Repeated chance experiments and simulations

A fair coin has model probability one half for Heads. An actual twenty-toss run contains twelve Heads. Which statement is correct?

Define one complete trial and one run

For this topic, one trial is one draw from a bag, one coin toss or one spinner turn

A run contains a stated number of complete repeated trials, including successes and other outcomes

Define success before collecting results. For example, success may mean selecting a red token

Replace and remix a token after each draw so the next trial uses the same original bag model

Record every completed trial. Repeating unsuccessful trials until success and discarding the failures changes the experiment

Observed relative frequency is the number of successes divided by all completed trials, including unsuccessful results

A student draws one token, records its colour, replaces it and remixes. Of twenty draws, seven are red. What is the observed relative frequency of red?

A digital model must preserve the original chances

A simulation uses a random device to imitate the defined chance experiment

If a device chooses a whole number uniformly from one to D, inclusive, each integer has probability

Assign integers to labels in the same proportions as the target outcomes

For three red tokens and one blue token, eight equally likely integer results can be assigned six red and two blue

Use a stated mapping that includes every integer once, with no gaps or overlaps. Keep the same mapping for every trial

Construct two inclusive blocks

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

Simulate one draw by generating one uniformly chosen whole number from 1 to 8, including both endpoints

Use contiguous blocks in the displayed label order. The first block starts at 1 and each next block starts one above the previous endpoint

Every generated integer must belong to exactly one category. Reset to the same model for every new trial

Complete the integer counts and inclusive block endpoints for this simulation

Keep exact integers or simplified fractions for probabilities and distances; do not round

The original bag has 4 equally likely tokens

The digital device has 8 equally likely integer results, so multiply every category count by 2

red: \(3\times2=6\) integers, from 1 to 6 inclusive

Its mapped chance \(\frac{6}{8}\) equals the original \(\frac{3}{4}\)

blue: \(1\times2=2\) integers, from 7 to 8 inclusive

Its mapped chance \(\frac{2}{8}\) equals the original \(\frac{1}{4}\)

The last endpoint is 8; the blocks leave no gaps or overlaps

Uniform integer selection and the fixed reset rule preserve the intended one-stage probabilities

Scale both category counts equally

Count both endpoints in an integer block

An integer block from one to six contains six integers, not five

For an inclusive block from a to b, the number of integers is b − a + 1

If one block ends at six, the next starts at seven. Using six again would create an overlap

The final endpoint must equal the upper bound of the random-number device

Preserve three category probabilities

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

Simulate one draw by generating one uniformly chosen whole number from 1 to 20, including both endpoints

Use contiguous blocks in the displayed label order. The first block starts at 1 and each next block starts one above the previous endpoint

Every generated integer must belong to exactly one category. Reset to the same model for every new trial

Complete the integer counts and inclusive block endpoints for this simulation

Keep exact integers or simplified fractions for probabilities and distances; do not round

The original bag has 4 equally likely tokens

The digital device has 20 equally likely integer results, so multiply every category count by 5

red: \(2\times5=10\) integers, from 1 to 10 inclusive

Its mapped chance \(\frac{10}{20}\) equals the original \(\frac{2}{4}\)

blue: \(1\times5=5\) integers, from 11 to 15 inclusive

Its mapped chance \(\frac{5}{20}\) equals the original \(\frac{1}{4}\)

green: \(1\times5=5\) integers, from 16 to 20 inclusive

Its mapped chance \(\frac{5}{20}\) equals the original \(\frac{1}{4}\)

The last endpoint is 20; the blocks leave no gaps or overlaps

Uniform integer selection and the fixed reset rule preserve the intended one-stage probabilities

Complete every count and endpoint

Partition a larger random-number device

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

Simulate one draw by generating one uniformly chosen whole number from 1 to 40, including both endpoints

Use contiguous blocks in the displayed label order. The first block starts at 1 and each next block starts one above the previous endpoint

Every generated integer must belong to exactly one category. Reset to the same model for every new trial

Complete the integer counts and inclusive block endpoints for this simulation

Keep exact integers or simplified fractions for probabilities and distances; do not round

The original bag has 8 equally likely tokens

The digital device has 40 equally likely integer results, so multiply every category count by 5

red: \(1\times5=5\) integers, from 1 to 5 inclusive

Its mapped chance \(\frac{5}{40}\) equals the original \(\frac{1}{8}\)

blue: \(2\times5=10\) integers, from 6 to 15 inclusive

Its mapped chance \(\frac{10}{40}\) equals the original \(\frac{2}{8}\)

green: \(3\times5=15\) integers, from 16 to 30 inclusive

Its mapped chance \(\frac{15}{40}\) equals the original \(\frac{3}{8}\)

yellow: \(2\times5=10\) integers, from 31 to 40 inclusive

Its mapped chance \(\frac{10}{40}\) equals the original \(\frac{2}{8}\)

The last endpoint is 40; the blocks leave no gaps or overlaps

Uniform integer selection and the fixed reset rule preserve the intended one-stage probabilities

Check the complete four-label mapping

The target bag contains three red tokens and one blue token. Which uniformly generated integer mapping reproduces that model exactly?

