Spot It! and balanced block designs: keys to better debate architecture for a plethora of candidates in presidential primaries?

Abstract

U.S. presidential primary debates are influential but under-researched. Before 2015, all of these debates, both Democratic and Republican, had 10 candidates or fewer. The first Republican debate in 2015, however, abided 17 candidates. They were split into two segments, with the 10 best-polling candidates in the main (prime-time) segment and the others in an ‘undercard’ session. A comparable pattern applied for the next six Republican debates. Concern arose not only because many candidates were crowded into a session but also because the undercard candidates were seen as receiving inferior exposure. The Democratic presidential primary debates that started four years later encountered similar difficulty. An authorized policy caused their candidates in each of the first two debates to be limited to 20, randomly divided into two groups of 10 appearing on successive nights. For remedy, this paper examines innovative debate plans, for different numbers of candidates, that feature symmetry among all candidates and entail many short segments with relatively few candidates in each. We apply combinatorial designs—balanced incomplete block designs and regular pairwise balanced designs, which are analogous to the games Spot It Jr.! Animals and (full-fledged) Spot It!, respectively.

1. Introduction

It seems reasonable to say that the heyday of incomplete block designs was around the middle of the last century. At that time, of course, attention was devoted mainly to the construction and analysis of such designs, and to their application, especially in agriculture. The present paper veers radically from earlier work on incomplete block designs. It involves no statistical analysis at all; and its application, based chiefly on balanced incomplete block and regular pairwise balanced designs, is aimed at a most unlikely target, that of presidential primary debates.

The debate season for primaries for the 2016 U.S. presidential election opened in 2015 and saw 17 Republican candidates at the outset, a record number at that time for either party. Only 5 Democratic candidates emerged, a number that may have been curbed by the predominance of Hillary Clinton. In any case, the issue of how to organize a debate for 17 candidates in a suitable manner spawned various opinions, ideas, and controversies. This paper suggests that, curiously, the combinatorial (or experimental) designs found in the toolboxes of statisticians and mathematicians, along with an assist from a certain fast-moving game played by children of all ages, could provide keys to improving the architecture for debates that have rather large numbers of candidates.

The presidential primary debates linked to the 2020 election started in 2019. They attracted a field of more than 20 Democratic candidates—even larger than the group of Republican contenders in 2015. Thus the problems of debate architecture are not diminishing.

More extensive introductory discussion, on matters including the impact of presidential primary debates, appears in the online Supplemental Material, Notes pertaining to Section 1.

2. Presidential primary debates in 2015-2016

Democratic candidates had nine debates in the 2015-2016 primary season [1]. The number of contestants was 5 in the first debate, 3 in the next three, and just 2 (Clinton and Bernie Sanders) in the last five. Each debate lasted roughly two hours. Because the Democrats had to accommodate far fewer participants than the Republicans, they avoided some of the problems that afflicted the GOP.

The Republicans had 12 debates [1]. The first seven had two parts, an ‘undercard’ segment and a later main segment in the evening on the same day. Assigned to the undercard segments were the candidates with the least favorable polling results. Among other issues, questions arose as to whether the undercard candidates received inferior exposure and were thus treated unfairly. The respective numbers of candidates for the 12 debates were

$$\begin{array}{cccccccccccc}17,& 15,& 14,& 12,& 13,& 10,& 11,& 7,& 6,& 5,& 4,& 4.\end{array}$$

For the first seven debates, these numbers broke down as

$$\begin{array}{ccccccc}7+10,& 4+11,& 4+10,& 4+8,& 4+9,& 3+7,& 4+7,\end{array}$$

where the first number in each pair refers to the undercard session and the second one to the main session. These numbers for the Republicans include every integer, other than 16, from 3 to 17, thus underscoring the obvious point that numbers of participants can vary widely. Our results below include options for innovative debate architecture for each number of contenders up to 21.

Most of the undercard segments lasted about an hour, with a couple somewhat longer. The seven corresponding main segments ran roughly two hours each, except for one that was about three hours. The five final debates were around two hours each.

Problems with splitting the 2015-2016 Republican candidates between the two sessions were manifold. For related details, see the online Supplemental Material, Notes pertaining to Section 2.

3. Presidential primary debates in 2019-2020

No Republican presidential primary debates occurred in connection with the 2020 election. The Democratic debates that took place, starting in June 2019 and ending in March 2020, numbered 11. The respective numbers of candidates in those 11 were

$$\begin{array}{ccccccccccc}20,& 20,& 10,& 12,& 10,& 7,& 6,& 7,& 6,& 7,& 2.\end{array}$$

The first two of these, however, were each split into two segments of 10 candidates each, coming on two consecutive nights. Candidate speaking time in each of the 13 resulting events consumed about two hours or a bit more.

The candidate field was even larger than the above numbers indicate, because qualification requirements prevented some candidates from participating. In addition, maximums on the numbers of candidates were imposed (20 in the case of the first two debates). At the outset, the total number of aspirants was thus in excess of 20.

For the initial debate, the 20 candidates were divided into the two groups of 10 by first assigning each contender to one of two tiers based on poll standings and then conducting a random drawing within each tier. The procedure for the second debate was the same except for a change from two tiers to three, with 4, 6, and 10 candidates in the top, middle, and bottom tiers, respectively (and with half the entrants in each tier assigned to each night). The shift from two tiers to three apparently came about [2] in reaction to the outcome of the drawing for the first debate, which resulted in only one of the five top-polling candidates appearing on the first night (an occurrence whose reasonable possibility any statistician could have recognized in advance).

With the first two debates both split into two segments, the existence of many pairs of candidates who did not appear together in either debate was inevitable. In particular, Joe Biden and Elizabeth Warren, whose respective poll rankings were then first and third (or thereabouts), did not face each other either time.

Assorted problems thus occurred with the debates in both 2015-2016 and 2019-2020. A large number of candidates has continued to spawn difficulties for presidential primary debates.

4. Framework

For the debate designs that we are about to present, for a given debate we denote the number of candidates by C and the number of segments by S. A segment (or session) embraces a subset of the C candidates who are together on the debate stage at the same time for a given (continuous) portion of the debate. Debate planning needs to be adaptable regarding the total number of candidates (C), which typically changes from one debate to the next, mainly as weaker candidates progressively drop out. Although C can thus change between debates, all S segments within any debate are obviously formed from the same set of C candidates.

As an axiomatic condition for fairness, we require that all candidates are to appear in the same number of segments, to be denoted by R. We use K to denote the number of candidates in a segment, if it is the same for all segments (in which case CR = SK).

See the online Supplemental Material, Notes pertaining to Section 4, for a few illustrations related to the above (which refer to some items in the Notes pertaining to Section 1).

5. Enter Spot It Jr.! Animals

We turn now to a surprising analogue of architecture for presidential primary debates, examining first the easier game of Spot It Jr.! Animals (said to be playable by four–year–olds) and later the full-fledged Spot It! The former has 31 circular cards with 6 animals pictured on each card. A total of 31 different animals appear on the cards, with each animal shown on 6 of the cards. On any given round (for some versions of the game), a card is in the center and a card is in front of each player. The cards are designed so that any two cards have one and only one animal in common. The player who is first to find the animal that is common between the center card and that player's card is the winner of the round upon calling out the name of the animal.

