Supplementary material
Proof of the model in Section 3.2

We now prove the result. To prove the result, we have toconsider six deviation patterns: (i) the candidate locating at y relocates at x < y; (ii) the candidate
locating at y relocates at x ∈ (y, 1 − y); (iii) the candidate locating at y relocates at x ≥ 1 − y; (iv)the candidate locating at 1 − y relocates at x ≤ y; (v) the candidate locating at 1 − y relocates atx ∈ (y, 1 − y); and (vi) the candidate locating at 1 − y relocates at x > 1 − y. We now discuss the six
deviation patterns.
In the first case, the locations of the candidates are depicted in Figure 1A. The shaded circle (c1 )is the relocating candidate.
0y1/21−y 1

c1 c20y

1/2

c31−y 1

c1 c20

c3y

1/2

1−y 1

c1 c2

(m = y)c3(m = 1 − y)Figure 1A: Case 1 (x < y)First, we suppose that m = y. In the first round, the shares of the candidates (c1 , c2 , and c3 ) are
1/2 − (y − x)/2, (y − x)/2 + 1/2 − y, and y. c2 , who locates at y, takes a share higher than c3 , wholocates at 1 − y, and then wins in the first round if y < 1/4. In the second round, c2 takes a sharehigher than 1/2, that is, c1 loses. Second, we suppose that m = 1 − y. In the first round, the shares of
c1 , c2 , and c3 are 0, y, and 1 − y, respectively. c1 loses in the first round. Therefore, c1 (the deviating
candidate) wins with probability 0, and the deviation is not beneficial.
In the second case, the locations of the candidates are depicted in Figure 2A. The shaded circle(c2 ) is the relocating candidate.
00y

1/2

c1c2y

1/2

1−y 1

c1c2(m = y)

c3

1−y 1
c30y
c1

1/2

1−y 1c2c3(m = 1 − y)Figure 2A: Case 2 (x ∈ (y, 1 − y))First, we suppose that m = y. In the first round, the shares of the candidates (c1 , c2 , and c3 ) are
1/2 + (x − y)/2, (x − y)/2 + (1 − x − y)/2, and (3y − x)/2, respectively, if x < 3y, and 1/2 + (x − y)/2,i1/2 − (x − y)/2, and 0 otherwise. c1 , who locates at y, takes a share higher than 1/2. In the secondround, c1 wins, that is, c2 loses. Second, we suppose that m = 1 − y. The inverse result holds,
that is, c3 wins in the second round, and c2 loses. Therefore, the deviating candidate, c2 , wins withprobability 0, and the deviation is not beneficial.
In the third case, the locations of the candidates are depicted in Figure 3A. The shaded circle (c3 )is the relocating candidate.
y01/21−y 1

c1

1/2y0

c2 c31−y 1

c1

1/2y0

c2 c3

c1(m = y)

1−y 1

c2 c3
(m = 1 − y)Figure 3A: Case 3 (x > 1 − y)The locations are similar to those in the first case. Therefore, the deviating candidate wins withprobability 0, and the deviation is not beneficial.
In the fourth case, the locations of the candidates are depicted in Figure 4A. The shaded circle(c1 ) is the relocating candidate.
y0
c10c1y1/2c2c3(m = y)1/21−y 1c2c3

1−y 1y0c1c2
c3

1/2

1−y 1(m = 1 − y)Figure 4A: Case 4 (x < y)Suppose that the deviating candidate, c1 , wins in the first round. Under both cases (m = y andm = 1 − y), the rival candidate in the second round locates at y. The rival (c2 or c3 ) can take a shareof more than half and then wins in the second round. Therefore, the deviating candidate wins withprobability 0, and the deviation is not beneficial.
In the fifth case, the locations of the candidates are depicted in Figure 5A. The shaded circle (c3 )is the relocating candidate.

ii

0y
c1
c20y

1/2

c1c2

c3
(m = y)

