Can You Pass a Test by Rolling a Die? 10,000 People and All of Humanity | What If, Calculated #04

Yes: 10 of 10,000 people passed our four-choice, 20-question test using only a die.

Passing required 12 correct answers, with no penalty for mistakes; 10 was the middle result from three simulation runs.

The formula gives odds of about 1 in 1,100, or an average of 9.4 passes per 10,000 people.

The result in 30 seconds

A lucky pass is possible. Whether you should expect to see one depends on both the test and how many people take it.

• Eliminate one wrong option on every question, then guess: 141 people passed, about 14 times as many as with pure guessing.
• Know one answer in every four for certain: 575 passed, more than 50 times as many.
• Make three wrong answers cancel one right answer: pure guessing passes about 1 in 34,000 times, according to the formula.
• Extend the test to 100 questions: nobody passed in our simulation, even with about 8.3 billion people taking it.

That last result still leaves a small chance of someone passing. To understand the difference between an unlikely event and an impossible one, start with the rules.

The test we actually modelled

Rolls of 1, 2, 3 and 4 select A, B, C and D. A 5 or 6 means roll again until an answer is selected. Rerolls do not count as wrong answers. Every option is equally likely, so each question has a 1/4 chance of being answered correctly.

We treat each person's choices as independent: another person's roll, or an earlier answer, does not affect the next choice. The questions, answer key, people and exam hall are fictional. This is a test of what random guessing can do under specified rules.

How to calculate a lucky pass

First count exactly how many answers are right

The binomial distribution describes the number of successes in repeated, independent attempts with the same chance of success. Here, a success is a correct answer.

Let n be the number of questions, p the chance of getting one right, and X the number answered correctly. The chance of exactly k correct answers is:

P(X = k) = C(n, k) × p^k × (1 − p)^(n − k)

P means probability. C(n, k) counts the ways to choose which k of the n questions are right. The two powers account for the correct and incorrect answers; multiplying by C(n, k) includes all their possible arrangements.

For this test, n = 20 and p = 1/4. A pass includes exactly 12 right, exactly 13 right, and every higher score. Add those probabilities:

q = P(X ≥ 12) = Σ[k=12…20] C(20, k) × (1/4)^k × (3/4)^(20 − k)

The Σ symbol means to add the terms across the stated range. The result is about 0.0935%, or about 1 in 1,100. Multiplying the individual probability q by the number of people gives the expected number of passes:

Expected passes = 10,000 × q ≈ 9.4

Why the simulation produced 10 rather than 9.4

An expected count is an average over repeated tests. A particular run must have a whole number of people passing, and that number varies.

Tamelabo's own program rolled a die for each person and each question, including rerolls. Three runs with different random seeds—settings that let us reproduce the rolls—produced 14, 10 and 5 passes. We used the median, the middle of those three results, for the video. The figure of 10 is not the average of those runs.

In the featured run, 25 people still had a path to passing before the final question. Seven had already secured a pass; the other 18 needed one more right answer. Three of those 18 got it, bringing the total to 10.

Test-taker #3777 was one of them. Their final answer was their twelfth correct answer: 12 right, eight wrong, exactly 60%. A very unusual answer sheet becomes less surprising when thousands of people each get a chance to produce it.

What partial knowledge changes

Keep the test fixed, but give the test-takers some reliable information.

Eliminating one wrong option changes p from 1/4 to 1/3 in the same formula. That improvement applies to every question, raising the chance of reaching the pass mark substantially.

If K answers are known for certain, bank those correct answers first. Use the binomial formula on the remaining 20 − K questions, and add the probabilities of getting at least 12 − K of them right. A known answer is always correct in this model.

The increases of about 14 times and more than 50 times refer to these particular strategies. They do not mean that any amount of knowledge always multiplies the chance of passing by ten.

Is narrowing everything to two options enough?

With two options left on every question, each guess has a 1/2 chance of being right. But the pass mark is 60%, and the chance of reaching it is 25.2%: about one person in four.

Knowing some answers for certain changes the calculation. This table uses the formula, with the original 20 questions, a 12-answer pass mark, no penalty, and an answer to every question.

Percentages are rounded for display. With nine answers secured and four options on the rest, you need at least three successful guesses from the remaining 11 questions. That happens 54.5% of the time. Knowing 11 answers raises the chance to 92.5%.

The video also compares these approaches using a rough measure: the number of wrong options you can eliminate. Narrowing all 20 questions to two options removes 40 wrong options. Count a known answer as eliminating all three wrong options, and knowing nine removes 27. Under that measure, the second approach uses fewer eliminations and gives better odds.

This does not measure study time. Recognising one wrong option and knowing the correct answer can require different amounts of work. The comparison describes the model; it cannot tell you how long to study.

What if wrong answers cost points?

A penalty raises the number of correct answers needed

Give one unit for a correct answer and subtract 1/3 of a unit for each wrong answer. The pass mark stays at 12 units. Since everyone answers all 20 questions, a person with c correct answers scores:

Score in correct-answer units = c − (20 − c)/3

With 14 right and six wrong, 14 − 6/3 = 12. Passing now requires at least 14 correct answers. Regrade #3777's answer sheet under this rule and the score becomes 46.7 out of 100: a fail.

