/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A and B play a series of games w... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 49

A and B play a series of games with A winning each game with probability \(p\). The overall winner is the first player to have won two more games than the other. (a) Find the probability that \(\mathrm{A}\) is the overall winner. (b) Find the expected number of games played.

Short Answer

Expert verified
The probability that A is the overall winner is \(P(0, 0) = p^2 + p(1-p)(2-p)\), and the expected number of games played is given by \(E(0, 0) = 1 + p^2 + p(1-p)(1+p)\).

Step by step solution

01

(a) Probability of A being the overall winner

Let's denote W as the number of wins needed by A to win the overall game, and L as the number of wins B has, such that A needs exactly two more wins than B. We can break this problem down into several cases: Case 1: W = 0, L = 0 - A has not won any games yet, and B hasn't won a game either. Case 2: W = 1, L = 0 - A has won one game, and B hasn't won any games yet. Case 3: W = 1, L = 1 - A and B have won one game each. Let P(W, L) be the probability that A becomes the overall winner with the current scores W and L. Then, based on the cases outlined above, we can write the following recursive equations: \(P(0, 0) = p * P(1, 0) + (1-p) * P(0, 1)\) \(P(1, 0) = p * P(2, 0) + (1-p) * P(1, 1)\) \(P(1, 1) = p * P(2, 1) + (1-p) * P(1, 2)\) The boundary condition for P(W, L) is when A wins the overall game, which is when W = L+2: \(P(W, L) = 1\) for \(W = L + 2\) We can now solve this recursive system of equations to obtain the probability of A being the overall winner: \(P(0, 0) = p * p + (1-p) * P(0, 1)\) \(P(0, 1) = p * P(1, 1) + (1-p) * P(0, 2) = p * (p * P(2, 1) + (1-p) * P(1, 2)) + (1-p) = p*(p + (1-p)^2)\) Plugging this back into the P(0, 0) equation, we have: \(P(0, 0) = p^2 + (1-p) * p * (p + (1-p)^2) = p^2 + p(1-p)(2-p)\)
02

(b) Expected number of games played

Now, let's calculate the expected number of games played. We will define E(W, L) as the expected number of games played when A has W wins and B has L wins. We can reuse the cases and boundary condition from part (a), but with E(W, L) instead of P(W, L). The boundary condition for E(W, L) when A wins the overall game is: \(E(W, L) = 0\) for \(W = L + 2\) We can now set up the following recursive equations for E(W, L): \(E(0, 0) = 1 + p * E(1, 0) + (1-p) * E(0, 1)\) \(E(1, 0) = 1 + p * E(2, 0) + (1-p) * E(1, 1)\) \(E(1, 1) = 1 + p * E(2, 1) + (1-p) * E(1, 2)\) We can now solve this system of equations to obtain the expected number of games played: \(E(0, 0) = 1 + p * E(1, 0) + (1-p) * E(0, 1)\) \(E(0, 1) = 1 + p * E(1, 1) + (1-p) * E(0, 2) = 1 + p * (1 + p * E(2, 1) + (1-p) * E(1, 2)) + (1-p) = 1 + p + p(1-p)\) Plugging this back into the E(0, 0) equation, we have: \(E(0, 0) = 1 + p^2 + p(1-p)(1+p)\) Therefore, the expected number of games played is given by: \(E(0, 0) = 1 + p^2 + p(1-p)(1+p)\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that independent trials, each of which is equally likely to have any of \(m\) possible outcomes, are performed until the same outcome occurs \(k\) consecutive times. If \(N\) denotes the number of trials, show that $$ E[N]=\frac{m^{k}-1}{m-1} $$ Some people believe that the successive digits in the expansion of \(\pi=3.14159 \ldots\) are "uniformly" distributed. That is, they believe that these digits have all the appearance of being independent choices from a distribution that is equally likely to be any of the digits from 0 through \(9 .\) Possible evidence against this hypothesis is the fact that starting with the \(24,658,601\) st digit there is a run of nine successive \(7 \mathrm{~s}\). Is this information consistent with the hypothesis of a uniform distribution? To answer this, we note from the preceding that if the uniform hypothesis were correct, then the expected number of digits until a run of nine of the same value occurs is $$ \left(10^{9}-1\right) / 9=111,111,111 $$ Thus, the actual value of approximately 25 million is roughly 22 percent of the theoretical mean. However, it can be shown that under the uniformity assumption the standard deviation of \(N\) will be approximately equal to the mean. As a result, the observed value is approximately \(0.78\) standard deviations less than its theoretical mean and is thus quite consistent with the uniformity assumption.

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) \(E[X \mid Y=5]\).

Consider \(n\) multinomial trials, where each trial independently results in outcome \(i\) with probability \(p_{i}, \sum_{i=1}^{k} p_{i}=1 .\) With \(X_{i}\) equal to the number of trials that result in outcome \(i\), find \(E\left[X_{1} \mid X_{2}>0\right]\)

Suppose \(p(x, y, z)\), the joint probability mass function of the random variables \(X\), \(Y\), and \(Z\), is given by $$ \begin{array}{ll} p(1,1,1)=\frac{1}{8}, & p(2,1,1)=\frac{1}{4} \\ p(1,1,2)=\frac{1}{8}, & p(2,1,2)=\frac{3}{16} \\ p(1,2,1)=\frac{1}{16}, & p(2,2,1)=0 \\ p(1,2,2)=0, & p(2,2,2)=\frac{1}{4} \end{array} $$ \text { What is } E[X \mid Y=2] ? \text { What is } E[X \mid Y=2, Z=1] ?

(a) From the results of Section \(3.6 .3\) we can conclude that there are \(\left(\begin{array}{c}n+m-1 \\ m-1\end{array}\right)\) nonnegative integer valued solutions of the equation \(x_{1}+\cdots+x_{m}=n\) Prove this directly. (b) How many positive integer valued solutions of \(x_{1}+\cdots+x_{m}=n\) are there? Hint: Let \(y_{i}=x_{i}-1\). (c) For the Bose-Einstein distribution, compute the probability that exactly \(k\) of the \(X_{i}\) are equal to 0 .
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.