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arXiv:1709.00625v2 [math.CO] 18 Feb 2019 Power Index Rankings in Bicameral Legislatures and the US Legislative System Victoria Powers ∗ February 19, 2019 Abstract In this paper we study rankings induced by power indices of players in sim- ple game models of bicameral legislatures. For a bicameral legislature where bills are passed with a simple majority vote in each house we give a condi- tion 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, regard- less 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 repre- sentative 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. 1 Introduction A power index assigns a numerical measure of power to each player in a simple game and thus yields a ranking of the players. In this paper we look at power index rankings of the players in simple game models of bicameral legislatures and similar legislative systems. 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 ∗Department of Mathematics, Emory University, Atlanta, GA 30322. Email: vpow- ers@emory.edu 1
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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 easily to the multi-cameral situation. We apply our results and techniques to study power index rankings in the standard simple game model of the US legislative system, which has four types of players: senators, representatives, the president, and the vice president. In the case of the US legislative system, we show that regardless of the power index chosen, the president always has more power than the other players, a senator always has more power than a representative, and a senator always has at least as much power as the vice president. For “reasonable” power indices, including the Banzhaf and Shapley- Shubik indices, the ranking from most power to least power is: president, senator, vice president, representative. These results apply to more general systems which are similar to the US legislative system, with a bicameral legislature plus a president or a bicameral legislature plus a president and vice president. Power indices for this simple game model of the US legislative system have been studied previously, in the context of calculating power with a specific index. For example, in the book by Taylor and Pacelli [7], the authors discuss calculating Banzhaf and Shapley-Shubik power in a model of the US legislative system (without the vice president). Brams, Affuso, and Kilgour [2] study Banzhaf and Johnston power in the same model (no vice president). In both of these cases, the authors are interested in the percent of power held by the players rather than power index rankings. Acknowledgements Thanks to Bruce Reznick for several helpful conversations and to several anonymous referees who provided extensive comments and suggestions that greatly improved the paper. 2 Preliminaries 2.1 Simple Games and Power Indices A (monotonic) simple game is a pair (N, W) where N = {1, 2, … , n} is the set of players and W is a set of subsets of N, called the winning coalitions, such that • ∅̸∈W. 2
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• N ∈W. • If S ∈W and S ⊆T, then T ∈W. The minimal winning coalitions are the winning coalitions for which no proper subset is winning. The set of winning coalitions is determined by the minimal ones since a subset S ∈W if and only if S contains a minimal winning coalition. A simple game is a model of a yes-no voting system in which the players are deciding on a single alternative such as a motion, bill, or amendment. The winning coalitions are precisely the sets of players that can force a bill to pass if they all support it. Given a simple game (N, W) and S ∈W containing player i, we say i is critical in S if S is winning and S \ {i} is losing. For i ∈N and 1 ≤k ≤n, let Ci = {S ∈W | i is critical in S}, Ci(k) = {S ∈Ci | |S| = k}. Let ci(k) = |Ci(k)|, the number of coalitions of size k in which i is critical. The numbers ci(k) are called critical numbers. Definition 1. Define a binary relation on N by i ⪰j iffci(k) ≥cj(k) for all k such that 1 ≤k ≤n, and write i ≻j if ci(k) > cj(k) for all k such that ci(k) and cj(k) are not both zero. Following [3], we call ⪰the weak desirability relation. 2.2 Power Indices Power indices are a way to measure the relative power of the players in a simple game. The most famous of these are the Shapley-Shubik index [6] and the Banzhaf index [1]. Semivalues were introduced in 1979 by Weber [8] as a generalization of the notion of a power index to general cooperative games. Dubey et al. [4] show that semivalues can be characterized in terms of a weighting vector (λ1, … , λn) such that λk ≥0 for all k and Pn k=1 λk �n−1 k−1 = 1. Given a power index Φ with weighting vector (λ1, … , λn), the Φ-power of a player i is defined by Φ(i) := n X k=1 λk ci(k). Thus the λi’s give a weighting of a player’s contribution to coalitions of size k. The Shapley-Shubik power index is defined by weighting coefficients λk = 1/ n �n−1 k−1 and the Banzhaf power index is defined by weighting coefficients λk = 1/2n−1. Any power index Φ defines a ranking on the set of players in a simple game and we write i ≥Φ j to denote that Φ(i) ≥Φ(j) and i >Φ j if Φ(i) > Φ(j). Clearly, different power indices can lead to different rankings for the same game. In [5], Saari and 3