But how is it that exactly one animal is in common between any two cards? The answer is that the set of cards corresponds to a balanced incomplete block (BIB) design, a type of experimental design with a long history that includes the pioneering statistician R. A. Fisher and has origins in agriculture. With traditional notation, a BIB design has v varieties (or, comparably, t treatments) and b blocks, with each variety appearing in r of the b blocks (r < b) and each block containing k of the v varieties (so that vr = bk). The key point for a BIB design, though, is that any two varieties appear together (in the same block) exactly λ times, where $\lambda =\frac{bk(k-1)}{v(v-1)}=\frac{r(k-1)}{v-1}$ For Spot It Jr.! Animals, we take the varieties to correspond to the cards and the blocks to the animals, so that we have v = 31 cards, b = 31 animals, r = 6 animals on each card, k = 6 cards containing each animal, and any two cards with exactly λ = 1 animal in common. (It is also true that any two animals have exactly one card in common, but that happens only because v and b are equal. For Spot It Jr.! Animals we could have taken the blocks to correspond to the cards and the varieties to the animals rather than vice versa, but such a match would not work for the illustration related to full-fledged Spot It! that appears later.)

But how does all of this relate to design of debate architecture? We let the v cards (or varieties) correspond to the C candidates; the b animals (or blocks) to the S segments; the r animals (out of b) on a given card to the R segments (out of S) in which the associated candidate appears; and the k cards (out of v) that contain a given animal to the K candidates (out of C) who are in the associated segment. The r animals on a card are thus translated as the R segments in which the associated candidate will participate. Corresponding to λ, we use $L=\frac{R(K-1)}{C-1}$ to denote the number of segments in which any two candidates appear together. The property that any two candidates face each other in the same number of segments holds not only for BIB designs but also for all the other debate designs that we present below. This attribute should ensure especial fairness by providing for a certain neutrality and symmetry among candidates.

Observe that the parameters (C, S, R, K, L) that we use for the debate situation correspond, respectively, to (v or t, b, r, k, λ) in the traditional experimental-design literature. We point out that a BIB design cannot have S < C (i.e. b < v), a restriction that unfortunately limits flexibility in finding useful BIB designs. (The inequality b ≥ v or S ≥ C as applied to BIB designs is Fisher's inequality, but in Proposition 3 below we provide a more general version.) A BIB design can have S > C as well as S = C, and also can have L > 1 as well as L = 1 (but $L=\frac{R(K-1)}{C-1}$ must be an integer). A BIB design with S = C (i.e. b = v) is called a symmetric BIB design.

Of course, the set of C = 31 candidates to which Spot It Jr.! Animals corresponds is larger than what one would generally expect to encounter for a presidential primary debate. But many BIB designs are available for values of C below (as well as above) 31. Suitable ones are presented below. Spot It Jr.! Animals is useful, though, for explaining and illustrating the general concept that we are looking at, as is true, even more so, for the more complex picture involving the full–fledged Spot It! that we examine later on.

6. Some BIB designs suitable for debates

Table 1 presents some BIB designs that could be applied for debate architecture, for various combinations of (C, S, R, K, L). Although the three reference works in the table show many of the designs (plans), it is not necessary to turn to those works to get a plan, because, as a further source, the final column of the table provides full details for obtaining each plan. The plans from the four sources may differ from one another, but generally only regarding insignificant matters such as labelings. In the table, if a reference does not show a given plan, then that reference either does not include the plan at all or else includes it without giving it an identification number.

Table 1.

Some BIB (balanced incomplete block) designs that could be applied for debates.

	Parametersa						Referencesb
Plan no.	C	S	R	K	L		CC	T	FY	How to construct
1-01	3	3	2	2	1		—	—	—	All 3 pairs
1-02	4	4	3	3	2		—	—	—	All 4 triples
1-03	4	6	3	2	1		11.1	—	—	All 6 pairs
1-04	5	10	4	2	1		11.2	—	—	All 10 pairs
1-05	5	10	6	3	3		—	—	—	All 10 triples
1-06	6	10	5	3	2		11.4	1	1	From #1-17, delete 1 segment and 5 candidatesc
1-07	6	15	5	2	1		11.3	—	—	All 15 pairs
1-08	7	7	3	3	1		11.7	2	—	Develop (1 2 4) mod 7
1-09	7	7	4	4	2		11.8	—	—	Develop (1 2 3 5) mod 7
1-10	7	21	6	2	1		—	—	—	All 21 pairs
1-11	8	14	7	4	3		11.10	3	6	From #1-22, delete 1 segment and 7 candidatesc
1-12	9	12	4	3	1		10.1	4	—	From #1-20, delete 1 segment and 4 candidatesc
1-13	9	18	8	4	3		11.11	5	11	Develop (1 2 3 5) mod 9 for first 9 segments; develop (1 2 5 7) mod 9 for last 9 segments
1-14	9	18	10	5	5		11.12	—	—	From #1-26, delete 1 segment and 10 candidatesc
1-15	10	15	6	4	2		11.16	7	3	From #1-23, delete 1 segment and 6 candidatesc
1-16	10	18	9	5	4		11.17	8	16	From #1-25, delete 1 segment and 9 candidatesc
1-17	11	11	5	5	2		11.19	11	2	Develop (1 2 3 5 8) mod 11
1-18	11	11	6	6	3		11.20	—	—	Develop (1 2 3 5 6 8) mod 11
1-19	12	22	11	6	5		—	14	34	From #1-28, delete 1 segment and 11 candidatesc
1-20	13	13	4	4	1		11.22	16	—	Develop (1 3 4 8) mod 13
1-21	13	26	6	3	1		11.21	15	4	Develop (1 2 5) mod 13 for first 13 segments; develop (1 3 8) mod 13 for last 13 segments
1-22	15	15	7	7	3		11.25	23	9	Develop (1 2 3 5 6 9 11) mod 15
1-23	16	16	6	6	2		11.27	27a	5	See text
1-24	16	20	5	4	1		10.2	25	—	From #1-27, delete 1 segment and 5 candidatesc
1-25	19	19	9	9	4		11.31	38	19	Develop (1 2 3 4 6 8 13 14 17) mod 19
1-26	19	19	10	10	5		11.32	—	—	Develop (1 2 3 4 6 8 13 14 16 17) mod 19
1-27	21	21	5	5	1		11.34	44	—	Develop (1 4 5 10 12) mod 21
1-28	23	23	11	11	5		—	52	35	Develop (1 2 3 4 6 8 9 12 13 16 18) mod 23

^aC = Number of candidates; S = Number of segments; R = Number of segments in which each candidate appears; K = Number of candidates who appear in each segment; L = Number of segments in which any two candidates appear together.

^bCC = Plans in Chapters 10 and 11 of Cochran and Cox [7]; T = Takeuchi [10]; FY = Fisher and Yates [11], pp. 90-93, but also see pp. 25-27.

^cPick any segment to delete, and then delete all candidates therein from all other segments also. Then renumber the remaining candidates.

For Table 1, as well as for later tables, the plans that we present were chosen selectively. The tables (collectively) give plans for every value of C (number of candidates) from 3 to 21. For a given value of C, a plan with a larger S (number of segments) may be excluded in favor of one with a smaller S, and a plan with more candidates in a segment (K in the case of Table 1) may be left out in favor of one whose segments are less crowded. In Table 1, some of the plans may be generally undesirable to use (because of high K) but appear in the table anyway because they serve to enable easy construction of other designs in the table.

