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This problem will present another proof of the ballot problem of Example \(3.27 .\) (a) Argue that \(P_{n, m}=1-P\\{A\) and \(B\) are tied at some point \(\\}\) (b) Explain why \(P\\{A\) receives first vote and they are eventually tied \(\\}\) \(=P\\{B\) receives first vote and they are eventually tied \(\\}\) Hint: Any outcome in which they are eventually tied with \(A\) receiving the first vote corresponds to an outcome in which they are eventually tied with \(B\) receiving the first vote. Explain this correspondence. (c) Argue that \(P\\{\) eventually tied \(\\}=2 m /(n+m)\), and conclude that \(P_{n, m}=(n-\) \(m) /(n+m)\)

Short Answer

Expert verified
In order to find the probability of candidate A winning (P(n,m)), we first establish a relationship between the probabilities of being tied and winning, given by \(P_{n,m} = 1 - P\{A \text{ and } B \text{ are tied at some point}\}\). We then demonstrate a one-to-one correspondence between the probabilities of A receiving the first vote and eventually being tied with B, and B receiving the first vote and eventually being tied with A. Next, we calculate the probability of eventually being tied as \(P\{\text{eventually tied}\} = 2 \times \frac{m}{n+m}\). Finally, using these results, we conclude that the probability of A winning \(P_{n,m} = \frac{n-m}{n+m}\).

Step by step solution

01

Part (a): Relationship between the probabilities of being tied and winning

The probability of a candidate A winning the election after being tied at some point with candidate B is equal to the total probability of Candidate A winning minus the probability that A and B are never tied, represented as: \[P_{n,m} = 1 - P\{A\text{ and }B\text{ are tied at some point}\}\]
02

Part (b): Correspondence between probabilities with different first voters

We need to show that: \[P\{A\text{ receives first vote and they are eventually tied}\} = P\{B\text{ receives first vote and they are eventually tied}\}\] Consider an outcome where A and B are eventually tied with A receiving the first vote. In this outcome, we can replace the first vote for A with a vote for B. The number of votes for each candidate remains the same, but now B gets the first vote instead of A. Likewise, we can do the same for any outcome in which B receives the first vote, and switch it to A. This demonstrates a one-to-one correspondence between the two probabilities, therefore proving their equality.
03

Part (c): Calculating the probability of candidates eventually being tied

Given the result from part (b), we can write down the probability of eventually being tied as \[P\{\text{eventually tied}\} = 2 \times P\{A\text{ receives first vote and they are eventually tied}\}\] Since there are (n+m) votes in total, the proportion of votes for each candidate is \[P\{A\text{ receives first vote}\} = \frac{n}{n+m}\] \[P\{B\text{ receives first vote}\} = \frac{m}{n+m}\] Using the result from part (b), we can find the probability of eventually being tied: \[P\{\text{eventually tied}\} = 2 \times \frac{m}{n+m}\] Putting it all together with our results from part (a), we can now conclude the probability of A winning: \[P_{n,m} = 1 - 2\times\frac{m}{n+m}\] Finally, we can simplify this equation to find the target probability expression: \[P_{n,m} = \frac{n-m}{n+m}\]

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Key Concepts

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

Probability Theory
When we talk about probability theory, we're delving into the mathematical study of randomness and uncertainty. It provides frameworks and tools to analyze situations where the outcome is not deterministic but rather has several potential results, each with a specific likelihood of occurrence. In the context of the ballot problem, this theory is important to determine the chances of candidate A or candidate B winning an election, assuming all possible outcomes.

For a simple example, let’s imagine flipping a fair coin. Probability theory helps us conclude that the chance of getting heads or tails is 50% each. Similarly, in our ballot problem, we use probability theory to calculate the likelihood of candidate A or B being in the lead after each vote is cast, which involves understanding subsequent events and their interrelationships.