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Sieberg look at rankings of players coming from power indices in cooperative games. They show that different indices can generate radically different rankings and that there can be many different rankings even for games with a relatively small number of players. This is in contrast to our results which show that there are only two possible rankings for a simple game model of the US legislative system. The following results follow easily from the definitions, see also [3, Theorem 3.4]. Proposition 1. Let i and j be two players in a simple game. (a) If i ⪰j, then for any power index Φ, i ≥Φ j. (b) Suppose there exist k, m such that ci(k) > cj(k) and ci(m) < cj(m). Then there exist power indices Φ and Ψ such that i >Φ j and j <Ψ i. 3 Bicameral Voting Systems We look at power in a bicameral legislative system where bills are passed with a simple majority in each house. We show that a member of the smaller house has more power than a member of the larger house regardless of the choice of power index used to measure power, apart from the following case: 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. In this exceptional case, the choice of power index will determine whether the members of the smaller house or the larger house have the most power. Recall that for n, k ∈N with 0 ≤k ≤n, the binomial coefficient �n k denotes the number of ways of choosing k elements from a set of n elements. For the simple games we study in this work, formulas for the critical numbers involve products of binomial coefficients. Results on binomial coefficients needed in this section can be found in the appendix. Suppose we have a simple game with two types of players, say that there are ms senators and mr representatives, and we want to count the number of coalitions that contain a specific senator and have exactly x senators and y representatives. Creating such a coalition consists of choosing x −1 players from ms −1 senators and choosing y players from the mr representatives. Thus there are �ms−1 x−1 · �mr y such coalitions. Let cs(k) denote the number of coalitions of size k in which a senator is critical and define cr(k) similarly for a representative. Lemma 1. Suppose that qs and qr are the the quotas for passage of a bill, i.e., the minimal winning coalitions consist of qs senators and qr representatives. Then 4
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(i) cs(k) ̸= 0 iffqs + qr ≤k ≤qs + mr and cr(k) ̸= 0 iffqs + qr ≤k ≤qr + ms. (ii) For k ∈N with qs + qr ≤k ≤qs + mr, cs(k) = ms −1 qs −1 · mr k −qs . (iii) For k ∈N with qs + qr ≤k ≤qr + ms, cr(k) = mr −1 qr −1 · ms k −qr . Proof. Since the minimal winning coalitions are exactly the coalitions with qs sena- tors and qr representatives, the coalitions in which a fixed senator is critical consist of the senator plus qs −1 of the ms −1 other senators along with between qr and mr of the mr representatives. The coalitions in which a particular representative is critical consist of the representative plus qr −1 of the mr −1 other representatives along with between qs and ms senators. The assertions follow easily from these observations. The lemma implies that to show cs(k) > cr(k) we must prove the following inequal- ity: ms −1 qs −1 · mr k −qs
mr −1 qr −1 · ms k −qr (1) This inequality is studied in the appendix. For the rest of this section we assume that ms < mr and the quotas correspond to simple majorities, so that qs = ⌈(ms + 1)/2⌉and qr = ⌈(mr + 1)/2⌉. Lemma 2. We have mr −qr ≥ms −qs. Hence, by Lemma 1, cs(k) = 0 implies cr(k) = 0. Proof. Suppose mr is odd, say mr = 2x + 1, then qr = x + 1. If ms = 2y + 1, then qs = y +1 and x > y, hence mr −qr = x > y = ms −qs. If ms = 2y, then qs = y +1 and x ≥y, hence mr −qr = x > y −1 = ms −qs. Now suppose mr is even, say mr = 2x, then qr = x + 1. If ms = 2y + 1, then x > y and mr −qr = x −1 ≥y = ms −qs. If ms = 2y, then x > y and mr −qr = x −1 > y −1 = ms −qs. Proposition 2. If qs ·mr > qr ·ms, then cs(k) > cr(k) for all k such that cs(k) ̸= 0. Proof. We need to show that Inequality (1) holds for all k with qr+qs ≤k ≤mr+qs. Case 1: If qr + qs ≤k ≤qr + ms, then cr(k) ̸= 0 and we use Proposition 10 (d). Since qs ·mr > qr ·ms, we need only show that (mr −qr)(qs +1) > (ms −qs)(qr +1). 5