Except for Plan #1-23, each plan in Table 1 is obtained through one of three methods. The first method consists simply of creating C(C – 1)/2 segments that consist of all possible pairs of the C candidates (as in Plan #1-01, which is the Canadian debate plan mentioned in the online Supplemental Material), or C(C – 1)(C – 2)/6 segments that consist of all possible triples. As an example of the latter type, Plan #1-05 has S = 10 segments that contain the candidates shown in the following 10 columns:

$$\begin{array}{cccccccccc}1& 1& 1& 1& 1& 1& 2& 2& 2& 3\\ 2& 2& 2& 3& 3& 4& 3& 3& 4& 4\\ 3& 4& 5& 4& 5& 5& 4& 5& 5& 5\end{array}$$

Candidates (1, 2, 3) appear together in segment 1, (1, 2, 4) are in segment 2, … , and (3, 4, 5) are in segment 10. Each of the C = 5 candidates appears in R = 6 segments, each segment has K = 3 candidates, and any two candidates appear together in L = 3 segments.

The second method is based on difference sets (e.g. [3], p. 39ff.). Except for Plans #1-13 and 1-21, this method is used in Table 1 for symmetric designs only. (The two exceptions have S = 2C and use two difference sets rather than one.) For a symmetric BIB design (S = C), the K integers that identify the candidates for the first segment, chosen in a special way so as to constitute a difference set and thus ensure the balance property, are all that is needed (besides the value of C) to construct the plan. One develops this initial segment modulo C (or mod C). This means that, in each successive segment, each candidate number is increased by 1, with (C + 1) changed to 1 when (C + 1) is reached. Plan #1–20 can provide an illustration. Its initial segment is the difference set (1 3 4 8) mod 13. The numbers of the candidates for its S = 13 segments are:

$$\begin{array}{ccccccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13\\ 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 1& 2\\ 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 1& 2& 3\\ 8& 9& 10& 11& 12& 13& 1& 2& 3& 4& 5& 6& 7\end{array}$$

Each of the C = 13 candidates appears in R = 4 segments, each segment has K = 4 candidates, and any two candidates appear together in L = 1 segment.

Spot It Jr.! Animals matches a (symmetric) BIB design where (C, S, R, K, L) is (31, 31, 6, 6, 1). Such a design can be obtained by developing the difference set (1 2 4 11 15 27) mod 31.

At this point we present two propositions related to symmetric BIB designs. The second one provides the third method for obtaining the BIB designs in Table 1.

Proposition 1.

In a symmetric BIB design, any two segments will have L candidates in common (just as any two candidates will appear together in L segments).

Proof

See (e.g.) Wallis [4], pp. 26-29 or Stinson [5], pp. 23-24.

One implication of Proposition 1 is that, with a symmetric BIB plan, no ordering of the segments can produce two successive segments with no candidates in common. In fact, exactly L of the K candidates in one segment will always appear in the following segment.

Proposition 2.

If one segment and all candidates therein are deleted from a symmetric BIB plan with parameters [C, S ( =  C), R ( =  K), K, L], the resulting plan will be BIB with parameters (C*, S*, R*, K*, L*), where C* = C – K, S* = S − 1, R* = R, K* = K – L, and L* = L.

Proof

See Wallis [4], p. 95ff.

To illustrate how the third method for getting the BIB designs in Table 1 uses Proposition 2, consider Plan #1-12. It can be derived by deleting, from Plan #1-20 (exhibited above), segment 13 and all candidates therein (candidates 13, 2, 3, and 7). The resulting plan has C* = 13 − 4 = 9 candidates, S* = 13 – 1 = 12 segments, and K* = 4 – 1 = 3 candidates per segment, with R* = 4 and L* = 1 (both unchanged). At first, the plan is:

$$\begin{array}{cccccccccccc}1& 4& 5& 4& 5& 6& 9& 8& 9& 10& 11& 12\\ 4& 5& 6& 6& 8& 8& 10& 10& 11& 12& 1& 1\\ 8& 9& 10& 11& 12& 9& 1& 11& 12& 4& 5& 6\end{array}$$

But then the candidates can be renumbered, with (e.g.) candidate 10 changed to 2, 11 to 3, and 12 to 7.

The BIB designs constructed through Proposition 2 are called residual designs, and some of them are resolvable designs (see, e.g. [6], sections 5.2, 4.6). A resolvable BIB design is one in which the S blocks (sessions) can be arranged into G > 1 groups of blocks such that each variety (candidate) appears exactly R/G times in each group of blocks. A BIB design that is resolvable provides a minor benefit for a debate, in that it enables the appearances of all candidates to be spread relatively evenly over the course of the debate. (Only a few such designs are available in Table 1, though.) To illustrate, the above configuration for Plan #1-12 becomes

$$\begin{array}{ccccccccccccccc}1& 5& 9& & 4& 8& 7& & 4& 5& 9& & 6& 2& 3\\ 4& 6& 3& & 5& 2& 1& & 6& 8& 2& & 8& 7& 1\\ 8& 2& 7& & 9& 3& 6& & 3& 7& 1& & 9& 4& 5\end{array}$$

after renumbering of the candidates and reordering of the columns (sessions). This (BIB) design is resolvable, as it has G = 4 groups in each of which every candidate appears exactly one time. (The resolvability of Plan #1-12 can also be shown in other ways; cf. Cochran and Cox [7], Plan 10.1.)

Of the eight residual designs that appear in Table 1, three (Plans #1-11, 1-12, and 1-24) are resolvable, as can be determined upon shuffling the order of the segments appropriately after obtaining the residual design from the inceptive design. They have G = R = 7, G = R = 4, and G = R = 5, respectively Although Plan #1-19 is not resolvable when constructed as a residual design as shown in Table 1, a resolvable BIB design with the same five parameters (and G = R = 11) does result with a different method of construction: Create a set of 11 segments by developing (1 2 4 5 6 10) mod 11, and match it with another set created by developing (3 7 8 9 11) mod 11 and appending candidate 12 to each segment.

Finally, Plan #1-23 is the only one in Table 1 that cannot be obtained with any of the above three methods for BIB designs. It is:

$$\begin{array}{cccc}2& 1& 1& 1\\ 3& 3& 2& 2\\ 4& 4& 4& 3\\ 5& 6& 7& 8\\ 9& 10& 11& 12\\ 13& 14& 15& 16\end{array}\begin{array}{cccc}6& 5& 5& 5\\ 7& 7& 6& 6\\ 8& 8& 8& 7\\ 1& 2& 3& 4\\ 9& 10& 11& 12\\ 13& 14& 15& 16\end{array}\begin{array}{cccc}10& 9& 9& 9\\ 11& 11& 10& 10\\ 12& 12& 12& 11\\ 1& 2& 3& 4\\ 5& 6& 7& 8\\ 13& 14& 15& 16\end{array}\begin{array}{cccc}14& 13& 13& 13\\ 15& 15& 14& 14\\ 16& 16& 16& 15\\ 1& 2& 3& 4\\ 5& 6& 7& 8\\ 9& 10& 11& 12\end{array}$$

The BIB designs in Table 1 offer many new prospects for improved debate plans. Nonetheless, the table has no designs at all for some values of C. In addition, some plans in the table may be rather unattractive because of S and/or K being relatively high. Although Table 1 shows 28 designs, the only ones that are relevant for a particular debate are, of course, those that have a value of C that coincides with the specific number of candidates in that debate. Are there suitable plans for debates that could enlarge the repertoire by going beyond BIB designs?

