Chapter 4: Problem 40
In the problem of points, discussed in the historical remarks in Section 3.2, two players, A and B, play a series of points in a game with player A winning each point with probability \(p\) and player \(\mathrm{B}\) winning each point with probability \(q=1-p\). The first player to win \(N\) points wins the game. Assume that \(N=3 .\) Let \(X\) be a random variable that has the value 1 if player A wins the series and 0 otherwise. Let \(Y\) be a random variable with value the number of points played in a game. Find the distribution of \(X\) and \(Y\) when \(p=1 / 2\). Are \(X\) and \(Y\) independent in this case? Answer the same questions for the case \(p=2 / 3\).
Short Answer
Step by step solution
Understanding the Problem
Case 1: Distribution for p = 1/2
Calculate Probability for 3 Points Played
Calculate Probability for 4 Points Played
Calculate Probability for 5 Points Played
Distribution of X when p = 1/2
Distribution of Y when p = 1/2
Independence Check for p = 1/2
Case 2: Distribution for p = 2/3
Probability Calculations for 3, 4, and 5 Points
Distribution of X when p = 2/3
Distribution of Y when p = 2/3
Independence Check for p = 2/3
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Random Variables
\( X \) is defined as the outcome of a player A's victory in the series. It takes the value 1 if player A wins and 0 otherwise. This means that \( X \) tells us the final result of the series in a binary form.
\( Y \) represents the total number of played points in each game. \( Y \) provides crucial insight into how the game unfolds, indicating how many rounds it takes for either player to win.
- \( X \) summarizes the event of player A winning the series.
- \( Y \) gives us details on the duration of the game based on the number of points played.
Binomial Distribution
In this scenario, the success in a trial is defined as player A winning a point. The distribution of \( X \) is affected by the probability \( p \), the likelihood of player A winning a point:
- For \( p = \frac{1}{2} \), every point played is independently won by either player with equal probability.
- For \( p = \frac{2}{3} \), player A is more likely to win any given point, hence changing the distribution over the total games.
Independence of Events
Statistics tell us that two random variables \( X \) and \( Y \) are independent if and only if \( P(X, Y) = P(X)P(Y) \) for any instance.
In the exercise, for both \( p = \frac{1}{2} \) and \( p = \frac{2}{3} \), we're asked to analyze the independence of \( X \) and \( Y \):
- The calculations show that joint probabilities don’t equal the product of their marginals, indicating dependence.
- This means the outcome of one variable (how many points were played) influences the probability of player A winning.