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This is equivalent to mrqs + mr −qr > msqr + ms −qs. Since mr · qs > ms · qr by assumption and mr −qr ≥ms −qs by Lemma 2, the inequality holds. Case 2: If ms + qr < k ≤mr + qs, then cr(k) = 0 and cs(k) ̸= 0, by Lemma 1 (a). Thus cs(k) > cr(k) is clear. Note that qs · mr > qr · ms is equivalent to qs/ms > qr/mr and thus the proposition says that as long ms < mr and the proportion of the smaller house needed to pass a bill is larger than the proportion needed in the bigger house, then for any senator S and any representative R, S ≻R. Theorem 1. If ms < mr and the quotas qs and qr correspond to simple majorities, then in each of the following cases, cs(k) > cr(k) for all k such that cs(k) ̸= 0: (i) ms and mr are both odd, (ii) ms and mr are both even, (iii) ms is even and mr is odd, (iv) ms is odd, mr is even, and mr > 2ms. Thus in these cases, for a senator S and a representative R, S >Φ R for any power index Φ that does not assign 0 power to both. Proof. By Proposition 2, we need only show that qs · mr > qr · ms, equivalently, qs · mr −qr · ms > 0. Case (i). If ms and mr are both odd, say ms = 2x + 1 and mr = 2y + 1 with x < y, then qs = x + 1, qr = y + 1 and ms · qr −mr · qs = y −x > 0. Cases (ii) and (ii) are similarly easy to check. Case(iv). Suppose ms = 2x + 1, mr = 2y, and mr > 2ms. Then qs · mr −qr · ms = y −2x −1, hence qs · mr −qr · ms > 0 iffy > 2x + 1 = ms iff2y > 4x + 2, i.e., iff mr > 2ms. 3.1 The Exceptional Case We now consider the remaining case not covered by Theorem 1: ms is odd, mr is even, and mr ≤2ms. In this case, the relationship between the cs(k)’s and cr(k)’s is more complicated. Assume ms = 2x + 1 and mr = 2y, so that qs = x + 1 and 6
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qr = y + 1. We first show that in this case, for the minimal winning coalitions (k = qs + qr) the critical number for a member of the larger house is greater than or equal to the critical number for a member of the smaller house. Lemma 3. With the above assumptions, cs(qs + qr) < cr(qs + qr) if mr < 2ms. If mr = 2ms, then cs(qs + qr) = cr(qs + qr) . Proof. By Lemma 5, cs(qs + qr) < cr(qs + qr) if qsmr < qrms and cs(qs + qr) = cr(qs + qr) if qsmr = qrms. If mr ≤2ms, then y < 2x + 1. We have qsmr −qrms = 2y(x + 1) −(2x + 1)(y + 1) = y −(2x + 1). If mr < 2ms, then y < 2x + 1 and qsmr −qrms < 0, hence qsmr < qrms. If mr = 2ms, then y = 2x + 1 and qsmr < qrms. The relationship in the case where mr = 2ms, i.e., the gap between the sizes of the two houses is as large as possible, is almost the same as in the non-exceptional cases: Proposition 3. Suppose ms = 2x + 1 and mr = 2ms = 4x + 2. Then (a) cs(k) ̸= 0 for 3x + 3 ≤k ≤5x + 3. (b) cs(3x + 3) = cr(3x + 3) (c) cs(k) > cr(k) for 3x + 4 ≤k ≤5x + 3. It follows that S >Φ R for any power index apart from the one that assigns weight λk = 0 for all k ̸= 3x + 3, which ranks S and R equally. Proof. We have qs = x + 1 and qr = 2x + 2. (a) follows from Lemma 1. (b) follows from Lemma 3. (c) We need to prove that inequality (1) holds for all k in the given range. By Corollary 1 in the appendix, it is enough to prove it for k = qr +qs +1 = 3x+4. By Proposition 10 (a) and using the fact that mr = 2ms and qr = 2qs, the inequality holds iff 4ms(ms −qs) 2qs(2qs + 1) > ms(ms −qs) qs(qs + 1) , which is equivalent to 2qs + 2 > 2qs + 1. We now look at the remaining “extreme” case within the exceptional case, i.e., the case where the gap between the two houses is as small as possible. 7
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Proposition 4. Suppose ms = 2x + 1 and mr = 2x + 2 so that qs = x + 1 and qr = x + 2. Then (a) cs(k) ̸= 0 for 2x + 3 ≤k ≤3x + 3 and cr(k) ̸= 0 iffcs(k) ̸= 0 (b) cs(k) < cr(k) for all k such that cs(k) ̸= 0. It follows that for any representative R and any senator S, R ≻S in this case. Proof. (a) The minimal winning coalitions have size qs +qr = 2x+3 and the largest coalition for which S or R is critical has size qs + mr = 3x + 3 = qr + ms. (b) cs(2x + 3) < cr(2x + 3) by Lemma 3. By Corollary 1 in the appendix, if cs(k) > cr(k) for some k > 2x+3, then cs(i) > cr(i) for all i ≥k. Thus it is enough to show that this fails for the largest possible value of k, i.e., it is enough to show that cs(3x + 3) < cr(3x + 3), which is easy: cs(3x + 3) = 2x x · 2x + 2 2x + 2 < 2x + 1 x + 1 · 2x + 1 2x + 1 = cr(3x + 3). Here is a summary of what happens in this exceptional case. Fix mr and ms and assume mr ̸= 2ms. As noted above, we have qs ·mr > qr ·ms so that the proportion of the smaller house needed for passage is less than the proportion that the bigger house needs. This shifts the advantage to the bigger house for the minimal winning coalitions. Then in the first extreme case, i.e., when the gap between the sizes of the two houses is as large as possible, the advantage only helps for minimal winning coalitions, so that for all other sizes of coalitions, the smaller house has the advantage. In the second extreme case, i.e., when the gap between the size of the two houses is as small as possible, the advantage stays with the bigger house for all coalition sizes. In