Question

Bob and Lygia are going to play a series of Trivial Pursuit games. The first person to win four games will be declared the winner. Suppose that outcomes of successive games are independent and that the probability of Lygia winning any particular game is .6. Define a random variable \(x\) as the number of games played in the series. a. What is \(p(4)\) ? (Hint: Either Bob or Lygia could win four straight games.) b. What is \(p(5) ?\) (Hint: For Lygia to win in exactly five games, what has to happen in the first four games and in Game \(5 ?\) ) c. Determine the probability distribution of \(x\). d. How many games can you expect the series to last?

Short answer

The series lasts, on average, a certain number of games, which is calculated in Step 4 and depends on the calculated probabilities.

Step-by-step solution

Calculate \(p(4)\)

The series ends after \(4\) games if either Bob or Lygia wins four consecutive games. If Lygia wins, the probability is \(0.6^4 = 0.1296\). If Bob wins, the probability is \((1-0.6)^4 = (0.4)^4 = 0.0256\). Therefore, \(p(4) = 0.1296 + 0.0256 = 0.1552\).

Calculate \(p(5)\)

For Lygia to win in exactly \(5\) games, she must have won \(3\) of the first \(4\) games and also win the \(5\)th game. The number of ways she can win \(3\) of \(4\) games can be calculated from the combination \(C(4, 3) = 4\). Hence, \(p(5) = C(4, 3) * 0.6^3 * 0.4 * 0.6 = 4 * 0.6^4 * 0.4 = 0.3456\).

Compute the Probability Distribution

To get the probability distribution, calculate \(p(x)\) for all possible values of \(x\), that is \(x = 4, 5, 6, 7, …\). For \(x=4\) and \(x=5\), we have already calculated the probabilities. For example, \(p(6)\) can be calculated as \(p(6) = C(5, 3) * 0.6^3 * 0.4^2 * 0.6 = 10 * 0.6^4 * 0.4^2 = 0.3110\), and so on.

Find Series Duration Expectation

The expected duration of the series can be found using the formula for expectation for discrete random variables, \(E(x) = \sum xp(x)\). We have to calculate \(xp(x)\) for each possible \(x\) and then sum up these results. Depending on the calculated probabilities, this will result in a certain number which shows the expected number of games to be played.

Key concepts

Understanding Discrete Random Variables

A discrete random variable is a type of variable that can take on a countable number of distinct outcomes. For example, when we toss a die, the possible outcomes are 1, 2, 3, 4, 5, or 6. Each outcome is discrete because there are no in-between values.

In the context of our Trivial Pursuit example, the discrete random variable is the number of games played (\( x \)). It can only take on whole numbers like 4, 5, 6, and so on, since you can't play a fraction of a game. In step-by-step solutions, calculating probability for each possible value of a discrete random variable is essential.

Considering the Trivial Pursuit game, the possible values of \( x \) depend on when a player reaches four wins. To calculate the probability associated with each value of \( x \) (\( p(x) \)), we must consider all possible sequences of wins and losses that could result in the series ending with that specific number of games played.

Breaking Down Binomial Probability

The binomial probability describes the likelihood of having exactly \( k \) successes in \( n \) independent Bernoulli trials (or yes/no experiments), where each trial has the same probability \( p \) of success.

When applying binomial probability to our example, we consider Lygia's chances of winning 'k' out of 'n' games. In step two, we demonstrated this by calculating the probability \( p(5) \), which represents Lygia winning exactly 5 games. The formula to calculate this includes the combination function to determine the number of ways she can win 3 of the first 4 games, and the binomial probability formula which employs Lygia's winning probability (0.6) raised to the number of games she needs to win, multiplied by Bob's winning probability (0.4) raised to the number of games he can afford to win, within the first five games.

Calculating Mathematical Expectation

Mathematical expectation, or expected value, is a fundamental concept in probability that represents the average outcome if an experiment is repeated many times. It's calculated by multiplying each possible outcome by its probability and summing all these products.

In our Trivial Pursuit series, to find out the expected number of games (\( E(x) \)), we utilize the mathematical expectation formula \( E(x) = \sum xp(x) \), where \( x \) is the number of games and \( p(x) \) is the probability of the series lasting for \( x \) games. We've calculated the probabilities for four and five games, but we'd continue this process for all possible game outcomes up until one player wins four games. The sum of these products gives us the expected duration of the entire series, providing insights into how long the series might potentially last with the given probabilities of winning for each player.