One useful concept in probability theory is the complementary probability, which is used in part (a) of the ballot problem solution. It states that the probability of an event happening is 1 minus the probability of it not happening. This concept allows us to relate the probability of A never being tied with B to the probability of A eventually winning the election. By taking the total probability (which is 1, representing a certain event) and subtracting the probability of A and B being tied at some point, we can find the probability of A winning without ever being tied with B.
Combinatorics
Combinatorics is a branch of mathematics concerned with counting, combination, and permutation of sets of elements, as well as the arrangement and selection of objects according to specific rules. It’s often used to solve problems associated with discrete structures and is pivotal in the study of probability, as seen in the ballot problem.

In probability problems, combinatory methods are employed to determine the number of ways certain outcomes can happen. This aids in calculating the probabilities of various scenarios.

Factorial Notation and Binomial Coefficients

A common tool in combinatorics is factorial notation, which is used to find the number of ways to arrange a set number of items. Another is the binomial coefficient, denoting the number of ways to choose a subset of items from a larger set, without regard to the order of selection.

In relation to the ballot problem, understanding the combinatorics behind vote arrangements provides a way to quantify the different possible sequences of A and B receiving votes. With the correspondence principle illustrated in part (b) of the solution, we effectively demonstrated that each possible scenario where candidate A receives the first vote and is tied with candidate B at some point has a matching scenario where B receives the first vote. Here, combinatorics underpins the idea that the structure and order of voting have impactful probabilities in political outcomes. Thus, being able to count these arrangements correctly is essential for determining the probability of each candidate winning.
Political Science Probability Models
We encounter political science probability models in scenarios where political outcomes, like election results or policy adoptions, are uncertain. These models use statistical tools to predict the probability of various political events based on historical data, polls, or other relevant information.

In the ballot problem, we see an example of a probability model applied to a political context—though in a simplified and theoretical form. Part (c) of the problem illustrates how we can calculate the probability of an election resulting in a tie using the model. The principle of correspondence from combinatorial theory helps us to equate the probability of a tie given that A or B receives the first vote, pointing out the symmetrical nature of the problem in terms of its probabilistic outcomes.

Political science often harnesses these models to make educated guesses about election results or to understand voter behavior. Although the real-world application of the ballot problem is limited due to complexities such as voter preferences and turnout, the underlying mathematical principles still provide valuable insights into the nature of competitive processes.

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Most popular questions from this chapter

Let \(Y\) be a gamma random variable with parameters \((s, \alpha) .\) That is, its density is $$ f_{Y}(y)=C e^{-\alpha y} y^{s-1}, \quad y>0 $$ where \(C\) is a constant that does not depend on \(y .\) Suppose also that the conditional distribution of \(X\) given that \(Y=y\) is Poisson with mean \(y\). That is, $$ P\\{X=i \mid Y=y\\}=e^{-y} y^{i} / i !, \quad i \geqslant 0 $$ Show that the conditional distribution of \(Y\) given that \(X=i\) is the gamma distribution with parameters (s \(+i, \alpha+1\) ).

\begin{aligned} &\text { (a) Show that }\\\ &\operatorname{Cov}(X, Y)=\operatorname{Cov}(X, E[Y \mid X]) \end{aligned} (b) Suppose, that, for constants \(a\) and \(b\), $$ E[Y \mid X]=a+b X $$ Show that $$ b=\operatorname{Cov}(X, Y) / \operatorname{Var}(X) $$

A coin that comes up heads with probability \(p\) is flipped \(n\) consecutive times. What is the probability that starting with the first flip there are always more heads than tails that have appeared?

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 individual traveling on the real line is trying to reach the origin. However, the larger the desired step, the greater is the variance in the result of that step. Specifically, whenever the person is at location \(x\), he next moves to a location having mean 0 and variance \(\beta x^{2}\). Let \(X_{n}\) denote the position of the individual after having taken \(n\) steps. Supposing that \(X_{0}=x_{0}\), find (a) \(E\left[X_{n}\right]\); (b) \(\operatorname{Var}\left(X_{n}\right)\)

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