cases in between the two extremes, what happens is that the bigger house starts out with an advantage, then at some point the advantage shifts to the smaller house and remains there as the sizes of the coalitions increase. For the biggest possible gap, this shift occurs as soon as the coalitions are no longer minimal. For the first extreme case, this shift never happens. The bigger the gap, the sooner the shift will happen. As an example, consider a case in between the two extremes: ms = 101 and mr = 150, so that qs = 51 and qr = 76. Then we have cs(k) < cr(k) for the two smallest values of k (k = 127 and k = 128) cs(k) > cr(k) for the remaining values of k for which cs(k) ̸= 0. What this means is that in the first extreme case (gap between the size of the houses is as large as possible), for all power indices φ, members of the smaller house have 8
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more power than members of the larger house. For the second extreme case (gap between the size of the houses is as small as possible), the members of the larger house have more power than the members of the smaller house for any power index. Finally, for the case in between the two extremes, there will be power indices for which members of the smaller house have more power and others for which member of the larger house have more power. 3.2 Generalization to Multi-cameral Legislatures The results on bicameral legislatures generalize relatively easily to a legislative body with any number of houses. Assume that we have n houses, denoted H1, … , Hn and that Hj has mj members. The relationship between the power of a member of Hj and Hl is the same as the relationship they would have if there were only two houses. Intuitively, this makes sense because being critical in a coalition for a particular member of a house is independent of the makeup of the coalitions in other houses. For each j, let qj denote the minimum number of votes needed in Hj to pass a motion. The following notation will be useful. Let [n] denote {1, 2, … , n}. Given I ⊆[n] and k ∈N, let UI(k) denote the number of different subsets S ⊆∪j∈IHj of size k such that for each j ∈I, |S ∩Hj| ≥qj. In other words, UI(k) is the number of ways of building a set of coalitions, one from each house in {Hj}j∈I, such that each coalition in Hj meets the threshold qj needed to pass legislation. Then 0 < UI(k) ≤P j∈I mj. Using this notation, notice that for a fixed j, a specific member r ∈Hj, and k ∈N we have cr(k) = mj −1 qj −1 · U[n]{j}(k −qj). Theorem 2. With notation as above, suppose that each qj represents a simple majority. Let r ∈Hj and s ∈Hl, where j ̸= l. Then (a) For each of the following cases, cs(k) > cr(k) for all k ∈N such that cs(k) ̸= 0: mj < ml and mj and ml are both odd, both even, mj is even and ml is odd, or mj is odd, ml is even and ml > 2mj. (b) If mj is odd and ml is even, then the relationship between cs(k) and cr(k) mirrors the relationship detailed in Section 3.1. Proof. We can build all coalitions of size k in which r is critical as follows: Choose qj −1 of the mj −1 members of Hj that are not r, then choose ql + d members of Hl, where d ranges from 0 to ml −ql, and finally choose, if possible, subsets of the 9
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remaining houses which meet the minimum so that the size of the resulting winning coalition is k. Thus cr(k) = ml−ql X d=0 mj −1 qj −1 · ml ql + d · U[n]−{i,j}(k −qj −ql −d), and similarly, cs(k) = mj−qj X d=0 ml −1 ql −1 · mj qj + d · U[n]−{i,j}(k −qj −ql −d). (a) It is easy to check in this case that ml −ql > mj −qj and thus, using the formulas above, to show cr(k) > cs(k) it is enough to show that for 0 ≤d ≤mj −qj, �mj−1 qj−1 · � ml ql+d
�ml−1 ql−1 · � mj qj+d . This follows from Proposition 10 exactly as in the proof of Theorem 1. (b) The same argument works in this case. 4 The US Legislative System We apply our results on bicameral legislative systems to study power in the US legislative system. We model the US legislative system as a simple game with 537 players: the president, vice president, 100 senators in the Senate, and 435 representatives in the House of Representatives. A bill passes if a majority of the senators and a majority of the representatives vote yes and the president signs the bill. If the president does not sign the bill, it can be passed with a supermajority of at least 67 senators and 290 representatives. The role of the vice president is to break ties in the Senate. For winning coalitions in which the president is critical, the vice president plays the same role as a senator; in these cases we can assume that the senate contains 101 players. We will call the set of senators plus the vice president the “full Senate”. There are two types of minimal winning coalitions: I. 51 from the full Senate, 218 representatives, and the president; II. 67 senators and 290 representatives. We look at critical instances for the four types of players in order to compare the critical numbers. Note that if a winning coalition contains exactly 51 senators, then every senator is critical and adding the vice president yields a coalition in which no senator is critical. Apart from this case, if a player who is not the vice president is 10
