Question

Four cards are drawn from a pack of $52$ playing cards. Find the probability of drawing exactly one pair.

Answer 1

To determine the number of ways of getting exactly one pair in a draw of four cards, we first break this task into four separate tasks and apply the Fundamental Principle of Counting. We have the following tasks:

1${}^{st}$ task is to select one face value.

2${}^{nd}$ task is to select two cards from the selected face value.

3${}^{rd}$ task is to select a card that is not of the selected face value.

4${}^{th}$ task is to select a card that is not of the face value of the 1${}^{st}$ task and is not of the face value of the 3${}^{rd}$ task.

Then by the Fundamental Principle of Counting, we obtain the following count of the number of ways of performing all four tasks:

(Number of ways to perform 1${}^{st}$ task) $\times$ (Number of ways to perform 2${}^{nd}$ task) $\times$ (number of ways to perform 3${}^{rd}$ task) $\times$ (number of ways to perform 4${}^{th}$ task)

$⟹=\left(13\right)\times {(}^4{C}_2)\times (48)\times (44)$

$⟹=\left(13\right)\times \frac{4!}{2!2!}\times \left(48\right)\times \left(44\right)$

$⟹=\left(13\right)\times \left(6\right)\times \left(48\right)\times \left(44\right)=164,736$

Now, we must remove any duplicate counts of four-card hands. Here, we have counted each of the four hands twice. To see this, consider one such hand, say 2 Aces, 1 King, and 1 Queen. In the above initial count we have counted this as different from a four-card hand consisting of 2 Aces, 1 Queen, and 1 King. (Note that we have already removed order of the two Aces by slecting combinations rather than permutations in their count.) Consequently, we must divide our initial count by the number of ways of permuting the results of the last two tasks, i.e., by 2!, that is, by 2.

Now, we may compute the probability of getting exactly one pair from a draw of four cards. We have:

The probability of getting exactly one pair from a draw of four cards without replacement =

the number of ways of getting exactly one pair in a draw of four cards by the number of ways of drawing four cards

$⟹=\frac{\frac{164,736}{2}}{{}^{52}{C}_4}$

$⟹=\frac{82,368}{\frac{52!}{(52-4)!4!)}}$

$⟹=\frac{82,368}{270725}$

$⟹=0.304$