A standard Deck of Cards and Probability of an Event

[Eli]

Question:.

Five cards are randomly drawn from a standard deck of cards. Find the probability of obtaining one pair.


Which among the two solutions below is correct ?

Solution One

Probability of an event = $\dfrac{\text{What you want (what you are interested in)}}{\text{All possibilities (possible outcomes)}} = \dfrac{\text{Number of events}}{\text{Total number of sample space}}$, usually abbreviated as

$P(E) = \dfrac{n(E)}{n(S)}$.

Things to know before attempting the question:

A standard deck of cards, contains:

- 52 cards (excluding Jokers (there are only 2 Jokers))

- 4 suits: Diamonds, Hearts, Clubs and Spades

- 13 cards in each suit

- 3 face cards in each suit: Jack, Queen and King

- 4 aces in a deck

Back to our question.

The total number of possible five-cards draws (total number of possibilities a person can draw a random combination of five cards from a deck), $n(S) = \begin{pmatrix} 52\\ 5 \end{pmatrix} = \dfrac{52!}{5!(52-5)!} = 2598960.$

1. Any pair (say of 2 A's, 2 2's, 2 K's , 2 J's, 2 8's and so on ) of cards can be chosen in $\begin{pmatrix} 4 \\ 2 \end{pmatrix} = 6 \ \text{ways}.$

2. The process of choosing a pair (in 1 above) can be done in $\begin{pmatrix} 13 \\ 1 \end{pmatrix} = 13 \ \text{ways}$.

3. So, after making the first draw of a pair in 1 above, next, we need to choose 3 more cards from the 12 possibilities left, none of which should match each other (i.e., no two cards can be drawn from the same suit). This can be done in $\begin{pmatrix} 12 \\ 3 \end{pmatrix} = 220 \ \text{ways}$.

4. However, each drawn card can be one of the 4 suits, and so there are $4^{3}$ ways to choose the suits.

Therefore, the number of possible draws with only one pair = $\begin{pmatrix} 4 \\ 2 \end{pmatrix} \begin{pmatrix} 13 \\ 1 \end{pmatrix} \begin{pmatrix} 12 \\ 3 \end{pmatrix}. 4^{3} = 1098240$, and the probability of obtaining one pair,

$P(\text{drawing one pair}) = \dfrac{1098240}{2598960} = \dfrac{352}{833}$.

See cards in the attachment which help to clarify the solution.


Sea also

[Eli]

Solution Two

To find the probability of obtaining one pair, we can break down the problem into a few steps:

Step 1: Choose the rank for the pair.

Since we need one pair, we need to choose one rank out of the 13 available ranks in a standard deck. This can be done in \(\binom{13}{1}\) ways.

Step 2: Choose the cards for the pair.

Once the rank is chosen, we need to select 2 cards from the 4 cards of that particular rank. This can be done in \(\binom{4}{2}\) ways.

Step 3: Choose the remaining 3 cards.

After selecting the pair, we are left with 3 remaining cards to choose. This can be done in \(\binom{52-4}{3} \) ways since we must choose 3 cards from the remaining 52-4 cards in the deck after excluding the rank from which a pair is chosen.

Step 4: Calculate the total number of possible selections.

To find the total number of possible selections, we need to choose 5 cards from a standard deck of 52 cards. This can be done in \(\binom{52}{5} \) ways.

Step 5: Calculate the probability of obtaining one pair.

Finally, we can calculate the probability by dividing the total number of ways to obtain one pair by the total number of possible selections:

\[\text{Probability} = \frac{\binom{13}{1} \times \binom{4}{2} \times \binom{52-4}{3}}{\binom{52}{5}}\]


Therefore, the probability of obtaining one pair from a randomly drawn 5-card from a standard deck of cards is given by

\[\frac{{\binom{13}{1} \times \binom{4}{2} \times \binom{48}{3}}}{{\binom{52}{5}}}\] .