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critical in a coalition that does not contain the vice president, then this player is still critical if the vice president is added to the coalition. For ease of exposition, we write P and V for the president and vice-president and let S be a fixed senator and R a fixed representative. We write cp(k) (resp. cv(k), cs(k), cr(k)) for the number of coalitions of size k in which the president (resp. the vice-president, a senator, a representative) is critical. The following table lists the different types of coalitions, along with their sizes, in which the president (P1 - P4), a senator (S1 - S3), a representative (R1 - R3), or the vice president (V) are critical, along with the possible sizes. Table 1: Critical numbers in the US system Type Members Size P1 P, 51 −66 from the Senate, 218 −435 representatives 270 −502 P2 P, V, 50 −66 from the Senate , 218 −435 representatives 270 −503 P3 P, 67 −100 from the Senate, 218 −289 representatives 286 −390 P4 P, V, 67 −100 from the Senate, 218 −289 representatives 287 −391 S1 P, S, 50 others from the full Senate , 218-435 representatives 270 −487 S2 S, 66 other senators, 290 −435 representatives 357 −502 S3 S, 66 other senators, 290 −435 representatives, V 358 −503 R1 P, R, 217 other representatives, 51 −101 from the full Senate 270 −320 R2 R, 289 other representatives, 67 senators 357 R3 R, 289 other representatives, 68 −101 from the full Senate 358 −391 V V, P, 50 from the Senate, and 218 −435 representatives 270 −487 Proposition 5. (a) If cp(k) = 0, then cv(k) = cr(k) = cs(k) = 0. (b) For all k ∈N such that cp(k) ̸= 0, cp(k) > cv(k) (c) For all k ∈N such that cp(k) ̸= 0, cp(k) > cs(k). Proof. (a) follows immediately from Table 1, the table of coalition sizes. Fix k such that cp(k) ̸= 0, then 270 ≤k ≤503. Let Cp(k) denote the set of coalitions of size k in which the president is critical and Cs(k) the set of coalitions of size k in which a senator is critical. (b): Every coalition in which V is critical contains P, and P is also critical, thus cp(k) ≥cv(k). In addition, given S ∈Cv(k), the coalition formed by removing V and adding a senator not already in S is in Cp(k) and not in Cv(k). Hence cp(k) > cv(k). (c): Define a function f : Cs(k) →Cp(k) as follows: Given S in Cs(k), if S is type S1, then P is critical in S and we define f(S) = S. If S is type S2 or S3, then the coalition S′ = (S \ {S}) ∪{P} is in Cp(k) since it contains only 66 senators, and we define f(S) = S′. Then f is clearly injective, hence cp(k) ≥cs(k). To show that 11
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the inequality is strict we need only show that f is not surjective. If 270 ≤k ≤356, then there are no type S2 or S3 coalitions in Cs(k). Thus any coalition in Cp(k) that does not contain S is not in Im f, and there are clearly many of these. Now suppose 357 ≤k ≤502 and let ˜S be a coalition in Im f of the form (S \ {S}) ∪{P} with S ∈Cs(k) of type S2 or S3. Then ˜S has exactly 66 senators and 434 or less representatives and we can construct a new coalition in Cp(k) by replacing any senator in ˜S by a representative not already in ˜S. This clearly yields a coalition in Cp(k) that is not in Im f. Hence f is not surjective in this case. For k = 503, coalitions in Cs(k) consist of S plus 66 other senators, 435 representa- tive, and V; while those in Cp(k) consist of P plus 66 senators, 435 representatives, and V. Then cp(503) = �100 66
�99 66 = cs(503). Therefore in all cases we have cp(k) > cs(k). Proposition 6. If cs(k) = 0, then cv(k) = 0. If cs(k) ̸= 0, then 270 ≤k ≤503 and we have (i) If 270 ≤k ≤356, then cs(k) = cv(k). (ii) If 357 ≤k ≤503, then cs(k) > cv(k). Proof. The first statement follows immediately from Table 1. Fix k with 270 ≤k ≤ 503. Coalitions in Cv(k) consist of V plus 50 senators, k −52 representatives, and the president, hence for 270 ≤k ≤487, cv(k) = 100 50 · 435 k −52 . (i) Since k ≤356, coalitions in Cs(k) are type S1 only, thus they consist of S plus 50 others from the full Senate, k −52 representatives, and P. Hence cs(k) = 100 50 · 435 k −52 = cv(k). (ii) For 488 ≤k ≤503, cs(k) > 0 and cv(k) = 0, so this is clear. For 357 ≤k ≤487, we note that in addition to the coalitions in Cs(k) of type S1 above, there are coalitions of type S2 or S3 and V is never critical in these, hence ck(s) > cv(k). Proposition 7. If cs(k) = 0, then cr(k) = 0. For all k ∈N such that cs(k) ̸= 0, cs(k) > cr(k). Proof. The first statement follows immediately from Table 1. If 321 ≤k ≤356 or 392 ≤k ≤503, then cr(k) = 0 and cs(k) ̸= 0, so there is nothing to prove. We break the remaining values of k into three cases: 270 ≤k ≤320, k = 357, and 358 ≤k ≤391. 12
