/*! 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 24 A coin is tossed repeatedly, hea... [FREE SOLUTION] | 91Ó°ÊÓ

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A coin is tossed repeatedly, heads turning up with probability \(p\) on each toss. Player A wins the game if heads appears at least \(m\) times before tails has appeared \(n\) times; otherwise player \(\mathrm{B}\) wins the game. Find the probability that A wins the game.

Short Answer

Expert verified
Probability that A wins is calculated as \( P_{0,0} \) using recursive relation.

Step by step solution

01

Identify Outcomes

When a coin is tossed, there are two possible outcomes: heads (H) and tails (T). The problem states that A wins if H appears at least \( m \) times before T appears \( n \) times.
02

Define Winning Condition for A

Player A needs at least \( m \) heads before player B can get \( n \) tails. We need to determine a sequence where heads reach \( m \) before tails reach \( n \).
03

Use Probability of Heads

For any single toss of the coin, heads appears with probability \( p \), and tails appear with probability \( 1-p \). We will calculate the probability of sequences where heads appear \( m \) times first.
04

Use Recursive Formula

Let \( P_{a,b} \) be the probability that A wins given the current numbers of heads, \( a \), and tails, \( b \). If \( a = m \), then A already wins, so \( P_{a,b} = 1 \). If \( b = n \), then B has won, so \( P_{a,b} = 0 \). Otherwise, after a toss, either heads appear (moving to \( a+1, b \)), or tails appear (moving to \( a, b+1 \)). Thus, \( P_{a,b} = p P_{a+1,b} + (1-p) P_{a,b+1} \).
05

Solve the Recursive Formula

To solve \( P_{0,0} \) for the probability of A winning starting from 0 heads and 0 tails, we recursively solve until all probabilities are determined starting from \( P_{m,0} = 1 \) and \( P_{0,n} = 0 \). By using base cases and iterating through the recursive formula, calculate \( P_{0,0} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Markov Chains
Markov Chains are a fascinating component of {probability} theory and are incredibly useful in situations where outcomes evolve over time based on current state characteristics. The crux of a Markov Chain is that future states depend solely on the present state and not on the sequence of events that preceded it. This makes them "memoryless."
In the exercise of determining the probability of A winning the coin toss game, we can think of the game as a Markov Chain. Each state in our chain is defined by the count of heads versus tails. We move between states every time the coin is tossed. The transition from one state to another is determined solely by whether the coin lands heads or tails, reducing the process to a simple yet powerful repeating pattern.
  • Each state transition has fixed probabilities: heads with probability \(p\) and tails with \(1-p\).
  • This memoryless property simplifies our calculations immensely.
  • Configuring the game's rules into this framework allows us to utilize established methods of calculating transitional probabilities.
Understanding Markov Chains illuminates the path from initial conditions "not enough heads or tails" to final conditions "A or B wins."
Recursive Formula
A recursive formula is a mathematical expression that defines each term of a sequence using the preceding terms. In the solution, the recursive formula is crucial to solving the probability of winning based on previous outcomes.
The formula we use, \( P_{a,b} = p P_{a+1,b} + (1-p) P_{a,b+1} \), involves two possible future states:
  • Moving to \( P_{a+1,b} \) when a head is tossed, thus increasing the count of heads by one.
  • Or moving to \( P_{a,b+1} \) when a tail is tossed, increasing the count of tails by one.
Defining base conditions is imperative for recursion:
  • \( P_{m,0} = 1 \) means if \( m \) heads are achieved, A wins instantly.
  • \( P_{0,n} = 0 \) indicates if player B accumulates \( n \) tails, A loses, and the game ends.
The recursive process continues, back-calculating from these known states, until we deduce \( P_{0,0} \), the probability of A's victory from the start.
Probability Theory
Probability Theory underpins all aspects of the coin toss game. It's the branch of mathematics that deals with the likelihood of different outcomes.
In our problem, we compute the likelihood that player A or B will win by counting the probabilistic paths to winning conditions:
  • The probability of heads, \( p \), and tails, \( 1-p \), are the elementary blocks guiding the next possible state in our chain.
  • Each sequence of tosses has a calculated probability, determined by the multiplicative law of independent events.
Probability theory allows us to structure these sequences in a manageable form. It leads directly to the calculation of motion between states in the Markov Chain.
Without probability theory, finding the correct outcome to ensure player A's win would be an endless endeavor. Starting with basic coin toss probabilities, we can model whole sequences and their likelihoods efficiently.

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Most popular questions from this chapter

A population of \(b\) animals has had a number \(a\) of its members captured, marked, and released. Let \(X\) be the number of animals it is necessary to recapture (without re-release) in order to obtain \(m\) marked animals. Show that $$ P(X=n)=\frac{a}{b}\left(\begin{array}{c} a-1 \\ m-1 \end{array}\right)\left(\begin{array}{l} b-a \\ n-m \end{array}\right) /\left(\begin{array}{l} b-1 \\ n-1 \end{array}\right) $$ and find \(\mathrm{EX}\). This distribution has been called negative hypergeometric.

Let \(X\) and \(Y\) be discrete random variables with joint mass function $$ f(x, y)=\frac{C}{(x+y-1)(x+y)(x+y+1)}, \quad x, y=1,2,3, \ldots . $$ Find the marginal mass functions of \(X\) and \(Y\), calculate \(C\), and also the covariance of \(X\) and \(Y\).

You roll a conventional fair die repeatedly. If it shows 1 , you must stop, but you may choose to stop at any prior time. Your score is the number shown by the die on the final roll. What stopping strategy yields the greatest expected score? What strategy would you use if your score were the square of the final roll?

\(N+1\) plates are laid out around a circular dining table, and a hot cake is passed between them in the manner of a symmetric random walk: each time it arrives on a plate, it is tossed to one of the two neighbouring plates, each possibility having probability \(\frac{1}{2}\). The game stops at the moment when the cake has visited every plate at least once. Show that, with the exception of the plate where the eake began, each plate has probability \(1 / N\) of being the last plate visited by the cake.

An urn contains \(N\) balls, \(b\) of which are blue and \(r(=N-b)\) of which are red. A random sample of \(n\) balls is withdrawn without replacement from the um. Show that the number \(B\) of blue balls in this sample has the mass function $$ \mathrm{P}(\boldsymbol{B}=k)=\left(\begin{array}{l} b \\ k \end{array}\right)\left(\begin{array}{c} N-b \\ n-k \end{array}\right) /\left(\begin{array}{l} N \\ n \end{array}\right) $$ This is called the hypergeometric distribution with parameters \(N, b\), and \(n\). Show further that if \(N, b\). and \(r\) approach \(\infty\) in such a way that \(b / N \rightarrow p\) and \(r / N \rightarrow 1-p\), then $$ \mathrm{P}(B=k) \rightarrow\left(\begin{array}{l} n \\ k \end{array}\right) p^{k}(1-p)^{n-k} $$ You have shown that, for small \(n\) and targe \(N\), the distribution of \(B\) barely depends on whether or not the balls are replaced in the um immediately after their withdrawal.

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