What Does “Fair” Mean in a Proportion Test?
When a coin is flipped many times, the number of heads may not be exactly half the flips. When a die is rolled many times, one face may appear more often than another. Some variation is expected by chance, so an uneven count by itself does not establish that an object is unfair. A hypothesis test compares the observed count with what a specified fair-chance model predicts.
For a coin, define success as landing heads. A fair coin has a heads probability of 0.5 on each flip, so its benchmark is \(p_0=0.5\). For a die, you can define success as rolling one specified face—for example, a six. A fair six-sided die has a probability of \(1/6\) of showing that face on each roll.
As in Defining the Parameter in Hypothesis Statements, the hypotheses describe a fixed, unknown probability, not the sample proportion. For a two-sided fairness question about a coin, write \(H_0:p=0.5\) and \(H_a:p\ne0.5\). A coin with a heads probability below 0.5 and a coin with a heads probability above 0.5 are both departures from the fair-coin model, so the alternative includes both directions.
For a die, a test of whether sixes occur with the probability expected for a fair die uses \(H_0:p=1/6\) and \(H_a:p\ne1/6\). This is a test about the probability of rolling a six. It is not, by itself, a complete assessment of whether all six faces have equal probabilities. Keep the precise outcome and parameter in view when stating the conclusion.
Plan the Test Around the Trial Process
A one-proportion \(z\)-test can be appropriate when each trial has two recorded outcomes: the specified success and everything else. For a coin, those outcomes are heads and tails. For the die example, they are a six and not a six. The null proportion \(p_0\) determines the expected counts and the standard error for the test.
For a fairness question, “random” is about how the flips or rolls are carried out. A consistent tossing or rolling method should give each trial a chance outcome, and trials should be reasonably independent. A deliberately controlled toss, a die that is not allowed to tumble, or a procedure that records only selected results may not support the intended chance model. The conclusion is about the probability under the particular conditions used; changing the surface, tossing method, or other conditions may change what is being studied.
The Large Counts check uses the null probability because the test asks whether the observed result is unusual if the fair model is true. As covered in Checking the Success-Failure Condition for Tests, calculate expected successes and expected failures under \(H_0\), rather than substituting the observed sample proportion.
The alternative for a fairness question is typically two-sided: \(H_a:p\ne p_0\). Therefore, results count as at least as extreme as the observed result when they are as far from \(p_0\) or farther in either direction. The p-value is the probability, assuming \(H_0\) is true, of results at least that extreme. A low p-value can provide convincing evidence against the fair-chance model; it is not the probability that the object is fair or unfair.
Worked Examples
Worked Example: 62 Heads in 100 Flips
A student uses one coin and a consistent tossing method, recording the result of 100 separate flips. The coin lands heads 62 times. Assess whether this provides evidence that the coin’s probability of landing heads differs from 0.5. Use \(\alpha=0.05\).
State: Let \(p\) be the true probability that this coin lands heads in a flip made using this tossing method. The hypotheses are \(H_0:p=0.5\) and \(H_a:p\ne0.5\). The alternative is two-sided because either a higher or a lower heads probability would differ from the fair-coin benchmark.
Plan: Use a one-proportion \(z\)-test. The student used a chance-based tossing method and recorded all 100 flips, so the Random condition is reasonable if the method gave each flip a chance outcome. The flips are treated as independent trials; this is an assumption about the tossing process, not a claim that the results must alternate or be evenly balanced. The 10% condition does not apply because the student is not sampling without replacement from a finite population. Under \(H_0\), the expected heads count is \(np_0=100(0.5)=50\), and the expected tails count is \(n(1-p_0)=100(0.5)=50\). Both are at least 10, so the Large Counts condition is met.
Do: There are \(x=62\) successes in \(n=100\) flips. The sample proportion, null standard error, and test statistic are:
Because the alternative is two-sided, the p-value is twice the area above \(|z|=2.40\): \(2\text{normalcdf}(2.40,1\text{E}99,0,1)\approx0.0164\), rounded. Since \(0.0164<0.05\), reject \(H_0\).
Conclude: The results provide convincing evidence that this coin’s probability of landing heads under the tossing method used differs from 0.5. The observed proportion is above 0.5, but the test provides evidence of a difference in either direction. It does not prove that the coin itself is physically defective or establish what would happen under every possible tossing method.
Worked Example: A Coin Result That Is Not Convincing Evidence
In a separate practice trial, a student records 56 heads in 100 flips of a coin. Assume the tossing method is consistent and chance-based. Test whether the coin’s heads probability differs from 0.5 at \(\alpha=0.05\).
State: Let \(p\) be the true probability of heads under this student’s flipping method. Use \(H_0:p=0.5\) and \(H_a:p\ne0.5\).
