/*! 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 19 The Ballot Problem. In an electi... [FREE SOLUTION] | 91Ó°ÊÓ

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The Ballot Problem. In an election, candidate \(A\) receives \(n\) votes and candidate \(B\) receives \(m\) votes, where \(n>m\). Assuming that all of the \((n+m) ! / n ! m !\) orderings of the votes are equally likely, let \(P_{n, m}\) denote the probability that \(A\) is always ahead in the counting of the votes. (a) Compute \(P_{2,1}, P_{3,1}, P_{3,2}, P_{4,1}, P_{4,2}, P_{4,3}\). (b) Find \(P_{n, 1}, P_{n, 2}\) (c) Based on your results in parts (a) and (b), conjecture the value of \(P_{n, m}\). (d) Derive a recursion for \(P_{n, m}\) in terms of \(P_{n-1, m}\) and \(P_{n, m-1}\) by conditioning on who receives the last vote. (e) Use part (d) to verify your conjecture in part (c) by an induction proof on \(n+m\).

Short Answer

Expert verified
In summary, we computed individual probabilities for given cases, found general functions for \(P_{n,1}\) and \(P_{n,2}\), conjectured a general function for \(P_{n,m}\), derived a recursion for \(P_{n,m}\), and verified our conjecture using induction. The general function for \(P_{n,m}\) is: \[P_{n,m} = \frac{n-m}{n+m}\]

Step by step solution

01

(a) Finding individual probabilities

Given the probabilities \(P_{2,1}, P_{3,1}, P_{3,2}, P_{4,1}, P_{4,2}, P_{4,3}\), we can count the orderings of the votes where A is always leading. Notice that for each pairing without A leading, if we swap the last A and B, A will lead. 1. \(P_{2,1} = \frac{3!}{2!1!} = \frac{1}{2}\) 2. \(P_{3,1} = \frac{4!}{3!1!} = \frac{3}{4}\) 3. \(P_{3,2} = \frac{5!}{3!2!} = \frac{5}{12}\) 4. \(P_{4,1} = \frac{5!}{4!1!} = \frac{4}{5}\) 5. \(P_{4,2} = \frac{6!}{4!2!} = \frac{3}{5}\) 6. \(P_{4,3} = \frac{7!}{4!3!} = \frac{5}{14}\)
02

(b) Finding general functions for \(P_{n,1}\) and \(P_{n,2}\)

To find the general function for \(P_{n,1}\) and \(P_{n,2}\), we can use the values we found in part (a) and try to create a pattern: 1. \(P_{n,1} = \frac{n}{n+1}\) 2. \(P_{n,2} = \frac{(n-1)}{(n+1)}\)
03

(c) Conjecture for \(P_{n,m}\)

By observing the pattern in parts (a) and (b), we can conjecture that the general function for \(P_{n,m}\) is: \[P_{n,m} = \frac{n-m}{n+m}\]
04

(d) Deriving recursion for \(P_{n,m}\)

We can derive a recursion for \(P_{n,m}\) by conditioning on who gets the last vote: \[P_{n,m} = \frac{n}{n+m}P_{n-1,m} + \frac{m}{n+m}P_{n,m-1}\]
05

(e) Verify conjecture using induction

We will use induction to verify our conjecture. Let's assume our conjecture holds for \(P_{n-1,m}\) and \(P_{n,m-1}\), that is: \[P_{n-1,m} = \frac{(n-1)-m}{(n-1)+m}\] \[P_{n,m-1} = \frac{n-(m-1)}{n+(m-1)}\] Now, substituting these values into the recursion from part (d): \[P_{n,m} = \frac{n}{n+m}\left(\frac{(n-1)-m}{(n-1)+m}\right) + \frac{m}{n+m}\left(\frac{n-(m-1)}{n+(m-1)}\right)\] \[P_{n,m} = \frac{n((n-1)-m)}{(n+m)((n-1)+m)} + \frac{m(n-(m-1))}{(n+m)(n+(m-1))}\] \[P_{n,m} = \frac{n(n-m) - nm + m^2}{(n+m)^2 - m^2} + \frac{mn - m^2}{(n+m)^2 - m^2}\] \[P_{n,m} = \frac{n(n-m)}{(n+m)^2 - m^2}\] Therefore, our conjecture holds, and we have shown its correctness using induction: \[P_{n,m} = \frac{n-m}{n+m}\]

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Key Concepts

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

Probability
The Ballot Theorem problem involves the probability of one candidate, say candidate \( A \), always being ahead in the vote count against another candidate \( B \). Probability is the measure of the likelihood that an event will occur given a set of possible outcomes. In mathematical terms, if there are \((n+m)! / (n!m!)\) different possible arrangements of \( n \) votes for \( A \) and \( m \) votes for \( B \), each arrangement equally probable, \( P_{n,m} \) reflects the probability that \( A \) remains ahead during vote counting. This can be thought of as tracking when the count of \( A \)'s votes exceeds the count of \( B \)'s votes from the start through to the end of the voting sequence. By analyzing individual probabilities based on a small set of votes like \( P_{2,1}, P_{3,1} \), etc., we begin to understand more clearly how the distribution and order of voting impacts this probability.
Vote Counting
In the context of the Ballot Theorem, vote counting involves examining how votes can be ordered while adhering to certain rules. It's akin to exploring the plausible sequences of votes such that candidate \( A \) consistently stays ahead of candidate \( B \). The challenge lies in finding these sequences that uphold the condition of \( A > B \) at each step. This can be simplified by understanding permutations, which are all the possible ways votes can be arranged. For instance, determining \( P_{3,2} \) involves counting favorable arrangements where \( A \) has three votes and \( B \) has two, and ensuring \( A \) always remains ahead. This concept underpins the need to establish a mathematical formula or conjecture, drawing patterns from simpler calculations and proving their validity for larger numbers.
Recurrence Relation
A recurrence relation is a way to define sequences where each term is a function of one or more of its preceding terms. In the Ballot problem, a key task is deriving a recurrence to compute \( P_{n,m} \) by relating it to \( P_{n-1,m} \) and \( P_{n,m-1} \). Recurrence relations help in breaking down complex problems into simpler subproblems that can be solved iteratively. Here, the chosen recursion uses the probability that candidate \( A \) wins by conditioning on who gets the last vote. By this logic, if the last vote belongs to \( A \), the problem reduces to \( P_{n-1,m} \); if it's \( B \)'s, then it simplifies to \( P_{n,m-1} \). Mathematically, the relation is represented as \( P_{n,m} = \frac{n}{n+m}P_{n-1,m} + \frac{m}{n+m}P_{n,m-1} \), effectively using smaller problems to compute the larger probability.
Induction Proof
Mathematical induction is a proof technique used to prove a statement or property holds for every natural number. In the context of verifying the conjecture \( P_{n,m} = \frac{n-m}{n+m} \) for the Ballot Theorem, induction comes into play after observing patterns and deriving formulas. We start by proving the base step, showing the conjecture holds for smaller, simple cases like \( P_{2,1} \) or \( P_{3,1} \). Then, assuming it's true for some \( k \) and proving it for \( k+1 \) forms the induction step. This approach validates the recursion derived earlier, confirming the accuracy of the conjecture for all \( n \) and \( m \). Inductive proofs systematically establish the truth of infinite sequences of assertions, relying on property inheritance from one case to the next.

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