Question

An unbiased die is successively rolled. Let \(X\) and \(Y\) denote, respectively, the number of rolls necessary to obtain a six and a five. Find (a) \(E[X]\) (b) \(E[X \mid Y=1]\), (c) \(E[X \mid Y=5]\).

Short answer

The short answers for the expected values are: (a) \(E[X] = 6\), (b) \(E[X \mid Y=1] = 6\), and (c) \(E[X \mid Y=5] = 10\).

Step-by-step solution

Find E[X]

First, let's find \(E[X]\). \(X\) follows a geometric distribution with probability of success \(\frac{1}{6}\). To find the expected value of \(X\), we will use the formula for the expected value of a geometric distribution: \(E[X] = \frac{1}{p}\) Where \(p\) is the probability of success, in our case, it is \(\frac{1}{6}\). Calculating \(E[X]\): \(E[X] = \frac{1}{1/6}\) \(E[X] = 6\)

Find E[X|Y=1]

Now, we need to find \(E[X | Y = 1]\). This means we need to find the expected value of \(X\) given that \(Y = 1\), or in other words, the expected number of rolls needed to get a six knowing that we got a five in the first roll. Since we know that it is not possible for \(X\) to take a value of 1 in this scenario, we can derive \(E[X | Y=1]\) by conditioning on the first roll not being a six: \(E[X | Y = 1] = E[X-1]+1\) Where \(E[X-1]\) is the expected number of remaining rolls to get a six, given that the first roll was not a six. As the die rolls are independent, \(E[X-1]\) is just the expected number of rolls to get a six in a modified version of the original problem where 6 is the only winning outcome. Calculating \(E[X | Y = 1]\): \(E[X | Y = 1] = E[X-1] + 1\) \(E[X | Y = 1] = \frac{1}{1/5} + 1\) \(E[X | Y = 1] = 6\)

Find E[X|Y=5]

Finally, we need to find \(E[X|Y=5]\). This means we need to find the expected value of \(X\) given that \(Y = 5\), or in other words, the expected number of rolls needed to get a six knowing that we got a five in the fifth roll. As in the previous step, we can condition on the first four rolls not being sixes: \(E[X|Y=5] = E[X-4] + 4\) Where \(E[X-4]\) is the expected number of remaining rolls to get a six, given that the first four rolls were not sixes. As the die rolls are independent, \(E[X-4]\) is just the expected number of rolls to get a six in a modified version of the original problem where 6 is the only winning outcome. Calculating \(E[X|Y=5]\): \(E[X|Y=5] = E[X-4] + 4\) \(E[X|Y=5] = \frac{1}{1/6} + 4\) \(E[X|Y=5] = 10\) So, the solutions are: a) \(E[X] = 6\) b) \(E[X|Y=1] = 6\) c) \(E[X|Y=5] = 10\)

Key concepts

Understanding Expected Value in Probability

At the heart of probability theory is the concept of expected value, often symbolized as 'E'. The expected value is a measure that gives an idea of the 'central' or 'typical' value that you would expect to get from a random variable after many trials. It's like the average outcome you would predict over a long period of time.

For example, consider an unbiased six-sided die. Since all outcomes are equally likely, the probability (\(p\)) of rolling a six is \frac{1}{6}\. If we define the random variable \(X\) as the number of rolls needed to get a six, the question arises: 'On average, how many rolls are needed to see a six?' To answer this, we utilize the expected value formula specifically designed for geometric distributions:

\[E[X] = \frac{1}{p}\]
By plugging in our probability value, we find that \[E[X] = \frac{1}{1/6} = 6\]
This value suggests that on average, it would take six rolls to get a six on the die. It's critical to note that the expected value doesn't predict individual outcomes, but instead the average over many trials which becomes a powerful tool when making predictions about random events.

Deciphering Conditional Probability and Expected Value

Conditional probability is another foundational concept in probability theory and statistics. It examines the probability of an event occurring given that another event has already taken place. When applied to expected value, this idea allows us to ask more targeted questions: 'How does our expectation change knowing that a certain condition is met?'

Referring to the given exercise, we delve into the realms of conditional expectation—here specifically, what's the expected number of rolls to get a six after we've already got a five? The variable \(Y\) equaling 1 indicates that a five was rolled on our very first attempt. This outcome affects our calculation of \(E[X | Y=1]\), the expected value of \(X\), with this new knowledge.

Since rolling a five in the first try doesn't affect the independence of subsequent rolls, we still expect to wait the same amount of time for a six as if we were starting from scratch, but with an additional roll already taken. Hence, the expected number of rolls remains unchanged, at 6 rolls, when \(Y=1\). Similarly, if \(Y=5\), meaning we got a five on the fifth roll, we add four to the expected value for the remaining rolls, yielding \(E[X|Y=5] = 10\).This elegantly showcases how conditional probability can adjust our expectations in the light of new information.

Decomposing Probability Models

Probability models provide us with a framework for quantifying the uncertainty associated with random events. These models include a range of tools and distributions, such as the geometric distribution used in this exercise. Looking at probability models aids in constructing a mental picture of the situation and aids us in calculations of expected values, probabilities, and other measures.

Geometric distribution is a probability model that assesses the likelihood of a specific number of independent trials needed until a first success occurs, given that each trial has the same probability of success. Its beauty lies in its simplicity and the straightforwardness of its associated formulas. For instance, the geometric distribution formula enables us to swiftly calculate the expected number of rolls to achieve the first success (rolling a six), and it also provides a scaffold for calculating more complex scenarios like conditional probabilities.

Understanding these models and their proper application ensures accurate predictions and assessments in real-world situations ranging from gaming strategies to scientific experiments and even business decision-making processes. Always remember, probability is not just about numbers; it's about understanding the structure and behavior of randomness in the world around us.