Question 1045969: If 5 cards are drawn from an ordinary deck of playing cards what is the probability that the cards will contain a pair or better?
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! Assumptions:
1) Assuming a 52 card deck. A deck has 13 ranks of each of the four suits: clubs, diamonds, hearts, and spades.
2) Assuming a hand of 5 cards, like your problem says.
:
Objective: Draw one pair (two of one face value, and 3 cards of different face
values, not matching the pair):
13 x 4C2 x 12C3 x 4^3
Prob(1 pair) = ---------------------- = 0.422569 or 42.257%
52C5
The denominator:
is the number of ways of selecting 5 cards from 52:
52C5 means 52 Choose 5.
:
The numerator:
13 is the number of ways of choosing the face value of the pair, e.g. two 10's or two queens or whatever.
4C2 is the number of ways of choosing the two cards from the four same-value cards the pack.
12C3 is the number of ways of choosing 3 different face values from the 12 still available - we have used up one face value for the pair. For argument's sake, suppose the 3 remaining face values are 3, 7, K
The 4^3 term arises in the following way. Suppose the face values are to be 3, 7, K. We have 4 ways of choosing the 3,since there are four 3's in the pack. Similarly, there are 4 ways of choosing the 7 and 4 ways of choosing the K, giving 4 x 4 x 4 = 4^3 ways of selecting the other 3 cards with a given set of face values.
Multiplying all these factors together gives the total number of ways
that we could choose exactly one pair. And this divided by 52C5 gives
the required probability.
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