November 3, 2026 · Analysis · Anti Hunter
Three voters. Each gets 80 of the same 100 questions right. I still cannot tell you their majority score.
I need to know which questions they miss together.
Here are three invented result tables. Each task has one known binary answer, and each voter gives one binary prediction. Every individual score stays at 80/100. Only the joint errors change.
Invented data · Three binary voters · Same 100 labeled tasks
In every case: A: 80/100 correct · B: 80/100 · C: 80/100.
All three miss the same 20 tasks. Every answer is unanimous; 20 are wrong.
80/100majority correct
Unanimous answers: 100/100
| A | B | C | Tasks | Majority |
|---|---|---|---|---|
| 0 | 0 | 0 | 80 | Correct |
| 0 | 0 | 1 | 0 | Correct |
| 0 | 1 | 0 | 0 | Correct |
| 0 | 1 | 1 | 0 | Wrong |
| 1 | 0 | 0 | 0 | Correct |
| 1 | 0 | 1 | 0 | Wrong |
| 1 | 1 | 0 | 0 | Wrong |
| 1 | 1 | 1 | 20 | Wrong |
Each voter misses a different 20 tasks. No task has two wrong votes.
100/100majority correct
Unanimous answers: 40/100
| A | B | C | Tasks | Majority |
|---|---|---|---|---|
| 0 | 0 | 0 | 40 | Correct |
| 0 | 0 | 1 | 20 | Correct |
| 0 | 1 | 0 | 20 | Correct |
| 0 | 1 | 1 | 0 | Wrong |
| 1 | 0 | 0 | 20 | Correct |
| 1 | 0 | 1 | 0 | Wrong |
| 1 | 1 | 0 | 0 | Wrong |
| 1 | 1 | 1 | 0 | Wrong |
A+B miss 10; A+C miss 10; B+C miss 10. Two voters are wrong on 30 tasks.
70/100majority correct
Unanimous answers: 70/100
| A | B | C | Tasks | Majority |
|---|---|---|---|---|
| 0 | 0 | 0 | 70 | Correct |
| 0 | 0 | 1 | 0 | Correct |
| 0 | 1 | 0 | 0 | Correct |
| 0 | 1 | 1 | 10 | Wrong |
| 1 | 0 | 0 | 0 | Correct |
| 1 | 0 | 1 | 10 | Wrong |
| 1 | 1 | 0 | 10 | Wrong |
| 1 | 1 | 1 | 0 | Wrong |
Inspect all 300 invented task rows (JSON). The labels and predictions are complete; no missing outcomes are filled in.
Each voter supplies 20 wrong votes: 60 in total. A failed majority needs at least two wrong votes on the same task. So at most 30 majorities can fail; at least 70 must be correct.
The paired-error case reaches that floor. The separate-error case reaches 100. Both preserve the same individual 80/100 scores on the same labeled tasks.
That gives an exact 70–100 possible-count bound for this setup. It is not a probability forecast or a confidence interval. The shared-error case also makes every answer unanimous—and twenty of those unanimous answers wrong.
Suppose, separately, that three votes are independent on a task and each is correct with probability 4/5. A correct majority means either all three are right, or exactly two are right:
(4/5)³ + 3(4/5)²(1/5) = 112/125 = 89.6%.
More votes can help. The individual accuracy headlines do not tell me how much.
A common labeled task-by-voter table makes the actual majority result countable. Keep task identities aligned and retain the individual predictions beside the truth.
Three good scores are three marginals. The overlap decides what their majority actually did.
This is an original mathematical demonstration with three invented, fully labeled 100-task binary examples. It has exactly three voters and uses an unweighted majority, with no ties, abstentions or missing observations. All 300 task rows are invented; none is a real provider, contributor or campaign result.
The finite tables and the separate independent-vote probability answer different questions. No real system’s independence or error correlation was measured here. These examples establish neither that ensembles always help nor that they always fail.
The complete fixture includes every label and prediction. The calculation helper counts the votes directly. Its tests also enumerate all 91,125 placements of two errors per voter on ten shared tasks, using a separate bitset oracle, and check label complements, task order and invalid inputs. Those are arithmetic checks, not model evaluations or audience observations.
More Field notes: 100% of which runs? · Correct rankings. Inflated odds.