7. Full-fledged Spot It! to the rescue

Regular Spot It! provides inspiration for debate architecture over and beyond that from Spot It Jr.! Animals. Although the two games are similar in most respects, full-fledged Spot It! corresponds not to a BIB design, but to a more general type of design. It has 55 circular cards with 8 symbols on each card. Any two cards have one and only one symbol in common. The total number of symbols is 57, but they do not all appear on the same number of cards: 42 of them are on 8 cards, 14 are on 7 cards, and one (a snowman) is on only 6 cards. The game is analogous to a debate that has C = 55 candidates, S = 57 segments, each candidate appearing in R = 8 segments, any two candidates appearing together in L = 1 segment, and differing segment sizes (42 segments with 8 candidates, 14 with 7, and one with 6). Of course, a debate could hardly have as many as 55 candidates, but, as before with Spot It Jr.! Animals, the game is useful for introducing and illustrating the concept.

The type of design to which full-fledged Spot It! corresponds is called a regular pairwise balanced (RPB) design ([5], p. 7) or (r, λ)-design ([8]; [4], pp. 21-22). An RPB design is the same as a BIB design except that the segment (block) sizes are allowed to differ. That is, each of C candidates still appears in the same number of segments (R) and any two candidates still appear together in exactly L segments, but the S segments need not all have the same number of candidates. RPB designs, of which BIB designs are a special case (or type), are thus useful for debate architecture because they provide added options.

Even more general (and less restrictive) than RPB designs are pairwise balanced (PB) designs ([3], p. 23; [5], p. 7; [4], p. 14ff.). They still require that any two candidates appear together in exactly L segments, but there can be differences among candidates as to how many segments they appear in, not just differences among segments as to how many candidates they include. (PB designs also can be non-binary—the number of times a given treatment occurs in the same block can be more than 1 as well as 0 or 1—but that option has no applicability here.) For debates, PB designs thus provide no added benefit over RPB designs if one requires that differing numbers of segments for different candidates are to be avoided. Although the added generality of RPB over BIB designs is valuable for enabling more debate plans, under our stipulations the added generality of PB over RPB designs provides no further advantage. RPB designs have not received the distinct attention that has been afforded to either BIB or PB designs, but they are especially useful for our purposes here.

The Spot It! game corresponds to an RPB design with C = 55, S = 57, R = 8, L = 1, and segment sizes of 8, 7, and 6. A debate plan with the Spot It! parameters can be constructed by starting with a BIB design where (C, S, R, K, L) is (57, 57, 8, 8, 1) [obtainable by developing the difference set (1 2 10 12 15 36 40 52) mod 57] and then deleting two of the 57 candidates. We provide below a number of debate plans that use RPB designs constructed in different ways (but having lower values of C, of course). First, though, we diverge by presenting four propositions that apply to RPB designs (which subsume BIB designs).

Proposition 3.

In any RPB design, the number of segments, S, must be at least as large as the number of candidates, C.

Proof

See Wallis [4], p. 24. The proof there refers to BIB rather than RPB designs, but with slight modification it works also for the latter.

The result given by Proposition 3 explains why neither Table 1 nor any of the later tables show any plans with S < C.

Proposition 4.

If c ≥ 1 candidates are deleted from a BIB design with parameters (C, S, R, K, L) and if each segment then retains at least two candidates, the resulting design is RPB with parameters C* = C – c, S* = S, R* = R, and L* = L (and with unequal segment sizes).

Proof

Clearly, the new design has C* = C – c, because c candidates have been deleted, and has S* = S, because no segments have been deleted. In addition, nothing changes for the non–deleted candidates, including their relationships with each other, so L* = L (i.e. any two non-deleted candidates still appear together in L segments) and R* = R.

Proposition 4 was applied to create all the RPB debate designs in Table 2 (using c = 1) and in Table 4 (with c > 1). The technique of deleting (what we here refer to as) candidate(s) from a BIB design to obtain an RPB design is mentioned by Anderson [3], p. 24 and by Wallis [4], p. 22.

Table 2.

Some RPB (regular pairwise balanced) designs usable for debates (with two segment sizes, within each of which all candidates appear equally; constructible by deleting one candidate from a BIB design).

	Parametersa
Plan no.	C	S	R	S1@K1; S2@K2	L	How to construct
2-01	6	7	3	3@2; 4@3	1	From Plan #1-08, delete candidate 7
2-02	6	7	4	4@3; 3@4	2	From Plan #1-09, delete candidate 7
2-03	8	12	4	4@2; 8@3	1	From Plan #1-12, delete candidate 9
2-04	8	12	8	8@5; 4@6	5	Can be constructed as complement of Plan #2-03
2-05	10	11	5	5@4; 6@5	2	From Plan #1-17, delete candidate 11
2-06	10	11	6	6@5; 5@6	3	From Plan #1-18, delete candidate 11—or can be constructed as complement of Plan #2-05
2-07	12	13	4	4@3; 9@4	1	From Plan #1-20, delete candidate 13
2-08	14	15	7	7@6; 8@7	3	From Plan #1-22, delete candidate 15
2-09	15	16	6	6@5; 10@6	2	From Plan #1-23, delete candidate 16
2-10	15	20	5	5@3; 15@4	1	From Plan #1-24, delete candidate16
2-11	18	19	9	9@8; 10@9	4	From Plan #1-25, delete candidate 19
2-12	20	21	5	5@4; 16@5	1	From Plan #1-27, delete candidate 21

^aC = Number of candidates; S = Number of segments; R = Number of segments in which each candidate appears; S₁ = Number of segments of the smaller size, each of which have K₁ candidates; similarly for S₂ and K₂ for the segments of the larger size; L = Number of segments in which any two candidates appear together.

Proposition 5 provides further detail regarding the designs in Table 2.

Proposition 5.

In Proposition 4, if c = 1, then the RPB design that results from deleting one candidate from the BIB design has S_1* = R segments each consisting of K_1* = K – 1 candidates and S_2* = S – R segments containing K_2* = K candidates each. Each (non-deleted) candidate appears L times (in total) in the R segments of size K – 1 and R – L times in the S – R segments of size K.

Proof

The deleted candidate is removed from all R of its segments, so those R (of the S) segments now have K – 1 candidates each. In them, the deleted candidate appeared a total of L times with each other candidate, so each non-deleted candidate now appears L times in those R segments. The results for the S_2* segments with all K candidates remaining are obtained by subtraction.

Proposition 6 pertains to complements. The complement of a design is the design obtained by transforming each segment of the original design so that it consists of all candidates not contained in the original segment. As an example, Plan #1-05 (exhibited above) has as its complement

$$\begin{array}{cccccccccc}4& 3& 3& 2& 2& 2& 1& 1& 1& 1\\ 5& 5& 4& 5& 4& 3& 5& 4& 3& 2,\end{array}$$

which happens to be equivalent to Plan #1-04.

Proposition 6.

Suppose an RPB design has parameters C, S, R, and L, with S₁ segments consisting of K₁ candidates, S₂ segments of size K₂, … , where Σ S_i = S and C − 1 > K_i > 1 for each K_i. Then its complement is also an RPB design, and has parameters C* = C, S* = S, R* = S – R, and L* = S – 2R + L, with S₁ segments of size (C − K₁), S₂ of size (C − K₂), … .

