Power index rankings in bicameral legislatures and the US legislative system

In this paper we study rankings induced by power indices of players in simple game models of bicameral legislatures. For a bicameral legislature where bills are passed with a simple majority vote in each house we give a condition involving the size of each chamber which guarantees that a member of the smaller house has more power than a member of the larger house, regardless of the power index used. The only case for which this does not apply is when the smaller house has an odd number of players, the larger house has an even number of players, and the larger house is less than twice the size of the smaller house. We explore what can happen in this exceptional case. These results generalize to multi-cameral legislatures. Using a standard model of the US legislative system as a simple game, we use our results to study power index rankings of the four types of players—the president, the vice president, senators, and representatives. We prove that a senator is always ranked above a representative and ranked the same as or above the vice president. We also show that the president is always ranked above the other players. We show that for most power index rankings, including the Banzhaf and Shapley–Shubik power indices, the vice president is ranked above a representative, however, there exist power indices ranking a representative above the vice president.

Thanks to Bruce Reznick for several helpful conversations and to several anonymous referees who provided extensive comments and suggestions that greatly improved the paper.

Authors and Affiliations

1. Department of Mathematics, Emory University, Atlanta, GA, 30322, USA

   Victoria Powers

Corresponding author

Correspondence to Victoria Powers.

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Appendix: Products of binomial coefficients

We prove the technical results on products of binomial coefficients that are needed to compare the numbers \(c_i(k)\) for different players in our simple game models. Here are some basic facts about binomial coefficients:

Recall that in order to prove that \(c_s(k) > c_r(k)\), we must prove that inequality (1) holds for k. For the convenience of the reader, the inequality is given again:

This holds iff

Define the following functions: For positive integers \(u < p\) and an integer i such that \(0 \le i < p - u\), define

Then inequality (1) holds iff inequality (4) holds iff

Lemma 4

With f and g as above,

for all i with \(0 \le i < p - u\),

Proof

Using basic properties of binomial coefficients, we have

which proves the claim. \(\square \)

Lemma 5

Let \(m_r, m_s, q_r, q_s \in {\mathbb {N}}\) such that \(1< q_r < m_r\) and \(1< q_s < m_s\). Then (1) holds for \(k = q_r + q_s\) if and only if \(q_s m_r > q_r m_s\). If \(q_s m_r = q_r m_s\), then (1) holds with > replaced by \(=\).

Proof

Inequality (1) holds for \(k = q_r + q_s\) iff \(f(m_r,q_r,0) > f(m_s,q_s,0)\). It is easy to check that \(f(p,u,0) = p/u\), hence (1) holds iff \(m_r/q_r > m_s/q_s\) iff \(q_s m_r > q_r m_s\). If \(q_s m_r = q_r m_s\), then \(f(m_r,q_r,0) = f(m_s,q_s,0)\) and we have equality in (1). \(\square \)

Proposition 10

Let \(m_r, m_s, q_r, q_s, \in {\mathbb {N}}\) such that \(m_s < m_r\), \(1< q_s < m_s\), and \(1< q_r < m_r\). Let \(N = \min \{m_s + q_s, m_r + q_r \}\). Then

1. (a)

   Inequality (1) holds for \(k = q_r + q_s + 1\) if and only if

   $$\begin{aligned} \frac{(m_r - q_r)}{(q_r + 1)} \frac{m_r}{q_r} > \frac{(m_s - q_s)}{(q_s + 1)} \frac{m_s}{q_s}. \end{aligned}$$
2. (b)

   Suppose \(d \in {\mathbb {N}}\) with \(0 \le d \le N\). If \(g(m_r,q_r,d) > g(m_s,q_s,d)\), then \(g(m_r,q_r,i) > g(m_s,q_s,i)\) for all i such that \(d \le i \le N\).

3. (c)

   Suppose \(d \in {\mathbb {N}}\) with \(0 \le d \le N\). If \(f(m_r,q_r,d) > f(m_s,q_s,d)\) and \(g(m_r,q_r,d) > g(m_s,q_s,d)\), then \(f(m_r,q_r,i) > f(m_s,q_s,i )\) for all i such that \(d \le i \le N\).

