Nice Bounds for the Generalized Ballot Problem

TL;DR

This paper establishes two precise bounds for the generalized ballot problem, which involves determining the probability that candidate A maintains a lead over candidate B by a factor of at least b5 during the counting process.

Contribution

It introduces two sharp bounds for the generalized ballot problem applicable to any real b5, extending previous results in election probability analysis.

Findings

Derived two tight bounds for the generalized ballot problem.

Applicable to any real b5, broadening previous special cases.

Enhances understanding of election probability dynamics.

Abstract

This paper gives two sharp bounds for the generalized ballot problem with candidate A receiving at least \mu times as candidate B for an arbitrary real number \mu.

Equations

$$P=\frac{a-\mu b}{a+b}.$$

$$P^{\ast }=\frac{a-\mu b+1}{a+1}.$$

$$P=\frac{a}{a+b}\sum _{j=0}^b{C}_j\frac{\left(\genfrac{}{}{0ex}{}{b}{j}\right)}{\left(\genfrac{}{}{0ex}{}{a+b-1}{j}\right)}$$

$$\sum _{j=0}^k{C}_j\frac{\left(\genfrac{}{}{0ex}{}{k}{j}\right)}{\left(\genfrac{}{}{0ex}{}{\lfloor k\mu \rfloor +k-1}{j}\right)}=0$$

$$\text{textsc}Theorem1.\frac{a-\lfloor \mu b\rfloor }{a+b}\le P\le \frac{a-\lfloor \mu \rfloor b}{a+b}$$

$$\text{textsc}Theorem2.\frac{\lfloor a-\mu b+1\rfloor }{a+b}\le P^{\ast }\le \frac{a+1-\mu b}{a+1}.$$

$$\left(\genfrac{}{}{0ex}{}{a_r+b_r-1}{b_r-1}\right)=\frac{b_r}{a_r+b_r}\left(\genfrac{}{}{0ex}{}{a_r+b_r}{a_r}\right)\ge \frac{1}{\lfloor \mu \rfloor +1}\left(\genfrac{}{}{0ex}{}{a_r+b_r}{a_r}\right)$$

$$P\cdot \left(\genfrac{}{}{0ex}{}{a+b}{a}\right)\le \left(\genfrac{}{}{0ex}{}{a+b}{a}\right)-(\lfloor \mu \rfloor +1)(\genfrac{}{}{0ex}{}{a+b-1}{b-1})=\frac{a-\lfloor \mu \rfloor b}{a+b}(\genfrac{}{}{0ex}{}{a+b}{a}).$$

$$\left(\genfrac{}{}{0ex}{}{a_r+b_r}{b_r-1}\right)=\frac{b_r}{a_r+1}\left(\genfrac{}{}{0ex}{}{a_r+b_r}{a_r}\right)\le \frac{1}{\mu }\left(\genfrac{}{}{0ex}{}{a_r+b_r}{a_r}\right)$$

$$P^{\ast }\cdot \left(\genfrac{}{}{0ex}{}{a+b}{a}\right)\le \left(\genfrac{}{}{0ex}{}{a+b}{a}\right)-(\mu )\left(\genfrac{}{}{0ex}{}{a+b}{b-1}\right)=\frac{a-\mu b+1}{a+1}\left(\genfrac{}{}{0ex}{}{a+b}{a}\right).$$

$$S_{r_{i+1}}^{\prime }\le S_{r_i+1}^{\prime }\le S_{r_i}^{\prime }+1$$

$$\frac{a-\lfloor \mu b\rfloor }{a+b^{\prime }}\le P\le \frac{a}{a+b^{\prime }}$$

Full text

Introduction

Suppose in an election candidate $A$ received $a$ votes and candidate $B$ received $b$ votes. We count the votes one at a time in any of the $\binom{a+b}{a}$ possible sequences. Let $a_{r}$ and $b_{r}$ denote the number of votes $A$ and $B$ have after counting the $r^{th}$ vote where $1\leq r\leq a+b$ (notice that $a_{r}+b_{r}=r$). Let $\mu$ be any positive real number. We call a sequence desirable if $a>\mu b$ and $a_{r}>\mu b_{r}$ for all $r$. We call a sequence cute if $a\geq\mu b$ and $a_{r}\geq\mu b_{r}$ for all $r$. Let $P$ denote the probability that a sequence is desirable and $P^{*}$ denote the probability that a sequence is cute.