Proof

See Wallis [4], p. 30. Although the proof there is for BIB rather than RPB designs, it is easily embellished to cover the latter by allowing for unequal segment sizes.

Every design in our tables has a complement, but few complements are shown, mainly because the segment sizes would generally be too large. Nonetheless, some debate designs that are not shown could be of interest on occasion, so Proposition 6 can be of help in finding such designs. The only instances of (two mutually) complementary design-parameter sets that both appear in Table 1 are those for the pairs consisting of Plans #1-04 and 1-05, 1-08 and 1-09, 1-13 and 1-14, 1-17 and 1-18, and 1-25 and 1-26.

8. Some RPB designs suitable for debates

Tables 2–4 show RPB designs that could be used for debates. The plans in Tables 2 and 4 can be derived by deleting, respectively, one candidate and more than one from a BIB design, as indicated in Proposition 4 above. Per Proposition 5, the plans in Table 2 have two segment sizes, with every candidate appearing the same number of times within each segment size. The designs in Table 3 are constructed in varying ways, but all of them likewise have two segment sizes with all candidates appearing equally often within each segment size. In Table 4, though, the plans have more than two segment sizes with candidates appearing unequal numbers of times within a segment size (though the total number of appearances is, of course, still the same for all candidates). Remember that, in any RPB design, the number of times (equal to L) that any two candidates appear together in the same segment is the same for every pair.

Table 3.

Some RPB (regular pairwise balanced) designs usable for debates (with two segment sizes, within each of which all candidates appear equally; constructed by different means).

	Parametersa
Plan no.	C	S	R	S1@K1; S2@K2	L	How to construct
3-01	4	10	6	6@2; 4@3	3	All 6 pairs of candidates; all 4 triples
3-02	5	10	5	5@2; 5@3	2	Develop (1 2) mod 5; develop (1 2 4) mod 5
3-03	8	12	6	4@2; 8@5	3	Use sub-design D6 (below); develop (1 2 3 4 6) mod 8
3-04	8	12	6	8@3; 4@6	3	Complement of Plan #3-03
3-05	8	16	6	8@2; 8@4	2	Develop (1 3) mod 8; develop (1 2 4 5) mod 8
3-06	8	18	8	8@3; 10@4	3	Develop (1 2 4) mod 8; develop (1 2 4 5) mod 8, then D10
3-07	9	15	6	6@3; 9@4	2	Use D5; use D8
3-08	9	15	7	6@3; 9@5	3	Use D5; use complement of D8
3-09	10	20	7	10@3; 10@4	2	Develop (1 2 4) mod 10; develop (1 2 5 7) mod 10
3-10	12	15	7	3@4; 12@6	3	Use D9; use D7
3-11	12	19	5	16@3; 3@4	1	Use D1; use D9
3-12	12	26	6	6@2; 20@3	1	Use D3; use D4
3-13	14	21	5	7@2; 14@4	1	Use D2; develop (1 2 5 7) mod 14
3-14	17	34	9	17@4; 17@5	2	Develop (1 2 6 8) mod 17; develop (1 2 5 8 10) mod 17

^aSee footnote to Table 2.

Sub-designs:

$$\begin{array}{l}\begin{array}{ccccccccccccccccc}\mathrm{D}1:& 1& 7& 5& 4& 1& 3& 6& 2& 1& 3& 7& 2& 1& 2& 3& 4\\ & 2& 8& 9& 6& 11& 5& 8& 4& 5& 4& 9& 10& 8& 6& 10& 5\\ & 3& 12& 10& 11& 12& 7& 10& 9& 6& 8& 11& 12& 9& 7& 11& 12;\end{array}\\ \begin{array}{cccccccc}\mathrm{D}2:& 1& 2& 3& 4& 5& 6& 7\\ & 8& 9& 10& 11& 12& 13& 14;\end{array}\begin{array}{ccccccc}\mathrm{D}3:& 1& 2& 3& 4& 5& 6\\ & 7& 8& 9& 10& 11& 12;\end{array}\\ \begin{array}{ccccccccccccccccccccc}\mathrm{D}4:& 1& 2& 3& 6& 1& 2& 3& 4& 5& 1& 2& 3& 8& 4& 4& 7& 8& 1& 2& 1\\ & 3& 7& 5& 9& 6& 4& 7& 9& 7& 10& 5& 6& 9& 5& 6& 11& 10& 2& 3& 5\\ & 4& 9& 12& 10& 8& 11& 8& 12& 10& 11& 6& 11& 11& 8& 7& 12& 12& 12& 10& 9;\end{array}\\ \begin{array}{ccccccc}\mathrm{D}5:& 1& 4& 7& 1& 2& 3\\ & 2& 5& 8& 4& 5& 6\\ & 3& 6& 9& 7& 8& 9;\end{array}\begin{array}{ccccc}\mathrm{D}6:& 1& 2& 3& 4\\ & 5& 6& 7& 8;\end{array}\\ \begin{array}{ccccccccccccc}\mathrm{D}7:& 1& 1& 2& 3& 1& 1& 2& 3& 1& 1& 2& 3\\ & 2& 4& 5& 6& 2& 5& 4& 4& 2& 5& 4& 4\\ & 3& 8& 7& 7& 3& 6& 6& 5& 3& 6& 6& 5\\ & 4& 9& 9& 8& 7& 7& 8& 9& 10& 8& 7& 7\\ & 5& 11& 10& 10& 8& 11& 10& 10& 11& 9& 9& 8\\ & 6& 12& 12& 11& 9& 12& 12& 11& 12& 10& 11& 12;\end{array}\\ \begin{array}{cccccccccc}\mathrm{D}8:& 2& 1& 1& 1& 2& 3& 1& 2& 3\\ & 3& 3& 2& 5& 4& 4& 4& 5& 6\\ & 4& 5& 6& 6& 6& 5& 8& 7& 7\\ & 7& 8& 9& 7& 8& 9& 9& 9& 8;\end{array}\begin{array}{cccc}D9:& 1& 2& 3\\ & 4& 5& 6\\ & 7& 8& 9\\ & 10& 11& 12;\end{array}\begin{array}{ccc}D10:& 1& 2\\ & 3& 4\\ & 5& 6\\ & 7& 8.\end{array}\end{array}$$

Note: Sub-designs D1, D4, D7, and D8 above are equivalent, respectively, to partially balanced incomplete block designs SR21 (p. 146), R16 (p. 189), SR25 (p. 147), and LS1 (p. 241) of Bose, Clatworthy, and Shrikhande [9].

Table 4.

Some RPB (regular pairwise balanced) designs usable for debates (with more than two segment sizes, within which candidates do not all appear equally; constructed by deleting two or more candidates from a BIB design).