4. (d)

   If \(f(m_r,q_r,0) > f(m_s,q_s,0)\) and \(g(m_r,q_r,0) > g(m_s,q_s,0)\), then inequality (1) holds for all k such that \(q_r + q_s \le k \le N\). It follows that inequality (1) holds for all k such that \(q_r + q_s \le k \le N\) if the following two inequalities hold:

   $$\begin{aligned} q_s m_r> q_r m_s, \quad (m_r-q_r)(q_s + 1) > (m_s -q_s)(q_r + 1). \end{aligned}$$

Proof

As noted above, inequality (1) holds for k iff \(f(m_r,q_r, k - q_r - q_s) > f(m_s,q_s, k - q_r - q_s)\).

1. (a)

   Inequality (1) holds for \(k = q_r + q_s + 1\) iff \(f(m_r,q_r,1) > f(m_s,q_s,1)\). By Lemma 4, \(f(p,u,1) = g(p,u,0) \cdot f(p,u,0)\), hence we need

   $$\begin{aligned} g(m_r,q_r,0) \cdot f(m_r,q_r,0) > g(m_s,q_s,0) \cdot f(m_s,q_s,0), \end{aligned}$$

   which yields the claimed inequality.

2. (b)

   The proof is by induction on i. By definition, \(g(m_r,q_r,i) > g(m_s,q_s,i)\) is

   $$\begin{aligned} \frac{m_r - q_r - i}{q_r + i + 1} > \frac{m_s - q_s - i}{q_s + i + 1} \end{aligned}$$

   which is equivalent to \(m_r(q_s + i) + m_r - q_r > m_s(q_r + i) + m_s - q_s\). By assumption this holds for \(i =d\). Assume it is true for some i with \(d \le i < N\). Then

   $$\begin{aligned} m_r(q_s + i + 1) + m_r - q_r= & {} m_r(q_s + i) + ( m_r - q_r ) + m_r \\ > m_s(q_r + i) + m_s - q_s + m_s= & {} m_s(q_r + i + 1) + m_s - q_s, \end{aligned}$$

   since \(m_r > m_s\). It follows that \(g(m_r,q_r,i+1) > g(m_s,q_s,i+1)\) and we are done by induction.

3. (c)

   Since \(f(p,u,i+1) = g(p,u,i) \cdot f(p,u,i)\) by Lemma 4, the claimed result follows easily from (b) by induction.

4. (d)

   This follows from (c) with \(d = 0\), noting that \(g(m_r,q_r,0) > g(m_s,q_s,0)\) iff the second inequality holds.

\(\square \)

Corollary 1

Let \(m_r, m_s, q_r, q_s, \in {\mathbb {N}}\) such that \(m_s < m_r\), \(1< q_s < m_s\), \(1< q_r < m_r\) and let \(N = \min \{ m_s-q_s, m_r - q_r\}\). Suppose we have a such that \(q_r + q_s + 1 \le a \le N\) and (1) holds with \(k = a\). Then (1) holds for all k such that \(a \le k \le N\).

Proof

It is enough to assume that a is minimal such that (1) holds for \(k = a\). For ease of exposition, let \(i = a - q_r - q_s\). Then, by minimality of a, we have \(f(m_r,q_r, i - 1) \le f(m_s,q_s, i - 1)\) and \(f(m_r,q_r, i ) > f(m_s,q_s,i )\). By Lemma 4, \( g(m_r,q_r, i - 1) \cdot f(m_r,q_r, i - 1) > g(m_s,q_s, i - 1) \cdot f(m_s,q_s,i - 1)\). The two inequalities imply \(g(m_r,q_r,i - 1) > g(m_s,q_s,i - 1)\). Then, by Proposition 10(b), \(g(m_r,q_r,i) > g(m_s,q_s,i)\). The result now follows from Proposition 10(c). \(\square \)

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