V63.0233, Theory of Probability Solutions
Worksheet for Section 2.5 : Sample spaces having equally likely outcomes July 6, 2009
A poker deck consists of 52 cards. Each card is marked with a suit—one of hearts (♡), spades (♠),
clubs (♣), or diamonds (♢), and a rank —a number from 2 through 10, or Jack (J), Queen (Q),
King (K), or (A). For purposes of ordering, J is equivalent to 11, Q to 12, and K to 13. A can be
considered 1 or 14 as necessary.
1. Suppose five cards are dealt from the deck. Find the probabilities of the following hands. Each
hand should be assumed to be no better than stated (e.g., a hand of three twos is also a hand of a
pair of twos, but that’s not useful)
(i) a pair of twos
Hint. Consider as the equally likely outcomes the set of five-card hands dealt from the 52 in
the deck. If these hands are counted without regard to order, then the number of favorable
hands can also be counted without regard to order. So, for instance, in this problem you
could assume the “favorable” hands have the first two cards being twos, and the last three
non-twos, with no pair among the last three, either.
(ii) any pair
(iii) two pair
(iv) three sixes
(v) three of any kind
(vi) a straight (five cards of consecutive ranks. Notice according to the rules that A-2-3-4-5 and
10-J-Q-K-A are straights, but not Q-K-A-2-3. Also, five cards of consecutive ranks and the
same suit is another hand; read on)
(vii) a flush (five cards of the same suit)
(viii) a full house (three of one kind and two of another)
(ix) four of a kind
(x) a straight flush (five cards of the same suit in consecutive ranks)
Solution. In all of these, the denominator is the number of five-card deals from a deck of 52:
( )
52
= 2, 598, 960
5
(i) There are four twos, of which we need to choose two. The other three cards must come from
three distinct ranks (otherwise we might get a better hand, such as two pair or a full house),
and from any of four suits. So the probability is
(4) (12)
2 × 3 ×4
3
(52) ≈ 3.25%
5
(ii) The probability is just 13 times the probability above, because we can choose any of the 13
ranks to pair. So the probability is
() ( )
13 × 4 × 12 × 43
2
(52) 3 ≈ 42.26%
5
1

(iii) We need to choose the two ranks which will be paired; for each of these, we need to choose
their suits. The last card can have any one of the remaining 11 ranks and any of the 4 suits.
Thus the probability is
(13) (4)2
2 × 2 × 4 × 11
(52) ≈ 4.75%
5
(iv) We need to choose three of the four sixes, then two cards from different ranks and any suit.
So the probability is (4) (12)
3 × 2 ×4
2
(52) ≈ 0.16%
5
(v) The probability is just 13 times the probability above, because we can choose any of the 13
ranks to triple. So the probability is
() ( )
13 × 4 × 12 × 42
3
(52) 2 ≈ 2.11%
5
(vi) The ranks in a straight are determined by the lowest card. For instance, if we know the
lowest is 2, the others are 3-4-5-6. The lowest card can be anything from A to 10, giving 10
possibilities. The suits can be chosen any way, except for the four possibilities of choosing all
the ranks the same. So the probability is
10(45 − 4)
(52) ≈ 0.39%
5
(vii) We can choose a suit and then five out of the 13 cards in that suit. But again we have to
subtract the 10 straight flushes in each suit. So we get
[( ) ]
4 13 − 10
(52)
5
≈ 0.20%
5
(viii) We choose one rank from which to draw three of the four cards, then another rank from which
to draw two more. The probability is
(13)(4)(12)(4)
1
(52)1
3 2
≈ 0.14%
5
(ix) We choose any rank and must take all four cards from that rank. The other card can be any
of the remaining ranks and any suit. So the probability is
(13)(4)(12)(4)
1
(52)1
4 1
≈ 0.024%
5
(x) As we remarked above, there are 40 straight flushes. The probability is
40
(52) ≈ 0.0015%
5
2