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Chapter 7: Problem 18

Cards from an ordinary deck of 52 playing cards are turned face up one at a time. If the 1 st card is an ace, or the 2 nd a deuce, or the 3 rd a three, or \(\dots,\) or the 13 th a king, or the 14 an ace, and so on, we say that a match occurs. Note that we do not require that the \((13 n+1)\) th card be any particular ace for a match to occur but only that it be an ace. Compute the expected number of matches that occur.

Short Answer

Expert verified
The expected number of matches that occur when drawing cards from an ordinary deck of 52 playing cards, with a match occurring if a card's rank matches its position in the sequence, is \(\boxed{1}\).

Step by step solution

01

Probability of a match at position 1

At the first position, since there are four aces in the deck and the deck has 52 cards in total, the probability of getting an ace at the first position is given by the ratio of the number of aces to the total number of cards in the deck: \[P(X_1) = \frac{4}{52} = \frac{1}{13}.\]
02

Probability of a match at position 2

At the second position, there are four deuces (2s) in the deck. The probability of getting a deuce at the second position depends on the first card that was drawn. Since there are 48 non-deuces remaining in the 51 cards, the probability of getting a deuce at the second position is given by the ratio of the number of deuces to the total number of cards remaining in the deck without deuces: \[P(X_2) = \frac{4}{51} \times \frac{48}{52} = \frac{1}{13}.\]
03

Probability of a match at position \(13n\)

Consider the general case for the \(X_{13n}\)th card. There are 4 cards of a particular rank in the deck, where \(1\leq n \leq 4\). We want to find the probability of getting a card of this rank at the \(13n\)-th position. The probability of getting a card of the desired rank at this position is given by the ratio of the number of cards of this rank remaining in the deck to the total number of cards remaining in the deck. Since there will be \(52 - 13n\) cards remaining in the deck after we draw \(13n - 1\) cards, the probability of getting a card of the desired rank at the \(13n\)-th position is given by: \[P(X_{13n}) = \frac{4}{52 - 13n}.\]
04

Calculate the expected number of matches

Now we will use the linearity of expectation to calculate the expected number of matches. The linearity of expectation states that the expected value of the sum of random variables is equal to the sum of the expected values of those random variables. In this case, the expected number of matches is the sum of the probabilities of a match at each of the 13 positions: \[\text{Expected number of matches} = \sum_{n=0}^3 P(X_{13n+1}).\] Using the probability of a match at position \(13n\), we can find the expected number of matches: \[\text{Expected number of matches} = \sum_{n=0}^3 \frac{4}{52 - 13n} = \frac{4}{52} + \frac{4}{52-13} + \frac{4}{52-26} + \frac{4}{52-39} = \frac{4}{52} + \frac{4}{39} + \frac{4}{26} + \frac{4}{13}.\] Now we can simplify and evaluate the expected number of matches: \[\text{Expected number of matches} = 4\left(\frac{1}{52} + \frac{1}{39} + \frac{1}{26} + \frac{1}{13}\right) = 4\left(\frac{1}{13}\right) = \boxed{1}.\] So the expected number of matches that occur is 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is a measure of the likelihood that a particular event will occur. It ranges from 0 to 1, where 0 indicates that the event cannot occur and 1 indicates certainty. In our example, we are calculating the probability of getting a specific card at a particular position when drawing cards one by one from a deck.

For instance, the probability of drawing an ace as the first card from a standard deck of 52 playing cards is calculated as follows:
  • There are 4 aces in a deck.
  • The probability is the ratio of favorable outcomes to total outcomes: \( P(X_1) = \frac{4}{52} = \frac{1}{13} \).
These calculations help us determine how likely events are, like drawing a matching card at specific positions as described in the exercise. Understanding these probabilities is crucial in computing expected outcomes.
Deck of Cards
A standard deck of cards consists of 52 cards, which include 4 suits: hearts, diamonds, clubs, and spades. Each suit contains 13 ranks: ace through king. In problems involving cards, knowing this structure helps in identifying the total number of possible cards for a particular rank or suit, which is essential for calculating probabilities.

For our problem, knowing that there are:
  • 4 cards of each rank (aces, deuces, threes, etc.) guides us on how we can calculate the possibility of drawing a specific card at its corresponding position (e.g., drawing a 2 at the second position).
  • The concept of 13 ranks reflects in calculating probabilities at each position, with cyclic repetitions after every 13 cards.
This understanding is foundational as it sets the stage for calculating matches and subsequently the expected value.
Linearity of Expectation
Linearity of expectation is a powerful concept in probability and statistics. It states that the expected value of a sum of random variables is the sum of the expected values of each individual random variable. The beauty of this principle is that it holds true regardless of whether the variables are independent.

In our exercise, we computed the expected number of matches in the card game:
  • Each possible match is treated as a random variable with a specific probability of occurring.
  • The expected number of matches is the sum of these probabilities, \( \sum_{n=0}^3 P(X_{13n+1}) \), leading to a straightforward addition of constant probabilities.
This approach simplifies complex calculations, allowing us to focus on each position independently then add results for a quick computation. Linearity of expectation makes handling sums of probabilities more manageable, providing an intuitive understanding of expected outcomes.

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Most popular questions from this chapter

Consider an urn containing a large number of coins, and suppose that each of the coins has some probability \(p\) of turning up heads when it is flipped. However, this value of \(p\) varies from coin to coin. Suppose that the composition of the urn is such that if a coin is selected at random from it, then the \(p\) -value of the coin can be regarded as being the value of a random variable that is uniformly distributed over \([0,1] .\) If a coin is selected at random from the urn and flipped twice, compute the probability that (a) the first flip results in a head; (b) both flips result in heads.

Repeat Problem 68 when the proportion of the population having a value of \(\lambda\) less than \(x\) is equal to \(1-e^{-x}\)

If \(X\) and \(Y\) are independent and identically distributed with mean \(\mu\) and variance \(\sigma^{2},\) find $$ E\left[(X-Y)^{2}\right] $$

If 101 items are distributed among 10 boxes, then at least one of the boxes must contain more than 10 items. Use the probabilistic method to prove this result.

An urn contains 30 balls, of which 10 are red and 8 are blue. From this urn, 12 balls are randomly withdrawn. Let \(X\) denote the number of red and \(Y\) the number of blue balls that are withdrawn. Find \(\operatorname{Cov}(X, Y)\) (a) by defining appropriate indicator (that is, Bernoulli) random variables $$ X_{i}, Y_{j} \text { such that } X=\sum_{i=1}^{10} X_{i}, Y=\sum_{j=1}^{8} Y_{j} $$ (b) by conditioning (on either \(X\) or \(Y\) ) to determine \(E[X Y]\)
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