Several authors started several articles quoting several well-known results of the Ballot Problem. For brevity, Andre [7] and Barbier[8] discovered that if $\mu\in\mathbb{N}$, then

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Aeppli[9] showed that if $\mu\in\mathbb{N}$, then

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Finally in 1962 Takacs[4] took a giant leap and bravely proved that for an arbitrary $\mu\in\mathbb{R}$,

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where $C_{0}=1$ and $C_{j}$ satisfies the following recurrence formula:

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for all positive integers $k$. This formula gives the exact value for $P$. However we can hardly imagine how big this number really is. Therefore this article proves the following bounds:

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We prove the two upper bounds with the Pseudo-Reflection Principle and the two lower bounds with Penetrating Analysis.

Pseudo-Reflection Principle

Let’s start with Theorem 1. We look at the relationship between the undesirable sequence and the sequence with $a$ votes for $A$ and $b-1$ votes for $B$. We call this the Pseudo-Reflection Principle because the case $\mu=1$ is essentially the reflection principle. 111However we are not finding any bijections here. We are only counting the number of undesirable sequences. Both Goulden[5] and Renault[6] proved the equality case (when $\mu\in\mathbb{N}$). They both considered the smallest $r$ such that $a_{r}\leq\mu b_{r}$. However this approach does not generalize to the case when $\mu\in\mathbb{R}$.

Instead we consider the largest $r$ such that $a_{r}\leq\mu b_{r}$. When the $r^{th}$ vote is counted, we must have $a_{r}=\lfloor\mu b_{r}\rfloor\leq\lfloor\mu\rfloor b_{r}$. There are $\binom{a_{r}+b_{r}}{a_{r}}$ such undesirable sequences. Now consider the number of sequences with $a_{r}$ votes for $A$ but $b_{r}-1$ votes for $B$. For each $r$ there are $\binom{a_{r}+b_{r}-1}{b_{r}-1}$ such sequences.

Consider the operation of replacing the first $r$ votes in an undesirable sequence with these $\binom{a_{r}+b_{r}-1}{b_{r}-1}$ sequences. This operation yields all sequences with $a$ votes for $A$ but $b-1$ votes for $B$ because for any sequence with $a$ votes for $A$ but $b-1$ votes for $B$, there must exist an $r$ such that $a_{r}-1\leq\mu(b_{r}+1)<a_{r}$.

Since

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we deduce that the number of undesirable sequence is at most $\lfloor\mu\rfloor+1$ times the sequences with $a_{r}$ votes for $A$ but $b_{r}-1$ votes for $B$. Therefore

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*Remark * The conditions for equality to hold are not trivial. Dvoretzky [11] proved that that equality holds if and only if $\mu$ is sufficiently close to $\frac{a}{b}$ or $\mu$ is sufficiently close to an integer. See Dvoretzky [11] for the precise definitions of sufficiently close.

Now we move on to Theorem 2. Notice that the upper bound for Theorem 2 is a trivial consequence of theorem 1 when $\mu$ is an integer. (We can simply add one vote for $A$ in the beginning of the sequence and then $a_{r}>b_{r}$.) But such a correlation does not give the sharp bound in Theorem 2 for $\mu\in\mathbb{R}$.

Using the Pseudo-Reflection technique we similarly consider the largest $r$ such that $a_{r}<\mu b_{r}$. We have $a_{r}=\lceil\mu b_{r}\rceil-1$. This time, however, we compare the ugly (non-cute) sequence to the sequence with $a+1$ votes for $A$ and $b-1$ votes for $B$. We similarly replace the first $r$ votes with sequences of $a_{r}+1$ votes for $A$ and $b_{r}-1$ votes for $B$. This operation yields all possible sequences with $a_{r}+1$ votes for $A$ and $b_{r}-1$ votes for $B$.

