Problem 36 - A First Course in Probability

Question

Two cards are chosen at random from a deck of 52 playing cards. What is the probability that they (a) are both aces? (b) have the same value?

Step-by-Step Solution

Answer

(a) The probability of both cards being aces is \(\frac{1}{221}\). (b) The probability of both cards having the same value is \(\frac{13}{221}\).

Step 1: Total number of ways to choose two cards from a deck of 52

Using combinations, we can find the total number of ways to choose two cards from a deck of 52, denoted as C(52,2). The formula for combinations is \(C(n,r) = \frac{n!}{r!(n-r)!}\), where n is the total number of items and r is the number of items you are choosing. In our case, n = 52 and r = 2. Total combinations = \(C(52, 2) = \frac{52!}{2!(52-2)!} = \frac{52!}{2!50!} = \frac{52 \times 51}{2} = 1326\) There are 1326 ways to choose two cards from a deck of 52 cards.

Step 2: (a) Both cards are aces

We need to find the probability that both cards are aces. There are 4 aces in the deck, so we first need to find the total number of ways to choose two aces from the 4 available. Using combinations again, we can calculate this as: Total combinations of aces = \(C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2} = 6\) There are 6 possible ways to choose two aces from the deck. Now, we can find the probability by dividing the successful outcomes (both cards are aces) by the total possible outcomes: Probability(both aces) = \(\frac{Total \: combinations \: of \: aces}{Total \: combinations} = \frac{6}{1326}\) Probability(both aces) = \(\frac{1}{221}\)

Step 3: (b) Both cards have the same value

Here, we want to find the probability that both cards have the same value (e.g., two 2's, two 3's, ..., two Kings). There are 13 different card values in the deck. For each card value, we have 4 of that card, one in each suit. Let's find the number of ways to choose two cards with the same value for one card value (e.g., two 2's): Combinations for one value = \(C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2} = 6\) There are 6 possible ways to choose two cards with the same value for one card value. We have 13 card values, so we can multiply this by 13 to get the total number of successful outcomes: Total successful outcomes = 6 (combinations for one value) x 13 (card values) = 78 Now, we can find the probability by dividing the successful outcomes (both cards have the same value) by the total possible outcomes: Probability(same value) = \(\frac{Total \: successful \: outcomes}{Total \: combinations} = \frac{78}{1326}\) Probability(same value) = \(\frac{13}{221}\) In conclusion, (a) The probability of both cards being aces is \(\frac{1}{221}\). (b) The probability of both cards having the same value is \(\frac{13}{221}\).

Key Concepts

Combinatorial Probability

Combinatorial probability involves calculating the likelihood of an event when there are several possible combinations. It's central to understanding events where order doesn't matter, like drawing cards from a deck. In our card selection exercise, we first identify the total number of two-card combinations possible from a 52-card deck. Utilizing a mathematical approach called combinations, expressed as \( C(n, r) \), we simplify complex scenarios into understandable probabilities.

To frame this in a real-world context, imagine if instead of cards, we were choosing two favorite fruits out of a basket of several types. The core idea of combinatorial probability helps us understand how likely it is to pick, say, two apples out of a varied assortment, without worrying about which apple we pick first.

Factorial Notation

Factorial notation is a mathematical shorthand used to express the product of an integer and all the integers below it down to one. It's symbolized by an exclamation point (!). For example, \(5!\) is calculated as \(5 \times 4 \times 3 \times 2 \times 1 = 120\). When we look at the calculation of card combinations in our exercise, factorials play a crucial role in the combinations formula \(C(n, r) = \frac{n!}{r!(n-r)!}\).

This might seem daunting, but think of it as a recipe—in the same way that a series of cooking steps leads to a meal, factorial notation provides a process to reach a numerical result. In our card example, we use factorials to find out how many ways we can select two cards from fifty-two, which turns a seemingly complicated task into a manageable number.

Probability of Independent Events

The probability of independent events refers to the likelihood of two or more events occurring where the outcome of one event doesn't impact the others. In the context of cards, drawing one card and then another without replacement represents two independent events. The initial choice doesn't affect the subsequent one. This concept is pivotal when calculating probabilities like those asked in the exercise.

Analogously, it's like trying on hats independently of trying on shoes; your choice of shoes doesn't limit your hat options. In our card problem, we assessed the probability of drawing two aces independently by dividing the ways to draw aces by the total combinations possible. By understanding independence, we easily figure out how each card draw stands alone.