	Parametersa
Plan no.	C	S	R	Si@Ki	L	How to construct
4-01	8	11	5	6@3; 3@4; 2@5	2	From Plan #1-17, delete candidates 8, 10, 11b
4-02	8	11	5	1@2; 3@3; 6@4; 1@5	2	From Plan #1-17, delete candidates 9, 10, 11
4-03	8	11	6	2@3; 3@4; 6@5	3	From Plan #1-18, delete candidates 6, 10, 11b
4-04	8	11	6	1@3; 6@4; 3@5; 1@6	3	From Plan #1-18, delete candidates 9, 10, 11
4-05	9	11	5	2@3; 6@4; 3@5	2	From Plan #1-17, delete candidates 10, 11
4-06	9	11	6	3@4; 6@5; 2@6	3	From Plan #1-18, delete candidates 10, 11
4-07	10	13	4	3@2; 6@3; 4@4	1	From Plan #1-20, delete candidates 11, 12, 13
4-08	11	13	4	1@2; 6@3; 6@4	1	From Plan #1-20, delete candidates 12, 13
4-09	14	16	6	2@4; 8@5; 6@6	2	From Plan #1-23, delete candidates 15, 16
4-10	14	20	5	1@2; 8@3; 11@4	1	From Plan #1-24, delete candidates 15, 16
4-11	17	19	9	4@7; 10@8; 5@9	4	From Plan #1-25, delete candidates 18, 19
4-12	17	21	5	6@3; 8@4; 7@5	1	From Plan #1-27, delete candidates 18, 19, 20, 21
4-13	17	21	5	1@2; 3@3; 11@4; 6@5	1	From Plan #1-27, delete candidates 17, 19, 20, 21b
4-14	18	21	5	3@3; 9@4; 9@5	1	From Plan #1-27, delete candidates 19, 20, 21
4-15	18	21	5	1@2; 12@4; 8@5	1	From Plan #1-27, delete candidates 17, 20, 21b
4-16	19	21	5	1@3; 8@4; 12@5	1	From Plan #1-27, delete candidates 20, 21

^aSee footnote to Table 2.

^bOne of the remaining candidates needs to be renumbered.

Each of Tables 2–4 shows the values of (C, S, R, L) for each RPB design. For each plan, Tables 2 and 3 show, for the first segment size, the number of segments (S₁) and the number of candidates (K₁) appearing in each one (S₁@K₁); and similarly for the segments of the second size (S₂@K₂). Table 4 likewise shows, for each segment size, the number of segments along with the number of candidates in each, but there are three or more segment sizes.

To illustrate how the designs of Table 2 are constructed, we consider Plan #2-07. It is obtained by deleting candidate 13 from Plan #1-20 (whose full details were already presented earlier). The result is:

$$\begin{array}{ccccccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 2\\ 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 1& 1& 3\\ 4& 5& 6& 7& 8& 9& 10& 11& 12& 4& 5& 2& 7\\ 8& 9& 10& 11& 12& & 1& 2& 3& & & 6& \end{array}$$

Plan #2-07 has C = 13 – 1 = 12 candidates, S = 13 segments, and R = 4 total appearances for each candidate. Across the 13 segments, any two candidates appear together in the same segment L = 1 time. The plan has R = 4 segments consisting of 4 – 1 = 3 candidates each and S – R = 13 – 4 = 9 segments with 4 candidates each. Each candidate appears in L = 1 of the former and in R – L = 4 – 1 = 3 of the latter (cf. Proposition 5 concerning all of the foregoing results).

Except for Plan #3-01 (which is a combination of Plans #1-02 and 1-03), the construction of any design in Table 3 is basically through the use of sub-design(s) (shown below the table) or difference set(s), or both. Note that, in the ‘How to construct’ column of the table, the statements before and after each semicolon refer, respectively, to the first (smaller) and second (larger) segment sizes. For illustration, we show how Plan #3-06 is obtained. It has S₁ = 8 segments with K₁ = 3 candidates each and S₂ = 10 segments of size K₂ = 4, as follows:

$$\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 2& 3& 4& 5& 6& 7& 8& 1\\ 4& 5& 6& 7& 8& 1& 2& 3\end{array}\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 2& 3& 4& 5& 6& 7& 8& 1\\ 4& 5& 6& 7& 8& 1& 2& 3\\ 5& 6& 7& 8& 1& 2& 3& 4\end{array}\begin{array}{cc}1& 2\\ 3& 4\\ 5& 6\\ 7& 8\end{array}$$

The 8 segments of size 3 are constructed by developing the difference set (1 2 4) mod 8. The first 8 of the 10 segments of size 4 are obtained by developing the difference set (1 2 4 5) mod 8. Then the final 2 segments come from copying sub-design D10, which is found below the table. Each of the C = 8 candidates appears in R = 8 of the S = 18 segments. Moreover, each one appears in 3 of the first 8 (smaller) segments and in 5 of the last 10 (larger) segments. Any two candidates appear together in L = 3 of the 18 segments.

We use Plan #4-07 to portray how designs in Table 4 are obtained. From Plan #1-20 (exhibited earlier), one deletes candidates 11, 12, and 13, yielding

$$\begin{array}{ccccccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 1& 1& 2\\ 3& 4& 5& 6& 7& 8& 9& 10& 3& 4& 5& 2& 3\\ 4& 5& 6& 7& 8& 9& 10& 2& & & & 6& 7\\ 8& 9& 10& & & & 1& & & & & & \end{array}$$

as Plan #4-07. Each of the C = 10 candidates appears in R = 4 of the S = 13 segments, which consist of S₁ = 3 segments with K₁ = 2 candidates each, S₂ = 6 of size K₂ = 3, and S₃ = 4 of size K₃ = 4. Across all 13 segments, any two candidates appear together in the same segment L = 1 time.

Unlike the designs of Tables 2 and 3, though, those in Table 4 do not have each candidate appearing equally in segments of a given size. In Plan #4-07, for instance, candidate 1 appears in 1 segment of size K₁ = 2, 1 segment of size K₂ = 3, and 2 segments of size K₃ = 4, whereas candidate 2 is in 0 segments of size 2, 3 of size 3, and 1 of size 4. (Of the remaining 8 candidates, 5 follow the same pattern as candidate 1 and the other 3 are like candidate 2.)

One implication of the unequal distribution among segment sizes concerns time allotments. Suppose that (for each i) M_i debate minutes are allocated to each of the S_i segments of size K_i. Then, for any design in Table 2 or 3, all candidates will receive equal exposure regardless of the ratio of M₁ to M₂ (or their values), because all C contenders appear equally within either segment size. (We assume for present purposes that the minutes within any segment are divided equally among the candidates.) For Plan #4-07, however, suppose that M₁ = M₂ = M₃ = 12. Then the number of minutes of exposure will be 1 × 12/2 + 1 × 12/3 + 2 × 12/4 = 16 for candidate 1 but 0 × 12/2 + 3 × 12/3 + 1 × 12/4 = 15 for candidate 2.

For plans in Table 4, unequal exposure times can be avoided by making M_i proportional to K_i. Thus, if (e.g.) in Plan #4-07 one sets M₁ = 8, M₂ = 12, M₃ = 16 (which are proportional to K₁ = 2, K₂ = 3, K₃ = 4), then every candidate receives 16 minutes of exposure (1 × 8/2 + 1 × 12/3 + 2 × 16/4 = 16 for candidates with the first pattern, 0 × 8/2 + 3 × 12/3 + 1 × 16/4 = 16 for those with the second).