Now we have

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which implies that the number of ugly sequence is at least $\mu$ times the number of sequences of $a_{r}+1$ votes for $A$ and $b_{r}-1$ votes for $B$. Therefore

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*Remark * We can translate these sequences into lattice paths from the origin the point $(b,a)$. A desirable path never touches the line $y=\mu x$, and a cute path never go below the line. The inequality $(\ref{notouch})$ shows that the number of undesirable paths is at least $\lfloor\mu\rfloor$ times the number of paths from the point $(1,0)$ to $(b,a)$. The inequality $(\ref{touch})$ shows that the number of ugly paths is at least $\mu$ times the number of path from the point $(1,-1)$ to $(b,a)$. But intuition does not help much in this problem since it involves calculations and one to $\lfloor\mu\rfloor$ correspondence rather than a pure bijection.

Penetrating Analysis

We first prove the lower bound for Theorem 2. We claim that at least $\lfloor a-\mu b\rfloor+1$ of the $a+b$ cyclic permutations of any given sequence of votes are cute. This method is called penetrating analysis in Mohanty [1].

For any given sequence, define the weighted partial sum as $S_{r}=a_{r}-\mu b_{r}$. Note that the sequence is cute if and only if $S_{r}\geq 0$ for all $r.$ Suppose $S_{i}$ is the minimum (if there are multiple $i$s then we can take any of them). We cyclically permute the first $i$ terms of the sequence to the end of the sequence. In other words we erase the first $i$ terms and attach them to the end of the sequence. Now let $S^{\prime}$ be the weighted partial sum for this new sequence. We finish the proof with three lemmas.

*Lemma 1. This new sequence is cute.

*Proof: * If $r\leq a+b-i$, then $S^{\prime}_{r}=S_{a+b-i}-S_{i}\geq 0$. If $r>a+b-i$, then $S^{\prime}_{r}=S_{r-(a+b-i)}+S_{a+b}-S_{i}\geq 0$ because $S_{a+b}\geq 0$. Therefore $S^{\prime}_{r}\geq 0$ for all $r$.

*Lemma 2. A cyclic permutation that begins with the $r^{th}$ term of this sequence is cute if $S^{\prime}_{r}\leq S^{\prime}_{t}$ for all $r+1\leq t\leq a+b.$ For convenience, we call such an $r$ and also the $r^{th}$ vote cute.

*Proof: * Let ${S^{\prime\prime}}$ denote the weighted partial sum for the cyclic permutation that begins with the $r^{th}$ term of this sequence. If $j\leq a+b-r$, then ${S^{\prime\prime}_{j}}=S^{\prime}_{a+b-j}-S^{\prime}_{r}\geq 0$. If $j>a+b-r$, then ${S^{\prime\prime}_{j}}=S^{\prime}_{j-(a+b-r)}+S^{\prime}_{a+b}-S^{\prime}_{r}\geq 0$ because $S^{\prime}_{r}\leq S^{\prime}_{a+b}$ and $S^{\prime}_{j-(a+b-r)}\geq 0$. Therefore ${S^{\prime\prime}_{r}}\geq 0$ for all $r$.

*Lemma 3. There exist at least $\lfloor a-\mu b\rfloor+1$ cute votes.

*Proof: * Let $r_{1}<r_{2}<\ldots<r_{k}=a+b$ denote all the cute votes. We have $S^{\prime}_{r_{k}}=a-\mu b$ and $S^{\prime}_{r_{1}}\leq 1$. Since $S^{\prime}_{r+1}\leq S^{\prime}_{r}+1$, we must have

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(If $S^{\prime}_{r_{i+1}}>S^{\prime}_{r_{i}+1}$, then there must exist another cute vote between $r_{i}+1$ and $r_{i+1}$, which contradicts the definition of $r_{i}$.) Therefore $k-1\geq S^{\prime}_{r_{k}}-S^{\prime}_{r_{1}}\geq a-\mu b-1$. Because $k$ is an integer, we have two cases:

If $a-\mu b-1$ is not an integer, then $k\geq\lceil a-\mu b\rceil=\lfloor a-\mu b+1\rfloor.$ 2. 2.