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Case 1. For 270 ≤k ≤320,the coalitions in Cs(k) are of type S1 and thus consist of S plus 50 others from the full Senate, k −52 representatives and the president. Coalitions in Cr(k) are of type R1 and hence consist of R plus 217 other represen- tatives, the president, and k −219 from the full Senate. It follows that cs(k) = 100 50 · 435 k −52 , cr(k) = 101 k −219 · 434 217 . We apply Proposition 10 (d) with ms = 101, mr = 435, qs = 51, and qr = 218. The conditions qsmr > qrms and (mr −qr)(qs + 1) > (ms −qs)(qr + 1) are easily checked. Hence, by the proposition, cs(k) > cr(k) for all k. Case 2. k = 357. Coalitions in Cs(357) consist of S, 66 other senators, and 290 representatives, while coalitions in Cr(357) consist of R, 289 other representatives, and 67 senators. Then cs(357) = 99 66 · 435 290
100 67 · 434 289 = cr(357). Case 3. For 358 ≤k ≤391, the coalitions in Cs(k) consist of S, 66 other senators and either k −67 representatives, or V and k −68 representatives. The coalitions in Cr(k) consist of R, 289 other representatives, and either k −290 senators or V and k −291 senators. Thus cs(k) = 99 66 · 435 k −67 + 99 66 · 435 k −68 cr(k) = 100 k −290 · 434 289 + 100 k −291 · 434 289 . (2) By Proposition 10(d) with ms = 100, mr = 435, qs = 67, and qr = 290, 99 66 · 435 k −67
100 k −290 · 434 289 , and with ms = 100, mr = 435, qs = 66, and qr = 289, 99 66 · 435 k −68
100 k −291 · 434 289 . Therefore cs(k) > cr(k). Finally, we compare the numbers cv(k) and cr(k). Apart from a narrow range of k’s, cv(k) is the larger of the two. 13
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Proposition 8. (a) If cv(k) = 0, then cr(k) = 0. (b) Suppose k ∈N such that cv(k) ̸= 0, so that 270 ≤k ≤487. If 270 ≤k ≤356 or 380 ≤k ≤487, cv(k) > cr(k). For the remaining k, i.e., 357 ≤k ≤379, cr(k) > cv(k). Proof. (a) This follows from Table 1. (b) Recall that coalitions in Cv(k) consist of V, 50 senators, the president, and k−52 representatives. Thus, for all such values of k, we have cv(k) = �100 50 � 435 k−52 . Case 1: For 270 ≤k ≤356, cv(k) = cs(k) > cr(k), by Proposition 6 and Proposition 7. Case 2: For k = 357, the only coalitions in which R is critical are of type R2 and consist of R plus 289 other representatives and 67 senators. Hence we have cv(357) = 100 50 435 305 < 434 289 100 67 = cr(357). Case 3: For 358 ≤k ≤390, coalitions in which R is critical are of type R3 and they consist of R plus 289 other representatives and k −290 members of the full Senate. Thus we have cr(k) = 434 289 101 k −290 (3) for these k. Using the computer algebra software Mathematica, we find that 434 289 101 k −290
100 50 435 k −52 iff358 ≤k ≤379. Case 4: For k = 391 we have cv(391) = 100 50 435 239
434 289 = cr(391), as claimed. Case 5: For 392 ≤k ≤487, cr(k) = 0 and cv(k) > 0, so cv(k) > cr(k). Theorem 3. In the simple game modeling the US legislative system, the weak de- sirability relation yields P ≻S ≻R and S ⪰V. It follows that (a) For any power index Φ for which Φ(p) ̸= 0 and Φ(s) ̸= 0, we have P >Φ S >Φ R and S ≥Φ V. (b) If Φ is the Banzhaf or Shapley-Shubik index, we have P >Φ S >Φ V >Φ R. 14
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Proof. Propositions 5, 6, and 7 imply that P ≻S ≻R and S ⪰V and (a) follows from this by Proposition 1. (b): By (a), we need only show that V >Φ R if Φ is the Banzhaf or Shapley-Shubik index. Recall that cv(k) ̸= 0 for 270 ≤k ≤487 and for these k, cv(k) = �100 50 � 435 k−52 , while cr(k) ̸= 0 for 270 ≤k ≤391. Suppose Φ is Banzhaf power, then Φ(V) = 1 2536 487 X k=270 100 50 435 k −52 . For R we must add up the contributions from the three types of critical instances. For type R1, cr(k) = �434 217 � 101 k−219 , type R2 corresponds to cr(357) = �434 289 �100 67 , and for type R3, cr(k) = �434 289 � 101 k−290 . Thus Φ(R) = 1 2536 320 X k=270 434 217 101 k −219 + 434 289 100 67 + 487 X 270 434 289 101 k −290 ! . It is easy to check that Φ(V) > Φ(R) using Mathematica. If Φ is Shapley-Shubik power then the calculation of Φ(V) and Φ(R) is as for Banzhaf power except that we must multiply each cr(k) and cv(k) by the weight λk = 1/ n �n−1 k−1 . Thus we need only use Mathematica to verify that 487 X k=270 λk 100 50 435 k −52
320 X k=270 λk 434 217 101 k −219
- λ357 434 289 100 67
487 X 270 λk 434 289 101 k −290 . Remark. The proofs of Propositions 5, 6, and 7 did not depend on the specific numbers of representatives and senators in the US system and the quotas in the sense that as long as the assumptions of Lemma 1 and of Proposition 10 hold for ms, mr, qs, and qr, then the conclusions of these propositions hold. However, comparing the ranking of a representative and the vice-president using Proposition 8 involves the specific numbers in the US system and thus does not generalize immediately. 4.1 Supermajority Rules For some bills in the US Senate a supermajority of 60 or more senators are required to vote yes in order to break a Filibuster. In this case, we still have S ≻R for S a 15
