Question

In a card game using a standard $52$ card deck, $4$-card hands are dealt. What’s the probability of being dealt $3$ diamonds ?

Answer 1

To find the probability of being dealt $3$ diamonds, we will use the concept of combination. We will first find the number of ways selecting $3$ diamonds cards from $13$ diamonds cards and then the number of ways in which fourth card will not be a diamond in order to find total ways in which $4$-card hands are dealt such that being dealt $3$ diamonds. We will then find the total number of ways in which $4$-card hands can be dealt from $52$ card deck. To find the probability of being dealt $3$ diamonds, we will take ratio of total ways in which $4$-card hands are dealt such that being dealt $3$ diamonds to total number of ways in which $4$-card hands can be dealt from $52$ card deck.

Complete step by step answer:
We have to find the number of possible $4$-card hands which contain exactly $3$ diamonds.As we know, in a standard deck there are $13$ ordinal cards i.e., Jack, Queen, King and $10$ aces and in each of $4$ suits i.e., Hearts, Diamond, Clubs and Spades.Therefore, we have total $13 \times 4$ cards i.e., $52$ cards.We will use the concept of Combination to determine the number of possible $4$-card hands which contain exactly $3$ diamonds as the order in which we are selecting the cards is not important here. So, we can use the concept of combination to solve this problem.As we know,
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
where $n$ is the number of items and $r$ is the number of items being chosen at a time.

Number of ways in which $3$ diamonds can be selected from $13$ diamonds card will be $^{13}{C_3}$ i.e.,
Number of ways of selecting $3$ diamonds cards from $13$ diamonds cards $=$ $^{13}{C_3}$
Putting in formula we get
${ \Rightarrow ^{13}}{C_3} = \dfrac{{13!}}{{3!\left( {13 - 3} \right)!}}$
On expanding we get
${ \Rightarrow ^{13}}{C_3} = \dfrac{{13 \times 12 \times 11 \times 10!}}{{3 \times 2 \times 10!}}$

On calculation we get
${ \Rightarrow ^{13}}{C_3} = 286$
So, the number of ways of selecting $3$ diamonds cards from $13$ diamonds cards = $$$$286
The fourth card can’t be a diamond. Therefore, we have to select from the remaining suits.
Number of ways in which fourth card will not be a diamond $=$ $52 - 13$
$= 39$
Putting all together we get, number of ways of being dealt $3$ diamonds $=$ 286 \times 39$$$$ = 11154
Number of ways in which $4$ card hands can be dealt from $52$ cards { = ^{52}}{C_4}$$$$ = \dfrac{{52!}}{{\left( {52 - 4} \right)!4!}}

On expanding, we get
Number of ways in which $4$ card hands can be dealt from $52$ cards $= \dfrac{{52 \times 51 \times 50 \times 49 \times 48!}}{{48! \times 4 \times 3 \times 2 \times 1}}$
On simplification, we get
Number of ways in which $4$ card hands can be dealt from $52$ cards $= 270125$
The probability of being dealt $3$ diamonds $= \dfrac{{{\text{number of ways of being dealt 3 diamonds}}}}{{{\text{Total number of ways in which 4 card hands can be dealt from 52 cards}}}}= \dfrac{{11154}}{{270125}}$
On simplification we get
The probability of being dealt $3$ diamonds $= 0.041$

Therefore, the probability of being dealt $3$ diamonds is $0.041$.

Note: Here, we are not interested in the order of the draw, that’s why we have used combination because a combination is a grouping or subset of items where the order does not matter. But if the order did matter then it will be solved using permutation as permutation is an arrangement in a definite order.