If $a-\mu b-1$ is an integer, then consider all $r$ such that $a_{r}-\mu b_{r}<0$. Since there are finitely many such negative values, there exist an $\epsilon$ such that $a_{r}-(\mu-\epsilon)b_{r}\geq 0$ implies $a_{r}-\mu b_{r}\geq 0$. Replacing $\mu$ with $\mu-\epsilon$ would not affect the number of cute sequences. Therefore $k\geq\lceil a-(\mu-\epsilon)b\rceil=a-\mu b+1.$

Now we have shown that that at least $\lfloor a-\mu b+1\rfloor$ of the $a+b$ cyclic permutations of any given sequence are cute. Therefore $P^{*}\geq\dfrac{\lfloor a-\mu b+1\rfloor}{a+b}.$

For Theorem 1, we can similarly show that at least $a-\lfloor\mu b\rfloor$ of the $a+b$ cyclic permutations of any given sequence are desirable. Notice that $\lfloor a-\mu b+1\rfloor=a-\lfloor\mu b\rfloor$ if $a-\mu b-1$ is not an integer, and $\lfloor a-\mu b+1\rfloor=a-\lfloor\mu b\rfloor+1$ otherwise. We can imitate the proof for Theorem 2 until the last step. If $a-\mu b-1$ is an integer, there does not $\epsilon$ such that $a_{r}-(\mu-\epsilon)b_{r}>0$ can guarantee that $a_{r}-\mu b_{r}>0$ because $a_{r}-\mu b_{r}$ can be equal to [math]. Therefore we can only conclude that $P\geq\dfrac{a-\lfloor\mu b\rfloor}{a+b}.$

Remark 1 We can also prove the lower bound by induction on $b$. Again let’s prove the result of Theorem 2, and we can follow the same procedure for Theorem 1. Base case $b=1$ is trivial. Suppose that the bound is valid for all positive integers less than $b$. We first apply lemma 1 to permute any given sequence into a cute sequence. Then we consider two situations:

If there exist a cute $r$ such that $0<r<b$, then we can cut the sequence into two cute sequences. By the inductive hypothesis, the number of cute votes no less than $\lfloor a_{r}-\mu b_{r}+1\rfloor+\lfloor a-a_{r}-\mu(b-b_{r})+1\rfloor\geq\lfloor a-\mu b+1\rfloor.$ 2. 2.

If there does not exist a cute $r$ such that $0<r<b$, then for all cute $r$, we must have $b_{r}=0$ or $b_{r}=b$. The rest follows trivially from $(\ref{duh}).$

Remark 2 Notice that any cute sequence must start with a vote for $A$. We can thus treat all the votes for $B$ in between two votes for $A$ as a single block. Therefore the argument is still valid even if $B$ receives any number of weighted votes as long as the weights add up to $b$. We can reformulate the problem as follow.

Suppose in an election $A$ received $a$ votes all weighted 1. However $B$ received $b^{\prime}$ weighted votes whose sum is $b$. Both $a$ and $b$ are integers, and $\mu$ is a real number. We count the $a+b^{\prime}$ votes in a random order. Let $a_{r}$ and $b_{r}$ denote the sum of the weighted votes $A$ and $B$ have after counting the $r^{th}$ vote where $1\leq r\leq a+b^{\prime}$. Define $P$ as the probability that $a_{r}>\mu b_{r}$ for all $1\leq r\leq a+b^{\prime}$ (desirable sequences). Then

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The upper bound is because each desirable sequence must start with a vote for $a$. This bound, although achievable, is extremely weak compare to $(\ref{one})$ and $(\ref{two})$. Goulden [5] discussed the equality case for the lower bound when $\mu=1$ and all the weights are integers. In fact equality holds for the lower bound if $\mu$ and all the weights are integers. Therefore if all the weights are integers, then $P\leq\dfrac{a-\lfloor\mu\rfloor b}{a+b^{\prime}}$, but we still cannot find a sharp upper bound for arbitrary weights.

Further Thoughts

The inspiration of this paper is to search for a closed formula for any $\mu\in\mathbb{R}$. Intuitively such a formula probably doesn’t exist. Even if it does we must use more advanced techniques. We still have many unanswered questions in this paper. For example, can we find a sharper bound for Theorem 1 and Theorem 2? Can we find an upper bound if given specific weights in the last problem? Can we derive similar inequalities for a multi-candidate election?