# Dealt cards

## With and without replacement

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.

## Using conditional probabilities

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).

## Using counting arguments

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

campbell@math.uni.edu