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The negative binomial distribution is a probability distribution that models the number of trials needed to achieve a fixed number of successful outcomes. This is an extension of the geometric distribution.
In our exercise, we are dealing with two separate instances of this distribution. Each person rolls their die repeatedly until they achieve five doubles. Each double counts as a successful outcome.

• "Success" here means rolling a double.
• The random variable is the total number of rolls needed to achieve the five successes.

This distribution is characterized by two parameters:

• The number of successful outcomes needed, noted as \(k\).
• The probability of success in a single trial, noted as \(p\).

For each person in our problem, \(k=5\) (as they need five doubles), and the probability \(p=\frac{1}{6}\) (only one out of six possible rolls results in a double). This makes both individual rolls fit into a negative binomial distribution.

Expected value is a fundamental concept in probability and statistics. It is essentially the average or mean value we expect from a random variable if we could repeat an experiment a large number of times.
For the negative binomial distribution, calculating the expected value is straightforward. The formula for the expected value \(E\) of a negative binomial variable is
\[E(Y) = \frac{k}{p} \]where

• \(k\) is the number of successes we are looking for (in our case, five doubles).
• \(p\) is the probability of success on each trial (\(\frac{1}{6}\)).

In the given problem, both dice rollers act independently. For each, the expected total number of rolls is:
\[E(Y_A) = E(Y_B) = \frac{5}{\frac{1}{6}} = 30\]
Therefore, the expected value for the total rolls, \(X = Y_A + Y_B\), is the sum of the expected values for each roller:
\[E(X) = 30 + 30 = 60\]This means, on average, a total of 60 rolls is expected for both individuals to gather five doubles.

Variance provides insight into the spread of a set of values from their mean. In simpler terms, it explains how much the numbers vary from the average.
For the negative binomial distribution, the formula for variance \(V\) is:
\[V(Y) = \frac{k(1-p)}{p^2}\]where

• \(k\) is the number of required successes.
• \(p\) is the probability of success per trial.

Given each individual needs five doubles, and the probability of rolling a double is \(\frac{1}{6}\), the variance for each person is
\[V(Y_A) = V(Y_B) = \frac{5 \cdot \frac{5}{6}}{(\frac{1}{6})^2} = 150\]
Since the rolls of both individuals are independent, the total variance for \(X = Y_A + Y_B\) is the sum of the variances:
\[V(X) = 150 + 150 = 300\]
This large variance indicates that while the average number of rolls is 60, there could be a wide range of actual outcomes in practice.

Understanding independent random variables is crucial as it simplifies the analysis and calculations in many problems like this one.
Two random variables are independent if the occurrence of one does not impact the probability of the occurrence of the other. In our scenario:

• A's and B's rolls are independent because one player's roll outcome does not affect the other's results.

The concept of independence becomes especially handy when calculating the expected value and the variance for the sum of independent random variables.

• For independent variables, the expected value of their sum is the sum of their expected values.
• The same principle applies to variance: the variance of the sum is the sum of the variances.

In our problem

• This independence is why we could straightforwardly take the component values calculated for A and B separately and sum them to find the expected value and variance for their combined rolls.

Recognizing when random variables are independent can greatly enhance your approach to tackling statistical problems.