# Dealt cards

If you draw two cards from the deck, what is the probability both are aces?
There are two interpretations for this question, you can deal two cards off the
top of the deck, or you can deal one card, replace it reshuffle, and deal the
second (perhaps same card, card. In the former case, the probability that the
first card is an ace is 1/13, but if the first card is an ace, only three aces
remain among the remaining 51 cards, hence the probaility the second is an ace
(conditioned on the first being an ace) is 3/51. We may employ the product
rule, letting FA designate that the first card is an ace and SA designate that
the second card is an ace: P(FA and SA) = P(SA|FA) × P(FA) = 3/51 ×
1/13 = 3/(51 × 13). In the second case, with the second card drawn from
the restored and reshuffled deck, the two draws are independent and P(FA and SA)
= P(FA) × P(SA) = 1/13 × 1/13 = 1/169.

Exercise: What is the
probability that two cards drawn from the deck are the same suit? Different
suits? Answer these questions for both the case of choosing the cards with and
without replacement.

The above calculations have used the approach of
conditional probabilities. This can be extended to other problems such as the
probability that five cards drawn from a deck without replacement are of the
same suit (a flush). The first card designates the suit, 12/51 of that suit
remain in the deck, if the first two cards are of the same suit 11/50 of the
same suit remain in the deck, ... . this line of reasoning extends to give that
the probability of being dealt a flush is 1 × 12/51 × 11/50 ×
10/49 × 9/48. However it is much more difficult to use conditional
probabilities to calculate the probability of a full house (two of one kind and
three of another kind).An aternative way to calculate the probability of being dealt
a flush is to count the number of different hands which are flushes, and divide
that by the total number of different hands. There are C(52,5) different hands
that can be dealt from a deck. There are C(13,5) different flushes in each
suit. Hence the probability of being dealt a flush is (4 ×
C(13,5))/C(52,5) = (12 × 11 × 10 × 9)/(51 × 50 × 49
× 48) as found above. The same reasoning can be used to calculate the
probability of a full house: Ther are 13 choices for the denomination there are
three of, which leaves 12 denominations for the pair; There are C(4,3) ways to
choose the the cards within the denomination of the treesome and C(4,2) ways to
choose the pair from the denomination of the twosome. Hence the probability of
a full house is (13 × 12 × C(4,3) × C(4,2))/C(52,5).

**Competencies:** Calculate the probability that four cards dealt from a
deck without replacement are of different suits, both by conditional probability
and by counting arguments.

**Reflection:** Which problems are more easily
done as conditional probability and which problems are more easily done by
counting arguments.

**Challenge:** Calculate the probability that three
cards dealt from a deck without replacement are of different suits, both by
conditional probability and by counting arguments.

May 2003

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