1/2

1−y 1

c31−y 101/2yc1
c2

1−y 1c3(m = 1 − y)Figure 5A: Case 5 (x ∈ (y, 1 − y))First, we suppose that m = y. We also suppose that the deviating candidate, c3 , wins in the firstround. The rival candidate in the second round locates at y. The rival (c1 or c2 ) takes a sharehigher than 1/2. In the second round, c3 loses. Second, we suppose that m = 1 − y. The deviating
candidate, c3 , takes its share higher than 1/2 and then wins in both rounds. Therefore, c3 wins withprobability 1/2. The deviation does not improve the probability of winning the election.
In the sixth case, the locations of the candidates are depicted in Figure 6A. The shaded circle (c3 )is the relocating candidate.
0y

1/2

1−y 1
c3

c1
c2
0y

1/2

c1
c2

(m = y)1−y 10

c3

1/2

yc1
c2

1−y 1
c3

(m = 1 − y)Figure 6A: Case 6 (x > 1 − y)First, we suppose that m = y. We also suppose that the deviating candidate, c3 , wins in the firstround. The rival candidate in the second round locates at y. The rival (c1 or c2 ) takes a sharehigher than 1/2. In the second round, c3 loses. Therefore, at best, c3 wins with probability 1/2. The
deviation does not improve the probability of winning the election.
From the six cases, we find that the result in Section 3.2 holds.
Proof of the model in Section 3.2 (without the runoff system)

Q.E.D.
We first show that thelocation pattern in Figure 2 is not an equilibrium outcome in a one-round election.
First, we suppose that m = y. The shares of the candidates are (1 − y)/2, (1 − y)/2, and y.
(1 − y)/2 > y if and only if y < 1/3. We have assumed y ∈ (0, 1/4). Therefore, one of the left-handside candidates wins. Second, we suppose that m = 1 − y. The shares of the candidates are y/2,y/2, and 1 − y. y/2 < 1 − y if and only if y < 2/3. We have assumed y ∈ (0, 1/4). Therefore, theright-hand-side candidate wins. The probabilities of winning the election are 1/4, 1/4, and 1/2.
iii

Suppose that one of the left-hand-side candidates relocates at x(< y). We suppose that m = y.
The shares of the candidates are 1/2 − (y − x)/2, 1/2 − y + (y − x)/2, and y. If x is sufficiently close toy, the share of the relocating candidate is close to 1/2. We have assumed y ∈ (0, 1/4). Therefore, therelocating candidate wins. The probability is 1/2. We conclude that the location pattern in Figure
2 is not an equilibrium outcome when the runoff system does not exist.
We now show that pure strategy equilibria do not exist in a one-round election. We consider threelocation patterns: (1) three candidate locate at one point; (2) two candidates locate at one point,
and the other candidate locates elsewhere; (3) each candidate locates at a different point.
First, we consider the following three cases, in which three candidates locate at one point (seeFigure B1).
(1a)

0y

1/2

1−y 1(1b)

0y

1/2

1−y 1

(1c)

0y1/2

1−y 1

Figure B1: Three candidates locate at one point
In cases 1a, 1b, and 1c, if the shaded candidate relocates at the point indicated by the triangle, theprobability of winning the election is at least 1/2 (it wins when m = 1 − y), that is, the relocation
enhances the winning probability of the shaded candidate. Therefore, the location patterns do notappear in equilibrium.
Second, we consider the following three cases, in which two candidates locate at one point theother candidate locates elsewhere (see Figure B2).

iv

(2a)

0y

1/2

1−y 1(2a’)

0y

1/2

1−y 1(2b)0y1/2

1−y 1(2c)0y1/2

1−y 1(2c’)