One may not feel, though, that a small difference in exposure times (16 versus 15 minutes in the case of Plan #4-07) is sufficient reason to depart from unequal M_i's. On the other hand, larger M_i's for larger K_i's may be deemed desirable anyway, on the grounds that a segment with more candidates deserves a greater time allocation. In fact, an argument can be made in favor of M_i's that vary even more from each other than those that are proportional to the K_i's. Suppose for the moment that time within any segment is divided equally not only among the K_i candidates in the segment but also among its K_i(K_i −1)/2 pairs of candidates [where we imagine that, hypothetically at least, any two candidates within a segment are to be engaged in debate between themselves for an identical amount of time, M_i divided by K_i(K_i −1)/2]. Plan #4-07 demonstrates that, with the M_i's proportional to the K_i's, candidate pairs in the different segment sizes receive different exposures. Exposure minutes are 8/1 = 8, 12/3 = 4, and 16/6 = 2 $\frac{2}{3}$ for pairs in segments of sizes 2, 3, and 4, respectively—highly disparate values, though the corresponding ones of 12/1 = 12, 12/3 = 4, and 12/6 = 2 for M₁ = M₂ = M₃ = 12 are even more so.

But now suppose that we choose the M_i's so that they are proportional to the values of K_i(K_i −1)/2. Then, for any plan in Table 2, 3, or 4, the time allocation for any pair of candidates included within any segment will be the same regardless of the size (K_i) of the segment that the pair is in. Additionally, for the plans in Tables 2 and 3, setting the M_i's proportional to the K_i(K_i −1)/2 values will give the same number of minutes of total exposure to every candidate because, for those plans, all candidates receive the same exposure regardless of the values of M₁ and M₂.

Per Proposition 7 that follows, however, it turns out that all candidates in any other RPB design, including those in Table 4, will also receive the same number of minutes of exposure if the M_i and K_i(K_i −1)/2 values are proportional. For illustration, suppose in the case of Plan #4-07 that one sets M₁ = 3, M₂ = 9, M₃ = 18 [which are proportional to K₁(K₁ −1)/2 = 1, K₂(K₂ −1)/2 = 3, K₃(K₃ −1)/2 = 6]. Then every candidate receives 13 $\frac{1}{2}$ minutes of exposure (1 × 3/2 + 1 × 9/3 + 2 × 18/4 = 13 $\frac{1}{2}$ for candidates with the first pattern, 0 × 3/2 + 3 × 9/3 + 1 × 18/4 = 13 $\frac{1}{2}$ for those with the second). Of course, each of the C(C −1)/2 = 45 candidate pairs receives 3 minutes, regardless of the size of the segment in which it appears.

Proposition 7.

Suppose an RPB design has parameters C, S, R, and L, with S₁ segments consisting of K₁ candidates, S₂ segments of size K₂, … , where Σ S_i = S and each K_i > 1. Let M_i minutes be allotted to each segment of size K_i, such that the M_i's are proportional to the values of K_i(K_i −1)/2 with a proportionality factor of h [i.e. M_i = hK_i(K_i − 1)/2]. Assume that time within any segment is allocated equally among candidate pairs and equally among candidates. Then (i) every candidate pair receives the same number of minutes of exposure each of the L times that it appears (regardless of the segment size), and (ii) the total number of minutes of exposure across all S segments is the same for each of the C candidates.

Proof

Part (i) is obvious because the weights, M_i, are constructed to be proportional to the number of candidate pairs, K_i(K_i −1)/2 for a segment that has K_i candidates. For part (ii), note first that, for any segment of size K_i in which a candidate Z appears, Z is in K_i – 1 pairs (one for each other candidate) and has total exposure minutes of M_i/K_i = h(K_i − 1)/2, or an average of h/2 minutes per pair. Because overall Z is paired L times with each of the C – 1 other candidates, Z's total exposure is L(C – 1)h/2 minutes, a result that holds for any candidate Z.

The illustration above with M₁ = 3, M₂ = 9, M₃ = 18 for Plan #4-07 has L = 1, C = 10, and h = 3. Thus L(C – 1)h/2 = 13 $\frac{1}{2}$ minutes, which agrees with what was found before.

Although it may seem that our development, just above but also earlier, is heavily tied to candidate pairs, there is no implication that pairs would need to engage in one-on-one matches. Rather, the basic mathematical role of the pairs involves the construction and properties of the designs and serves to bring about symmetry.

In addition, although for mathematical simplicity it is assumed above that exposure is equal for each candidate and for each candidate pair, no implication is intended that such equality can easily be implemented, or necessarily should be. In the online Supplemental Material, though, we do allude to the issue of whether it is desirable for better–known candidates to receive more speaking time.

9. Summarization of the debate plans

Table 5 consolidates the debate plans of Tables 1–4 so that all plans for a given number of candidates, C, are shown together on one line. Although the table covers a total of 69 plans, obviously the only ones applicable to a specific debate would be those that appear on the line for the C-value that pertains to that debate. In a long series of presidential primary debates, the number of candidates would generally differ (but usually not increase) from one debate to the next, so that different C-values would apply and different plans would thus be needed.

Plan #1-28 (for C = 23) is the only one in Tables 1–4 that is omitted from Table 5. Because of high K, it and Plans #1-25 and 1-26 are useful mainly for providing means for constructing other designs, although the last two are still shown in Table 5.

Table 5.

Summary of the debate plans in Tables 1–4, by number of candidates, C, for C from 3 to 21.

		Number of plans in Table
C		1	2	3	4		Total	For each plan: Number of segments, S (before @ sign) and number(s) of candidates per segment (after @ sign) Plans in different tables are separated by the / sign
3		1	0	0	0		1	3@2 ///
4		2	0	1	0		3	4@3; 6@2 // 10@2,3 /
5		2	0	1	0		3	10@2; 10@3 // 10@2,3 /
6		2	2	0	0		4	10@3; 15@2 / 7@2,3; 7@3,4 //
7		3	0	0	0		3	7@3; 7@4; 21@2 ///
8		1	2	4	4		11	14@4 / 12@2,3; 12@5,6 / 12@2,5; 12@3,6; 16@2,4; 18@3,4 / 11@3,4,5; 11@2,3,4,5; 11@3,4,5; 11@3,4,5,6
9		3	0	2	2		7	12@3; 18@4; 18@5 // 15@3,4; 15@3,5 / 11@3,4,5; 11@4,5,6
10		2	2	1	1		6	15@4; 18@5 / 11@4,5; 11@5,6 / 20@3,4 / 13@2,3,4
11		2	0	0	1		3	11@5; 11@6 /// 13@2,3,4
12		1	1	3	0		5	22@6 / 13@3,4 / 15@4,6; 19@3,4; 26@2,3 /
13		2	0	0	0		2	13@4; 26@3 ///
14		0	1	1	2		4	/ 15@6,7 / 21@2,4 / 16@4,5,6; 20@2,3,4
15		1	2	0	0		3	15@7 / 16@5,6; 20@3,4 //
16		2	0	0	0		2	16@6; 20@4 ///
17		0	0	1	3		4	// 34@4,5 / 19@7,8,9; 21@3,4,5; 21@2,3,4,5
18		0	1	0	2		3	/ 19@8,9 // 21@3,4,5; 21@2,4,5
19		2	0	0	1		3	19@9; 19@10 /// 21@3,4,5
20		0	1	0	0		1	/ 21@4,5 //
21		1	0	0	0		1	21@5 ///

Among the plans for any given C, there will generally be tradeoffs. Thus, goals of a small S (number of segments), small K or K_i's (candidates per segment), and smaller number of segment sizes (which are 1, 2, 2, and at least 3 for plans from Tables 1–4, respectively) may conflict with one another and force tradeoffs, especially when C is greater. Because the relative importance of these three goals may differ depending on circumstances, the choice of a plan may also differ, but may be made easier by a larger menu of options.

