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.