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senator and R a representative In fact, S ≻R will hold regardless of the number of votes needed to pass a bill in the Senate, as long as this number is greater than 50. Proposition 9. Suppose qs votes are needed to pass legislation in the Senate, where 51 ≤qs ≤100, and everything else remains the same. Then we have P ≻S ≻R as before. Proof. The proof that P ≻S generalizes immediately. To show that S ≻R we first note that the conditions of Lemma 1 are still satisfied and thus we have cs(k) = 0 implies cr(0) = 0. We can then apply Proposition 10 (d) from the appendix as in the proof of Proposition 7. The conditions needed are (1) qsmr > qrms and (2) (mr −qr)(qs + 1) > (ms −qs)(qr + 1). In this case, we have ms = 100, mr = 435, and mr = 218. For (1) we have (a)(435) > (218)(100) iffa ≥51, and for (2) we have (435−218)(a+1) > (100−a)(219), which holds iffa ≥50. This proves S ≻R. The proof that the P ≻S does not depend on the number of senators needed to pass legislation when the president votes yes, hence P ≻S still holds in the this case. Suppose that the House of Representatives decided to raise the quota for passing bills in order to insure that members have more power than senators. Assume that only 51 votes are needed in the Senate. Using Table 1, we see that for winning coalitions containing the president, the range of k in which cs(k) ̸= 0 is 270 ≤k ≤ 503, thus in order to insure that there are no k such that cs(k) ̸= 0 and cr(k) = 0, there must be coalitions in which a representative is critical with sizes up to 503. It follows that the minimum number of representatives needed to pass a bill in the House would have to be at least 401 (503 −102). If qr = 401 and qs remains 51, then (401)(101) > (51)(435), so for k = qr + qs + 1 (minimal winning coalitions), by Lemma 5, cr(k) > cs(k). For the largest possible coalitions for which cs(k) ̸= 0, i.e., k = 503, we have cr(503) = 434 400 101 101
100 50 435 435 = cs(503). Since cr(k) > cs(k) for k the smallest and largest possible values, by Corollary 1, we have R ≻S in this case. Hence it is possible for the House to adopt a supermajority rule which would give their members more power than members of the Senate. Remark. The above discussion shows that in a bicameral legislative system mem- bers of a chamber can increase their power by adopting supermajority a higher quota for passage of bills. 16
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5 Concluding Remarks We have investigated rankings induced by power indices of players and the weakly desirable relation in simple game models of bicameral legislative systems. In three cases (depending on the relative parity of the sizes of the houses), a member of the smaller house is ranked above a member of the larger house, regardless of the power index used. We showed that a sufficient condition for the members of the smaller house to always have more power than the members of the larger house is that qs/ms > qr/mr , where ms < mr are the sizes of the houses and qr, qs are the thresholds needed to pass a bill, i.e., the proportion of the smaller house needed to pass a bill is greater than the proportion of the larger house needed to pass a bill. In the fourth (exceptional case), where the size ms of the smaller house is odd and the size mr of the bigger house is even, if mr ≤2ms, then the condition qs/ms > qr/mr fails. When the gap is as small as possible, i.e., mr = ms + 1, then members of the larger house will be ranked above members of the smaller house by all power indices. For the largest possible gap, i.e., mr = 2ms, the larger house has an advantage only for minimal winning coalitions. All of these results generalize to multicameral legislatures. Our main application of these results is to a standard simple game model of the US legislative system. We showed that the president always has the most power, a sena- tor always has more power than a representative, and that a representative has more power than the vice-president for most power indices, including the Banzhaf and Shapley-Shubik indices. Most of these results apply to similar legislative systems. 6 Appendix: Products of Binomial Coefficients We prove the technical results on products of binomial coefficients that are needed to compare the numbers ci(k) for different players in our simple game models. Here are some basic facts about binomial coefficients: n k
n! k!(n −k)!, n k + n k + 1
n + 1 k + 1 , n k + 1 / n k = n −k k + 1 Recall that in order to prove that cs(k) > cr(k), we must prove that inequality (1) holds for k. For the convenience of the reader, the inequality is given again: ms −1 qs −1 · mr k −qs
mr −1 qr −1 · ms k −qr (1) This holds iff mr k −qs / mr −1 qr −1
