# Thread: What is the probability of getting a exactly one pair when drawing 4 cards?

1. ## What is the probability of getting a exactly one pair when drawing 4 cards? var addthis_config = {"data_track_clickback":false};

13%

2. Four Cards are drawn at random without replacement from a standard deck of 52 cards. What is the probability of exactly one pair?

3. We begin with the observation that the probability of getting exactly one pair from a draw of four cards = the number of ways of getting exactly one pair in a draw of four cards / the number of ways of drawing four cards.

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:

1st task is to select one face value.
2nd task is to select two cards from the selected face value.
3rd task is to select a card that is not of the selected face value.
4th task is to select a card that is not of the face value of the 1st task and is not of the face value of the 3rd task.

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

(# of ways to perform 1st task) ? (# of ways to perform 2nd task) ? (# of ways to perform 3rd task) ? (# of ways to perform 4th task) =

(13)?(C(4,2))?(48)?(44) =

(13)?(4! / (2!2!))?(48)?(44)) =

(13)?((4?3?2?1) / ((2?1)?(2?1)))?(48)?(44) =

(13)?(6)?(48)?(44) = 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 / the number of ways of drawing four cards =

(164,736 / 2) / C(52,4) =
82,368 / (52! / ((52-4)!4!)) =
82,368 / ((52)?(51)?(50)?(49)?(48!) / (48!4!)) =
82,368 / (13)?(17)?(25)?(49) =
82,368 / 270,725 ?
.3042

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