Pair Coverage in Best–Worst Choice Sets

A practical mathematical explanation of how many choice screens are needed when a larger set of items is presented in smaller subsets.

 

1. What pair coverage means

Suppose a category contains v items, but only k items are shown on a screen at one time. A pair of items is covered when those two items appear together on at least one screen.

Pair coverage is therefore a property of the choice-set design. It does not mean that the respondent literally makes a separate binary choice for every pair. In a Best–Worst or MaxDiff task, one screen can contain several items, and the respondent chooses the Most and Least preferred items from that set.

Important distinction: if five items appear together, all ten unordered pairs among those five items have been exposed together. The Best–Worst response provides preference information about the set, but this should not be described as ten independent binary comparisons. When five items appear on the same screen, there are 10 different pairs of items within that group. For example, Item A appears alongside B, C, D and E; Item B also appears alongside C, D and E; and so on, giving 10 pairs altogether.

This means that one five-item screen provides pair coverage for 10 pairs: each of those pairs has appeared together at least once. However, the learner does not make 10 separate choices. The learner makes just two selections from the five items — one as Most and one as Least. Pair coverage therefore describes how thoroughly the items have been presented together across the instrument, not the number of individual choices made by the learner.

The purpose of a good design is to spread the screens so that no item or pair is systematically neglected, while keeping the number of screens manageable for the learner.

 

2. How many pairs exist?

With v items, the number of distinct unordered pairs is:

Ptotal = v(v − 1) 2 = C(v,2)

For example, with 10 items:

C(10,2) = 10 × 9 2 = 45

So a category containing 10 items contains 45 different pairs.

 

3. How many pairs are present on one screen?

If a screen shows k items, then the number of pairs appearing together on that screen is:

Pscreen = k(k − 1) 2 = C(k,2)

If 5 items are shown:

C(5,2) = 5 × 4 2 = 10

Each screen therefore contains 10 item-pairs.

 

4. The tempting calculation — and why it is only a lower bound

A natural first calculation is to divide the number of required pair appearances by the number of pair positions on each screen. If every pair must appear at least λ times, this gives:

b ≥ ⌈ λ × C(v,2) C(k,2) ⌉

Here b is the number of screens. The ceiling brackets ⌈ ⌉ mean “round upward”.

For 10 items shown 5 at a time, with every pair required once:

b ≥ ⌈45 / 10⌉ = 5
But 5 screens are not enough. The equation above counts pair positions, but it does not guarantee that those positions can be arranged without leaving some pair uncovered.
 

5. A stronger lower bound: every individual item must appear often enough

Consider one particular item. It must be paired with each of the other v − 1 items. Every time that item appears on a screen containing k items, it meets at most k − 1 other items.

Therefore, if every pair must occur at least λ times, each item must appear on at least:

r ≥ ⌈ λ(v − 1) k − 1 ⌉

Since there are v items, that requires at least v × r item-appearances in total. Each screen supplies exactly k item-appearances, so:

b ≥ ⌈ v × r k ⌉

Example: 10 items, 5 per screen, each pair once

r ≥ ⌈9 / 4⌉ = 3

Every item must therefore appear on at least 3 screens. Ten items then require at least 30 item-appearances. Five screens would provide only 25 item-appearances:

5 × 5 = 25 < 30

Therefore five screens are mathematically impossible. At least:

b ≥ ⌈30 / 5⌉ = 6 screens

A six-screen arrangement does exist, so in this case the lower bound is achievable.

 

6. The 10-item / 5-per-screen result

Applying the same argument gives the following minimums for the specific case used in PassionFinder: 10 items in a category, 5 shown per screen.

Required pair coverage (λ) Minimum appearances of each item (r) Minimum screens (b)
Every pair at least once ⌈9/4⌉ = 3 6
Every pair at least twice ⌈18/4⌉ = 5 10
Every pair at least three times ⌈27/4⌉ = 7 14
Every pair at least four times ⌈36/4⌉ = 9 18

These values are not merely rough estimates. For this particular v = 10, k = 5 case, arrangements can be constructed that achieve 6, 10, 14 and 18 screens respectively while meeting the stated pair-coverage requirement. Therefore these are true minima for λ = 1, 2, 3 and 4.

Practical consequence: increasing the number of screens gives more opportunities for each pair of items to occur together, but it also increases respondent workload. The instrument designer is therefore balancing measurement coverage against learner fatigue.
 

7. Where Balanced Incomplete Block Designs fit

A Balanced Incomplete Block Design (BIBD) is a particularly elegant special case. In a perfect BIBD:

  • every screen contains exactly k items;
  • every item appears exactly r times;
  • every pair appears together exactly λ times.

The BIBD parameters satisfy:

r(k − 1) = λ(v − 1)
bk = vr

Combining these gives:

r = λ(v − 1) k − 1
b = λv(v − 1) k(k − 1)

However, these equations describe a perfectly balanced design. Their arithmetic conditions are necessary but are not, by themselves, a guarantee that such a BIBD exists for every possible combination of parameters.

PassionFinder does not need a perfect BIBD in order to achieve useful pair coverage. A covering design or a nearly balanced algorithmic design is often more practical: the requirement is that every pair is covered at least the desired number of times, even if some pairs occur slightly more often.

 

8. Worked example: 7 items shown 3 at a time

With 7 items there are:

C(7,2) = 21 pairs

A screen of 3 items contains:

C(3,2) = 3 pairs

The pair-count lower bound is therefore:

b ≥ 21 / 3 = 7

In this special case the bound is achievable. A seven-screen BIBD exists: each item appears three times and every pair occurs exactly once. This is the classic 7-point, 3-per-block design.

 

9. Why this matters for respondents

Best–Worst tasks are useful because they force trade-offs. A conventional rating scale often allows a respondent to rate many desirable items highly. A Best–Worst task requires the learner to discriminate within a smaller set: which item is most like me, most important, most attractive, or most preferred — and which is least?

Smaller choice sets reduce the amount that must be considered at one moment. More screens, however, increase total workload. Pair coverage therefore provides a transparent way of discussing this design trade-off:

  • More items per screen create more pair exposures on each screen, but make each decision more cognitively demanding.
  • Fewer items per screen simplify each decision, but usually require more screens.
  • More screens allow stronger and more repeated coverage, but increase learner effort and the risk of fatigue.
 

10. The key takeaway

For a category containing 10 items with 5 items shown per screen, the minimum number of screens required to guarantee at least 1, 2, 3 and 4 appearances of every pair is:

6  →  10  →  14  →  18

one coverage     two coverages     three coverages     four coverages

(Back...)

Last modified: Tuesday, 8 September 2026, 8:33 AM