Plan: A one-proportion \(z\)-test is appropriate if the flips are independent chance trials. The 10% condition is not relevant to repeated flips. Under the null, the expected counts are \(100(0.5)=50\) heads and \(100(0.5)=50\) tails, so the Large Counts condition is met.
Do: The sample proportion is \(\hat p=56/100=0.56\). The null standard error is \(\sqrt{0.5(0.5)/100}=0.05\), giving:
For the two-sided alternative, the p-value is \(2\text{normalcdf}(1.20,1\text{E}99,0,1)\approx0.2301\), rounded. Since \(0.2301>0.05\), fail to reject \(H_0\).
Conclude: These data do not provide convincing evidence that the coin’s heads probability differs from 0.5. They do not prove that the coin is fair or that \(p=0.5\). A result of 56 heads in 100 flips is not so unusual under the fair-coin model that this test rejects it at the 0.05 significance level.
Worked Example: Checking Whether Sixes Occur at the Fair-Die Rate
A person rolls a six-sided die 120 times using a consistent rolling method. The die shows a six on 30 rolls. Test whether the probability of rolling a six differs from the \(1/6\) probability predicted by a fair die, using \(\alpha=0.05\).
State: Let \(p\) be the true probability that this die shows a six on one roll under the method used. The hypotheses are \(H_0:p=1/6\) and \(H_a:p\ne1/6\).
Plan: Use a one-proportion \(z\)-test, counting a roll of six as success. The rolls are treated as independent outcomes from a chance-based rolling method. The 10% condition does not apply to repeated rolls. Under \(H_0\), the expected success count is \(np_0=120(1/6)=20\), and the expected failure count is \(n(1-p_0)=120(5/6)=100\). Both are at least 10, so the Large Counts condition is met.
Do: There are \(x=30\) sixes in \(n=120\) rolls, so \(\hat p=30/120=0.25\). The null standard error and test statistic are:
Using the unrounded standard error gives the same test statistic to two decimal places. For the two-sided alternative, the p-value is \(2\text{normalcdf}(2.45,1\text{E}99,0,1)\approx0.0143\), rounded. Since \(0.0143<0.05\), reject \(H_0\).
Conclude: The results provide convincing evidence that the probability of rolling a six with this die under the method used differs from \(1/6\). The observed proportion of sixes is higher than \(1/6\). Because a fair die must give each face probability \(1/6\), evidence that the six probability differs is evidence against the fair-die model. However, this test assesses the six outcome; it does not separately estimate the probabilities of the other five faces.
Common Mistakes and AP Exam Tips
A fairness test is still a one-proportion test, but the fair-object model supplies the benchmark. Be precise about what one success means and what the test can support.
- Using the sample proportion as the null benchmark. The null value comes from the fair model: 0.5 for heads on a fair coin, or \(1/6\) for one specified face on a fair six-sided die. The sample proportion is evidence compared with that value.
- Using a one-sided alternative just because the sample is high. If the question is whether an object is fair, outcomes that are unusually high or unusually low both matter. Write a two-sided alternative before calculating the p-value.
- Calling any uneven count proof of unfairness. Chance variation can produce counts that differ from their expected values. A test assesses whether the difference is sufficiently unusual under the null model; it does not prove a physical cause.
- Claiming that a large p-value proves fairness. Failing to reject means the data do not provide convincing evidence against the null at the selected significance level. It is not proof that the null is true.
- Applying the 10% condition to repeated trials. The 10% condition concerns sampling without replacement from a finite population. For repeated flips or rolls, discuss whether the chance trials are reasonably independent instead.
- Overstating what a die test establishes. A one-proportion test of sixes addresses the probability of six. A failure to reject that particular null does not demonstrate that all six face probabilities equal \(1/6\).
Key Takeaway
A coin or die fairness question can be framed as a comparison between an observed outcome proportion and the probability predicted by a fair-object model. The one-proportion \(z\)-test measures whether the result is unusual enough under that model to provide convincing evidence of a difference. Its conclusion is limited to the outcome, object, and trial conditions represented by the test.
Check Your Understanding
For each question, identify the fair-model benchmark and explain what a one-proportion test can conclude.
- A student records 48 heads in 100 flips and asks whether the coin is fair. Define \(p\) and write the null and alternative hypotheses.
- For 90 rolls of a fair six-sided die, what are the expected counts of fours and non-fours under the null hypothesis?
- A two-sided coin test has a p-value of 0.03 and \(\alpha=0.05\). State the decision and describe the conclusion in context without claiming proof.
- A die test of the probability of rolling a one fails to reject \(H_0:p=1/6\). Explain why this does not establish that the die is fair.
- Why is the 10% condition generally not the relevant independence check for 100 repeated flips of one coin?