four diamonds, six hearts, two spades, and one club?

f) four cards of one suit, six cards of a second suit, two cards of a third suit, and one card of the fourth suit?

Short answer

Answer

a) The probability of a \(13\) card bridge hand contains all \(13\) hearts is \(\frac{1}{{C(52,13)}}\).

b) The probability of a \(13\) card bridge hand contains \(13\) cards of same suit is \(\frac{4}{{C(52,13)}}\).

c) The probability of a \(13\) card bridge hand contains seven spades and six clubs is \(\frac{{2,944,656}}{{C(52,13)}}\).

d) The probability of a \(13\) card bridge hand contains seven cards of one suit and six cards of second suit is \(\frac{{35,335,872}}{{C(52,13)}}\).

e) The probability of a \(13\) card bridge hand contains four diamonds, six hearts, two spades and one club is \(\frac{{1,244,117,160}}{{C(52,13)}}\).

f) The probability of a \(13\) card bridge hand contains four cards of one suit, six cards of a second suit, two cards of a third suit and one card of the fourth suit is \(\frac{{2,944,656}}{{C(52,13)}}\).

Step-by-step solution

Given data

A \(13\) card bridge hand contains four cards of one suit, six cards of a second suit, two cards of a third suit and one card of the fourth suit.

Concept of Combination

A combination is a selection of items from a set that has distinct members.

Formula:

\(_n{C_r} = \frac{{n!}}{{r!(n - r)!}}\)

Calculation for the probability of \(13\) hearts

a)

A \(13\) card bridge hand contains all \(13\) hearts.

The number of possibilities to chose \(13\) cards from a deck is \(C(52,13)\).

The number of possibilities to chose \(13\) hearts from a deck is \(C(13,13) = 1\).

Then, the required probability is \(\frac{{C(13,13)}}{{C(52,13)}} = \frac{1}{{C(52,13)}}\).

Therefore, the probability of a \(13\) card bridge hand contains all \(13\) hearts is \(\frac{1}{{C(52,13)}}\).

Calculation for the probability of \(13\) cards of the same suit 

b)

A \(13\) card bridge hand contains \(13\) cards of same suit.

We know that there are four suits in deck, the number of possibilities to chose \(13\) cards of the same suit is \(C(4,1) \times C(13,13) = 4\).

Then, the required probability is \(\frac{{C(4,1) \times C(13,13)}}{{C(52,13)}} = \frac{4}{{C(52,13)}}\).

Therefore, the probability of a \(13\) card bridge hand contains \(13\) cards of same suit is \(\frac{4}{{C(52,13)}}\).

Calculation for the probability of \(7\) spades and \(6\) clubs

c)

A \(13\) card bridge hand contains seven spades and six clubs.

There are \(13\) spades and \(13\) clubs in the deck, the number of possibilities to chose \(6\) spades from \(13\) is \(C(13,6) = \frac{{13!}}{{7!6!}} = 1716\).

Then, the required probability is \(\frac{{C(13,6) \times C(13,7)}}{{C(52,13)}} = \frac{{2,944,656}}{{C(52,13)}}\).

Therefore, the probability of a \(13\) card bridge hand contains seven spades and six clubs is \(\frac{{2,944,656}}{{C(52,13)}}\).

Calculation for the probability of \(7\) cards of one suit and \(6\) cards of a second suit

d)

The number of possibilities to chose a suit from four is \(C(4,1) = 4\).

The number of possibilities to chose \(7\) cards of the same suit is \(C(4,1) \times C(13,7) = 6864\).

The number of possibilities to choose a suit from remaining three is \(C(3,1) = 3\).

The number of possibilities to chose \(6\) cards of the same suit is \(C(3,1) \times C(13,6) = 5148\)

Then, the required probability is \(\frac{{C(4,1) \times C(13,7) \times C(3,1) \times C(13,6)}}{{C(52,13)}} = \frac{{35,335,872}}{{C(52,13)}}\).

Therefore, the probability of a \(13\) card bridge hand contains seven cards of one suit and six cards of second suit is \(\frac{{35,335,872}}{{C(52,13)}}\).

Calculation for the probability of \(4\) diamonds, \(6\) hearts, \(2\) spades, and \(1\) club

e)

There are \(13\) spades and \(13\) clubs in the deck.

The number of possibilities to chose \(4\) diamonds from \(13\) is given as:

\(\begin{aligned}{}C(13,4) = \frac{{13!}}{{9!4!}}\\C(13,4) = 715\end{aligned}\)

The number of possibilities to chose \(6\) hearts from \(13\) is given as:

\(\begin{aligned}{}C(13,6) = \frac{{13!}}{{7!6!}}\\C(13,6) = 1716\end{aligned}\)

The number of possibilities to choose \(2\) spades from \(13\) is given as:

\(\begin{aligned}{}C(13,2) = \frac{{13!}}{{11!2!}}\\C(13,2) = 78\end{aligned}\)

The number of possibilities to choose \(1\) club from \(13\) is given as:

\(\begin{aligned}{}C(13,1) = \frac{{13!}}{{12!1!}}\\C(13,1) = 13\end{aligned}\)

Then, the required probability is \(\frac{{715 \times 1716 \times 78 \times 13}}{{C(52,13)}} = \frac{{1,244,117,160}}{{C(52,13)}}\).

Therefore, the probability of a \(13\) card bridge hand is \(\frac{{1,244,117,160}}{{C(52,13)}}\).

Calculation for the probability of one suit, second suit, third suit, and fourth suit

f)

The number of possibilities to chose 4 cards one suit is given as:

\(\begin{aligned}{}C(13,4) = \frac{{13!}}{{9!4!}}\\C(13,4) = 715\end{aligned}\)

The number of possibilities to chose \(6\) cards of second suit is given as:

\(\begin{aligned}{}C(13,6) = \frac{{13!}}{{7!6!}}\\C(13,6) = 1716\end{aligned}\)

The number of possibilities to choose \(2\) cards of third suit is given as:

\(\begin{aligned}{}C(13,2) = \frac{{13!}}{{11!2!}}\\C(13,2) = 78\end{aligned}\)

The number of possibilities to choose \(1\) card of the fourth suit is given as:

\(\begin{aligned}{}C(13,1) = \frac{{13!}}{{12!!!}}\\C(13,1) = 13\end{aligned}\)

Then, the required probability is \(\frac{{715 \times 1716 \times 78 \times 13}}{{C(52,13)}} = \frac{{1,244,117,160}}{{C(52,13)}}\).

Therefore, the probability of a \(13\) card bridge hand contains is \(\frac{{2,944,656}}{{C(52,13)}}\).