0y

1/2

1−y 1

Figure B2: Two candidates locate at one point
In case 2a (2a’), two candidates locate at point x, which satisfies x < y. In this case, the othercandidate should locate at x0 , which satisfies x < x0 < (1 + 6y − 3x)/3 because its probability ofwinning is 1 due to the location (note that, if 9y > 2 + 3x, (1 + 6y − 3x)/3 is larger than 1 − y; seecase 2a’). Given the locations of the three candidates, if the shaded candidate relocates at the point
indicated by the triangle (1 − y), the probability of winning the election is at least 1/2, that is, therelocation enhances the winning probability of the shaded candidate. Therefore, the location patterndoes not appear in equilibrium.
In case 2b, two candidates locate at point y. In this case, the other candidate should locate atx0 , which satisfies y < x0 < (1 + 3y)/3 because its probability of winning is 1 due to the location.
Given the locations of the three candidates, if the shaded candidate relocates at the point indicatedby the triangle, the probability of winning the election is at least 1/2, that is, the relocation enhances
the winning probability of the shaded candidate. Therefore, the location pattern does not appear inequilibrium.
In case 2c (2c’), two candidates locate at point x, which satisfies y < x ≤ 1/2. If x < y + 1/6, theother candidate should locate at x0 , which satisfies x < x0 < (1 + 6y − 3x)/3 because its probabilityof winning is 1 by the location (see case 2c); otherwise, it should locate at point x00 , which induces its
probability of winning to be 1/2 (in case 2c, it wins when m = 1 − y; in case 2c’, it wins when m = y).
Given the locations of the three candidates, if the shaded candidate relocates at the point indicatedv

by the triangle, the probability of winning the election is 1/2; that is, the relocation enhances thewinning probability of the shaded candidate. Therefore, the location pattern does not appear inequilibrium.
Finally, we consider the following five cases, in which each candidate locates at a different point(see Figure B3).
(3a)

0

y

1/2

1−y 1(3b)0y1/2

1−y 1(3c)

0

y

1/2

1−y 1(3d)

0

y

1/2

1−y 1(3e)

0

y1/2

1−y 1

Figure B3: Each candidate locate at a different point
In case 3a, two candidates locate at points xi , which satisfies xi < y (i = 1, 2). In this case, the othercandidate should locate at x0 , which satisfies x0 < 4y − x1 − 2x2 , because its probability of winning is
1 due to the location. Given the locations of the three candidates, if the shaded candidate relocatesat the point indicated by the triangle, the probability of winning the election is at least 1/2; thatis, the relocation enhances the winning probability of the shaded candidate. Therefore, the location
pattern does not appear in equilibrium.
In case 3b, one candidate locates at point x, which satisfies x < y, and another candidate locatesat point y. In this case, the other candidate should locate at x0 , which satisfies x0 < 2y − x1 , because
its probability of winning is 1 due to the location. Given the locations of the three candidates, if
the shaded candidate relocates at the point indicated by the triangle, the probability of winningthe election is at least 1/2; that is, the relocation enhances the winning probability of the shadedcandidate. Therefore, the location pattern does not appear in equilibrium.
In case 3c, one candidate locates at point y, and the other two candidates locate at points xi ,which satisfies y < xi < 1 − y. The candidate locating at point y wins with probability 1/2 (it winsviwhen m = y). The right-hand-side candidate also wins with probability 1/2 (it wins when m = 1−y).
Given the locations of the three candidates, if the shaded candidate relocates at the point indicatedby the triangle, the probability of winning the election is 1/2; that is, the relocation enhances thewinning probability of the shaded candidate. Therefore, the location pattern does not appear inequilibrium.
In case 3d, two candidates locate at points y and 1 − y, and the other candidate locates at points
x, which satisfies y < x < 1 − y. The candidate locating at point y wins with probability 1/2 (itwins when m = y). The right-hand-side candidate also wins with probability 1/2 (it wins whenm = 1 − y). Given the locations of the three candidates, if the shaded candidate relocates at thepoint indicated by the triangle, the probability of winning the election is 1/2; that is, the relocation
enhances the winning probability of the shaded candidate. Therefore, the location pattern does notappear in equilibrium.
In case 3e, three candidates locate at points xi , which satisfies y < xi < 1 − y (i = 1, 2, 3).
This case is similar to case 3d. Given the locations of the three candidates, if the shaded candidate
relocates at the point indicated by the triangle, the probability of winning the election is 1/2; thatis, the relocation enhances the winning probability of the shaded candidate. Therefore, the location
pattern does not appear in equilibrium.
From the discussion, we find that stable location patterns do not exist in the setting, that is, purestrategy equilibria do not exist in a one-round election.

viiQ.E.D.