ms k −qr / ms −1 qs −1 . (4) 17
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Define the following functions: For positive integers u < p and an integer i such that 0 ≤i < p −u, define f(p, u, i) := p u + i / p −1 u −1 , g(p, u, i) := p −u −i u + i + 1. Then inequality (1) holds iffinequality (4) holds iff f(mr, qr, k −qr −qs) > f(ms, qs, k −qr −qs). Lemma 4. With f and g as above, f(p, u, i + 1) = g(p, u, i) · f(p, u, i) for all i with 0 ≤i < p −u, Proof. Using basic properties of binomial coefficients, we have f(p, u, i + 1)/f(p, u, i) = p u + i + 1 / p u + i = p −u −i u + i + 1, which proves the claim. Lemma 5. Let mr, ms, qr, qs ∈N such that 1 < qr < mr and 1 < qs < ms. Then (1) holds for k = qr +qs if and only if qsmr > qrms. If qsmr = qrms, then (1) holds with > replaced by =. Proof. Inequality (1) holds for k = qr + qs ifff(mr, qr, 0) > f(ms, qs, 0). It is easy to check that f(p, u, 0) = p/u, hence (1) holds iffmr/qr > ms/qs iffqsmr > qrms. If qsmr = qrms, then f(mr, qr, 0) = f(ms, qs, 0) and we have equality in (1). Proposition 10. Let mr, ms, qr, qs, ∈N such that ms < mr, 1 < qs < ms, and 1 < qr < mr. Let N = min{ms + qs, mr + qr}. Then (a) Inequality (1) holds for k = qr + qs + 1 if and only if (mr −qr) (qr + 1) mr qr
(ms −qs) (qs + 1) ms qs . (b) Suppose d ∈N with 0 ≤d ≤N. If g(mr, qr, d) > g(ms, qs, d), then g(mr, qr, i) > g(ms, qs, i) for all i such that d ≤i ≤N. (c) Suppose d ∈N with 0 ≤d ≤N. If f(mr, qr, d) > f(ms, qs, d) and g(mr, qr, d) > g(ms, qs, d), then f(mr, qr, i) > f(ms, qs, i) for all i such that d ≤i ≤N. 18
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(d) If f(mr, qr, 0) > f(ms, qs, 0) and g(mr, qr, 0) > g(ms, qs, 0), then inequality (1) holds for all k such that qr + qs ≤k ≤N. It follows that inequality (1) holds for all k such that qr +qs ≤k ≤N if the following two inequalities hold: qsmr > qrms, (mr −qr)(qs + 1) > (ms −qs)(qr + 1). Proof. As noted above, inequality (1) holds for k ifff(mr, qr, k −qr −qs) > f(ms, qs, k −qr −qs). (a) Inequality (1) holds for k = qr +qs +1 ifff(mr, qr, 1) > f(ms, qs, 1). By Lemma 4, f(p, u, 1) = g(p, u, 0) · f(p, u, 0), hence we need g(mr, qr, 0) · f(mr, qr, 0) > g(ms, qs, 0) · f(ms, qs, 0), which yields the claimed inequality. (b) The proof is by induction on i. By definition, g(mr, qr, i) > g(ms, qs, i) is mr −qr −i qr + i + 1 > ms −qs −i qs + i + 1 which is equivalent to mr(qs + i) + mr −qr > ms(qr + i) + ms −qs. By assumption this holds for i = d. Assume it is true for some i with d ≤i < N. Then mr(qs + i + 1) + mr −qr = mr(qs + i) + (mr −qr) + mr > ms(qr + i) + ms −qs + ms = ms(qr + i + 1) + ms −qs, since mr > ms. It follows that g(mr, qr, i + 1) > g(ms, qs, i + 1) and we are done by induction. (c) Since f(p, u, i + 1) = g(p, u, i) · f(p, u, i) by Lemma 4, the claimed result follows easily from (b) by induction. (d) This follows from (c) with d = 0, noting that g(mr, qr, 0) > g(ms, qs, 0) iffthe second inequality holds. Corollary 1. Let mr, ms, qr, qs, ∈N such that ms < mr, 1 < qs < ms, 1 < qr < mr and let N = min{ms−qs, mr −qr}. Suppose we have a such that qr +qs+1 ≤a ≤N and (1) holds with k = a. Then (1) holds for all k such that a ≤k ≤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 −qr −qs. Then, by minimality of a, we have f(mr, qr, i −1) ≤f(ms, qs, i −1) and f(mr, qr, i) > f(ms, qs, i). By Lemma 4, g(mr, qr, i−1)·f(mr, qr, i−1) > g(ms, qs, i−1)·f(ms, qs, i−1). The two inequalities imply g(mr, qr, i −1) > g(ms, qs, i −1). Then, by Proposition 10(b), g(mr, qr, i) > g(ms, qs, i). The result now follows from Proposition 10 (c). 19
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References [1] J.F. Banzhaf, Weighted voting doesn’t work: a mathematical analysis, Rutgers Law Review 19 (1965), 317–343. [2] S. Brams, P. Affuso, and D.M. Kilgour, Presidential power: a game-theoretic analysis, The Presidency in American Politics (P. Brace, C. Harrington, and G. King, eds.), New York University Press, 1989, pp. 55–74. [3] F. Carreras and J. Freixas, On ordinal equivalence of power measures given by regular semivalues, Math. Social Sci. 55 (2008), 221–234. [4] P. Dubey, P. Neyman, and R.J. Weber, Value theory without efficiency, Math. Oper. Res. 6 (1981), 122–128. [5] D. Saari and K. Sieberg, Some surprising properties of power indices, Games and Economic Behavior 36 (2000), 241–263. [6] L.S. Shapley and M. Shubik, A method for evaluating the distribution of power in a committee system, American Political Science Review 48 (1954), 787–792. [7] A. Taylor and A. Pacelli, Mathematics and politics, second ed., Springer, New York, 2008, Strategy, voting, power and proof. [8] R. J. Weber, Subjectivity in the valuation of games, Game Theory and Related Topics, North Holland, 1979, pp. 129–136. 20
Canonical Hub: CANONICAL_INDEX
Ring 2 — Canonical Grounding
- INDEX
- MASTER INDEX
- An induced electric current always tends to cancel the field change that caused it