For a pure four-option guess, the average score per question is:

(1/4) × 1 − (3/4) × (1/3) = 0

An average of zero does not give everyone a score of zero. Add the binomial probabilities for 14 or more correct answers, and the chance of passing is about 1 in 34,000, roughly one-thirtieth of the chance without the penalty.

Regrading the three simulated groups gave 1, 3 and 0 passes. The formula predicts about 0.3 passes per 10,000 people on average. When successful outcomes are rare, the count from one group can look quite different from the long-run average.

Eliminating a wrong option still helps

With three options remaining, the same penalty gives an average score per question of:

(1/3) × 1 − (2/3) × (1/3) = 1/9

That is positive. In the simulation, nine people out of 10,000 passed with this strategy and the penalty; the expected count is about 8.8. The penalty cancels the average score from a completely uninformed four-option guess. It does not cancel the benefit of reliably eliminating a wrong answer.

There is a historical example of a guessing penalty in the SAT. Through the January 2016 test, an incorrect multiple-choice answer cost a quarter point. Incorrect write-in math answers had no penalty. The redesigned SAT first given in March 2016 removed the guessing penalty. That historical quarter-point rule is distinct from the one-third penalty used in our experiment.

Could anyone pass a 100-question version?

Remove the penalty again and return to pure four-option guessing. Now make the test 100 questions long, with 60 correct needed to pass.

The same simulated group of 10,000 continued from its original first 20 answers. In the featured run, the last people lost their chance of passing at question 74. All three runs finished with zero passes.

The calculation still uses the same binomial formula: change n, then add probabilities from the required number of correct answers up to n. These are the results for four-option guessing, a 60% pass mark and no penalty:

The average score, expressed as a percentage, stays at 25% on both the 20-question and 100-question tests. With more questions, the proportion answered correctly clusters more closely around 1/4. A jump all the way to 60% becomes much rarer.

One person's odds versus the chance that somebody passes

For the worldwide run, we used 8,300,678,396 people: the UN's medium projection for 2026 from World Population Prospects 2024. It is a population projection, not a head count.

Let q be one person's chance of passing and N the number of independent test-takers. The probability that everybody fails is (1 − q)^N. Therefore:

P(at least one pass) = 1 − (1 − q)^N

Using the 100-question probability and that population gives about 0.11%, or about once in 900 worldwide runs. We calculate with the unrounded probability, rather than the rounded “1 in 7.5 trillion” figure.

All three worldwide simulations produced zero passes. In the featured run, one person remained after question 92, with 52 right and 40 wrong. They needed every one of the final eight answers to be correct. Question 93 was wrong, ending their chance of passing. For the large population, the simulation groups people together and uses the binomial distribution to calculate how many get an answer right.

Under the original 20-question rules, the worldwide simulation produced about 7.76 million passes. Giving more people a chance to try and giving each person more questions to answer have very different effects. Even so, the 100-question probability remains above zero.

Where this differs from a real exam

Every unknown answer in this model has a fixed chance of being right, and guesses are independent. A known answer is always correct, and an eliminated option is always wrong. We do not model mistaken confidence or accidentally eliminating the right answer.

We also do not model how long learning takes. The 54.5% result for nine known answers belongs to this fictional test. It cannot establish that learning nine answers is enough preparation for a real exam.

FAQ

With 10,000 people, is somebody guaranteed to pass?

No. Under the original 20-question rules, the formula puts the chance of zero passes at about once in 12,000 runs. In a separate check of 1,000 simulated groups, every group had between 2 and 19 passes, with none at zero. That observation does not make a zero-pass outcome impossible.

Do rolls of 5 and 6 count as wrong answers?

No. They trigger a reroll. Only the selected A–D answer is checked against the fictional answer key. The six die faces are not all assigned directly to the four options.

If I know nine answers, will guessing the rest guarantee a pass?

No. With no penalty and four options on the remaining questions, the chance is 54.5%. “More than half pass” is not a guarantee. Knowing 11 answers gives a 92.5% chance; securing 12 gives 100% under these rules.

With a penalty, should I answer every unknown question?

Improving your average score and improving your chance of passing are different goals. With our one-third penalty, someone who knows 12 answers can leave the rest blank and secure exactly 60%, guaranteeing a pass. Guessing all the remaining four-option questions instead gives a 63.3% chance of passing. Our experiments and comparison table assume everyone answers every question.

Try different rules in the simulator

The episode's Dice Exam Simulator lets you take the 20-question test yourself and see where your luck ranks among the same 10,000 people as in the video. You can then open Change the rules or the Scrape-a-Pass Table.

Start with New test-takers under the same rules to see how the count varies. Then change the pass mark or penalty and check how many answers you must know for at least half the group to pass. Finally, try the worldwide population and explore how many questions still allow a lucky pass.

The number of correct answers needed is rounded up to a whole number. As a result, adding a question can sometimes make passing easier. Check the displayed number of required correct answers alongside the result to see why.