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PR-Squared: UK 2001 results, and voting power

Julian D. A. Wiseman

Abstract: This short paper contains the results of the UK 2001 election under PR-Squared, and introduces the concept of voting power.

Publication history: Only here. Usual disclaimer and copyright terms apply.

[ This paper has been partially superseded by PR-Squared: A New Description, effective September 2001. ]

This short paper contains the results of the UK 2001 election under PR-Squared, and introduces the concept of voting power. It assumes that readers are already very familiar with PR-Squared (or with PR-Squared in the context of New Zealand), and are fully conversant with the worked example of the calculations. Readers might also be interested in various technical notes relating to PR-Squared.

On 7th June 2001 there was a general election in the United Kingdom, won by the Labour Party. The parties' votes, their squares, and hence the number of seats won under PR-Squared, are in the following table:



vote power
Liberal Democrat4,815,24923,187bn73.137337,031
UK Independence390,575153bn0.481406,178
Plaid Cymru195,89238bn0.120large
Sinn Fein175,93331bn0.100large
Scottish Socialist72,5185bn0.020large
Socialist Alliance57,5533bn0.010large
Socialist Labour57,2893bn0.010large
Alliance Party NI28,9991bn0.000large

So, under PR-Squared, with the three large parties splitting the bulk of the vote 40.74% : 31.69% : 18.26%, the largest party would have had a majority of 69 — workable, but not so large as to inspire contempt for the House.

The last column of the table introduces a new concept: voting power. How useful is one extra vote to, say, the Conservative Party. Well, because seats are rounded to a whole number, one extra vote would not have altered the result at all. So let's instead ask how much difference one extra vote would make to the unrounded seat numbers. Well, one extra vote would have increased the Conservative's unrounded seat score by about 0.0000350956, or one part in 28,493.59. In other words, at this level of votes the Conservatives gain or lose (unrounded) seats at the rate of approximately 28,494 votes per seat. Note that a Conservative vote has more 'power' than a Liberal Democrat vote, which is worth about 1 part in 37,031 of a seat. This is a general rule: a vote for the second-largest party has more power than a vote for the third-largest, which has more power than a vote for the fourth-largest, etc (a fact proved mathematically in the appendix).

And why design an electoral system that allows the power of a vote to vary with the size of the party for which it is cast? Of course, to incentivise parties to be large — in other words, to encourage parties to declare coalitions before the election rather than after. Such a system prevents power drifting from the ballot box to the negotiating table.

Julian D. A. Wiseman, June 2001


[ Effective April 2009 This appendix has been superseded by PR-Squared: the Majoritarian Incentives of Voters and of Parties, especially its †5. ]

We wish to calculate the marginal utility of a vote: how many extra seats does such-and-such a party get per additional vote. Unfortunately, the answer is almost always zero, as the seat totals usually round to the same value with or without the extra vote. To avoid this inconvenience, we instead consider the unrounded seat totals. This marginal utility is called the 'voting power' of a party, and it can be thought of as the probability that an extra vote will result in an extra seat.
Voting power could be defined using partial derivatives, or by assuming a whole additional vote. The following mathematics shows that, using either definition, the voting power for the second-largest party is greater than the voting power for the third-largest party, which is greater than the voting power for the fourth-largest party, etc. However, the ranking of the voting power of the largest party is unconstrained: it can have the greatest voting power, the least, or somewhere in between.
For simplicity, we start by using partial derivatives, and assume that the total number of seats is unity, so we are dealing in the proportion of seats won by each party. Let us assume that parties' votes are a, b, c, etc; that a >= b >= c >= 0; and that a > 0. Let
  s = a*a + b*b + c*c + d*d + ...  
  t = s - a*a - b*b - c*c = d*d + e*e + f*f +... etc  
Let A be the proportion of seats won by the party with a votes, B with b, etc. Then, because rounding is ignored,
  A = a*a / s, B = b*b / s, etc.  
Observe that
  dB/db = 2 b (s - b*b) / (s*s)  
and likewise that
  dC/dc = 2 c (s - c*c) / (s*s)  
These partial derivatives are the parties' voting power. Trivially, if b = c then dB/db = dC/dc, so let's assume that b > c, and now show that dB/db > dC/dc. So,
  dB/db > dC/dc  
<==> 2 b (s - b*b) / (s*s) > 2 c (s - c*c) / (s*s)  
<==> b (a*a + c*c + t) > c (a*a + b*b + t)  
<==> (b - c)*(a*a + t) > (b - c)*b*c  
<==> a*a + t > b*c  
which is true because a >= b > c >= 0 <= t.
Hence, as required, the voting power for a non-largest party is greater than the voting power for a still-smaller party. It remains to show that the voting power for the largest party can be big, small, or in between. We proceed by example, and assume that there are five parties: if the votes split 2:1:1:1:1 then the largest has the greatest voting power. If the votes split 3:2:1:1:1 then the largest party has the second-greatest voting power, if 4:2:2:1:1 third, if 5:2:2:2:1 fourth, and if 4:1:1:1:1 fifth.
In integer arithmetic, the result is the same as using partial derivatives, but the algebra is more complicated. Define the voting power of party B to be
  VP[B] = (b+1)*(b+1)/(s+2*b+1) - b*b/s  
and likewise for VP[C]. We wish to show that VP[B]-VP[C] > 0 assuming that a >= b+1 and b >= c+1.
Start with VP[B]-VP[C] and multiply by s*(1+2*b+s)*(1+2*c+s), and divide by (b-c), all of which are positive. It then remains to show that
  0 < 2*s*s + s*(-2*b*b -2*c*c -2*b*c -b -c) -4*b*c*c -4*b*b*c -4*b*c -2*b*b -2*c*c -b -c  
Substitute s with (a*a+b*b+c*c+t), a with (b+1+x), and b with (c+1+y). All of c, t, x and y are non-negative. With these substitutions it remain to show that
  0 < 32 + 72*y+64*y*y+26*y*y*y+4*y*y*y*y + x*(68+110*y+62*y*y+12*y*y*y) + x*x*(49+51*y+14*y*y) + x*x*x*(16+8*y) + 2*x*x*x*x + c * ( 48 + 88*y+56*y*y+12*y*y*y + x*(98+110*y+32*y*y) + x*x*(48+26*y) + 8*x*x*x ) + c*c * ( 25 + 37*y+14*y*y + x*(56+32*y) + 14*x*x ) + c*c*c * ( 4 + 6*y + 12*x ) + t * ( 17+19*y+6*y*y + x*(16+8*y) + 4*x*x + c*(16+8*x+10*y) + 6*c*c ) + 2*t*t  
which is trivially true as every term on the right-hand side is non-negative, and the first is positive.
Hence, as before, the voting power for a non-largest party is greater than the voting power for a still-smaller party. But what about the voting power of the largest party? Again, this is unconstrained, as illustrated by the following five 5-party examples in which the votes split 4:3:3:3:3, 4:3:2:2:2, 3:2:2:1:1, 3:2:2:2:1 and 2:1:1:1:1.
(The assumptions in this proof can be made slightly less generous. If the minimum value of a-b is m, and the minimum value of b-c is n, then the same proof works for any m and n such that 2/3 <= m <= 1 and (7-6*m)/3 <= n <= 1. For simplicity of typesetting this is not proved here.)

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