For plans without a uniform number of candidates per segment (Tables 2–4), we examined three methods for choosing the relative number of minutes to allocate to each segment: same number of minutes for every segment, regardless of segment size; minutes proportional to K_i, the number of candidates in a segment; and minutes proportional to K_i(K_i −1)/2, the number of its candidate pairs. The third method is the only one that provides equal exposure, in a certain sense, both for all candidates and for all candidate pairs. But, because it also results in the greatest discrepancy among minutes per segment, it might not always be deemed ideal.

10. Wider applications

One can ask if our work in this paper may have applications or ramifications beyond presidential primary debates. Possibilities would include other kinds of debates or similar events, either political or otherwise.

A different type of application could be to certain special mixer events, either business or social, in which a total of C individuals are to get to know each other better by meeting in S small groups (akin to our segments) such that each person is together in a group with each other person the same number (L) of times (could be L = 1). Such an application might even be fitting during pandemic times when smaller groups can meet safely but larger ones cannot. Of course, most of our designs would allow no more than one or two groups to meet simultaneously (to avoid overlap of participants), so that not everyone could meet at once.

Regular pairwise balanced designs, of which balanced incomplete block designs are a special case, have received little if any previous attention in the statistical literature. Our work here has benefited heavily by being able to use RPB designs for debate plans rather than being confined to BIB designs. At least under certain circumstances, might the more traditional applications that utilize BIB designs also benefit from an expanded menu that embraces RPB designs? Investigation of this question is beyond the scope of the present paper, but would involve study, for RPB designs, of such matters as practical advantages and disadvantages in addition to efficiencies of estimates of treatment differences.

11. Closing remarks

The types of debate plans that we have proposed involve a significant departure from customary recent practice. That is mainly because they entail a larger number of segments, necessarily accompanied by less time allocated per segment. But they also treat all candidates evenhandedly, and they enable fewer candidates per segment. Although their adoption would require overcoming inertia and forsaking the status quo, they deserve full consideration.

For an illustration of how our approach compares with the current one, we show how our Plan #2–12 contrasts with the plan used for the first two Democratic debates in the summer of 2019 (described earlier). Each of those Democratic debates consisted of two events (segments) on consecutive nights, with 10 of the 20 candidates appearing in each event. Two given candidates could appear together on the same night in both debates, in neither, or in one debate but not the other. Each candidate in each debate was on the stage 50% of the time (i.e. for the entire event on one of the two nights).

We suppose that our Plan #2-12 is also spread across two consecutive nights (though it could be carried out in a single night if shorter segments and/or a longer-than-usual total time were acceptable). The plan has the C = 20 candidates assigned to S = 21 segments, of which S₁ = 5 have K₁ = 4 candidates per segment and the remaining S₂ = 16 have K₂ = 5. Each candidate appears in R = 5 of the segments.

Any two candidates appear together in L = 1 segment, thus giving every candidate evenly matched ability to face each opponent. With this attribute, it may be easier for one candidate to challenge another on a special issue on which the two of them sharply diverge.

We can allow M₁ = 10 minutes of (speaking) time for each of the S₁ = 5 segments with K₁ = 4 candidates and M₂ = 12 $\frac{1}{2}$ minutes for each of the S₂ = 16 with K₂ = 5 contestants. (With these values, M_i is proportional to K_i.) Total time (for the 21 segments across the two evenings) is thus (5 × 10) + (16 × 12 $\frac{1}{2}$) = 250 minutes. If (e.g.) all S₁ = 5 segments with K₁ = 4 are scheduled on the first night along with 6 of the S₂ = 16 segments with K₂ = 5 (so that the second night consists of the remaining 10 segments with K₂ = 5), then (speaking) time on each night is 125 minutes.

With Plan #2-12, although all candidates have to be present on both nights, the total time that each one is on stage (in their S = 5 segments over the two evenings) is just (1 × 10) + (4 × 12 $\frac{1}{2}$) = 60 minutes. That is 24% (=  60/250) of the total debate time, far below the corresponding 50% figure (pointed out above) for the plan that was actually used. Thus the candidates have more time to rest, and perhaps also more to cogitate since their segments are spread out.

With the large number of segments in Plan #2-12 and most of our other plans, one might ask whether significant time could be lost while switching from one set of candidates to the next. One may find different ways to minimize transition time between segments. One way might be to use a two-part rotating stage that rotates 180° between segments. (Such a feature could attract extra interest from viewers in addition to providing quick transition.) Of course, any candidate who is in two consecutive segments would have to be able to move from one half of the stage to the other.

We do not try to deal here with the matter of randomizations in our plans. One may want to randomize in some way the order of the segments; the order of candidate appearances within a segment; and the candidates’ assignments to their identification numbers.

In some respects our plans differ little from what is done currently. They will not enable any participant to address every debate question, but that is no different from what happens now. As indicated in the online Supplemental Material, for a given total speaking time the average speaking time available per contestant is not affected by whether the segments are many or few: It is the same with our many short segments, and their fewer candidates per segment, as it is with one or two long and crowded segments.

One way in which our plans could force a minor change (which may be welcome anyway) is in regard to discipline. With short segments, it would be more important to enforce strict time limits to prevent some candidates in a segment from trying to get more than a fair share of speaking time.

In every plan of ours, any two candidates appear together in the same number, L, of segments. One may ask, is this condition essential? It forces restriction to plans for which the number of segments, S, equals or exceeds the number of candidates, C (Proposition 3 above), thus leading sometimes to more segments than one might like. But without it, one could face complaints about unfair treatment, such as from a candidate who appears against a front–runner in fewer segments (perhaps none at all) than do some rivals. Furthermore, Table 5 already presents menus of options that may generally provide adequate choice.

Nevertheless, we mention the possible use of partially balanced incomplete block (PBIB) designs with two associate classes, catalogued in Bose, Clatworthy, and Shrikhande [9]. Besides having S < C in some cases, those plans provide the same number of candidates in every segment and have every candidate appearing in the same number of segments. The number of segments in which two candidates appear together, however, is either L₁ or L₂ rather than a single value L. (Observe that PBIB designs do appear as sub-designs below Table 3 but are used there as components of larger plans.)

As seen in the online Supplemental Material, debates before and during U.S. presidential primaries have not received the academic attention they deserve even though they are more influential than general-election debates. Of these primary debates, moreover, the earlier ones not only have the greater impact, but also generally include a larger number of candidates (because of subsequent dropouts) and are thus the ones that face the most difficulty stemming from the size of the candidate field. Prior ways of setting up the debates become increasingly unsatisfactory as the number of candidates rises. If all candidates are together in a single session, the number on the stage at one time becomes unmanageable. If the debate is split into an undercard session and a main session, the candidates in the former are at a handicap. Although the candidates can be split into two sessions in some type of more balanced fashion, each candidate will still appear together with only some of the others. By exploiting the tools of combinatorial design, with focus on balanced incomplete block designs and (the lesser-known) regular pairwise balanced designs, we have developed a debate architecture that treats the entire group of candidates symmetrically and also avoids crowded arrays of too many candidates appearing simultaneously.

In 2015-2016 it was the Republican presidential primaries that attracted an unduly large number of candidates. Four years later the Democrats faced the same complication. In future presidential cycles, of course, the number of contenders could be cumbersome for either party.

Disclosure statement

No potential conflict of interest was reported by the author(s).