Calculate a proportion from the full observed record

Add all category frequencies to check the total number of completed trials

Observed relative frequency is observed successes divided by completed trials

Model probability instead comes from the stated chance mechanism, such as red tokens divided by all equally likely tokens

Expected successes equal model probability times the trial count. This expectation can be fractional and is not a guaranteed observed count

Keep exact fractions in these calculations rather than rounding before comparing

Compare proportions by their non-negative distance

To find how far an observed proportion is from the model probability, subtract the smaller value from the larger

For the first observed proportion and model probability, write the model probability in tenths

For the second observed proportion and model probability, write the model probability in fortieths

Distance is zero only when the two proportions are equal. It does not tell whether the observation was above or below the model

Compare exact distances when deciding which record is closer; rounded decimals can hide a difference

Compare an observed run with its model

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording red

These are supplied illustrative run records for analysis, not a claim that this learner collected them

Run 1: 20 completed trials

Complete the exact observed and predicted run record

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of red is \(\frac{3}{5}\)

Run 1: add every category frequency to check \(14+6=20\)

There are 14 successes, so observed relative frequency is \(\frac{14}{20}=\frac{7}{10}\)

Expected successes: \(\text{expected successes}=20\times\frac{3}{5}=12\)

Write \(\frac{7}{10}=\frac{35}{50}\) and \(\frac{3}{5}=\frac{30}{50}\); the distance is \(\frac{35}{50}-\frac{30}{50}=\frac{1}{10}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

Use the observed counts rather than the expected count

Retain an exact difference between fractions

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording red

These are supplied illustrative run records for analysis, not a claim that this learner collected them

Run 1: 12 completed trials

Complete the exact observed and predicted run record

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of red is \(\frac{1}{3}\)

Run 1: add every category frequency to check \(5+7=12\)

There are 5 successes, so observed relative frequency is \(\frac{5}{12}\)

Expected successes: \(\text{expected successes}=12\times\frac{1}{3}=4\)

Write \(\frac{5}{12}=\frac{15}{36}\) and \(\frac{1}{3}=\frac{12}{36}\); the distance is \(\frac{15}{36}-\frac{12}{36}=\frac{1}{12}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

Allow a fractional expected count

Zero successes is different from zero trials

With zero successes in ten completed trials, the observed relative frequency is 0/10 = 0

This does not prove the event is impossible under the model; a possible event can fail to appear in a short run

With zero completed trials, no observed relative frequency is defined because there is no trial total to divide by

Do not enter zero as an estimate before any observations have been made

A possible event may have no successes in one run

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording red

These are supplied illustrative run records for analysis, not a claim that this learner collected them

Run 1: 10 completed trials

Complete the exact observed and predicted run record

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of red is \(\frac{1}{5}\)

Run 1: add every category frequency to check \(0+10=10\)

There are 0 successes, so observed relative frequency is \(\frac{0}{10}=0\)

Expected successes: \(\text{expected successes}=10\times\frac{1}{5}=2\)

Write \(0=\frac{0}{5}\) and \(\frac{1}{5}=\frac{1}{5}\); the distance is \(\frac{1}{5}-\frac{0}{5}=\frac{1}{5}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

A simulation has been set up but no trial has run. What observed relative frequency should be reported?

Distinguish simulated records from supplied examples

The next records are generated by a digital random process from the displayed token weights

Each trial chooses one equally likely elementary token position and restores the same model before the next trial

A fixed run seed reproduces the same simulated record. This makes the example checkable; it does not turn the model prediction into the observed result

A supplied illustrative record is labelled differently and is used to practise analysis, not presented as this learner’s own experiment

Analyse an actual seeded three-category run

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording blue

These records were generated by seeded digital simulations of the stated model

Run 1: 100 completed trials

Complete the exact observed and predicted run record

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of blue is \(\frac{3}{6}=\frac{1}{2}\)

Run 1: add every category frequency to check \(28+54+18=100\)

There are 54 successes, so observed relative frequency is \(\frac{54}{100}=\frac{27}{50}\)

Expected successes: \(\text{expected successes}=100\times\frac{1}{2}=50\)

Write \(\frac{27}{50}=\frac{54}{100}\) and \(\frac{1}{2}=\frac{50}{100}\); the distance is \(\frac{54}{100}-\frac{50}{100}=\frac{1}{25}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

Recompute all summaries from a different simulated run

Use a large simulated record without assuming exact agreement

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording yellow

These records were generated by seeded digital simulations of the stated model

Run 1: 1000 completed trials

Complete the exact observed and predicted run record

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of yellow is \(\frac{4}{10}=\frac{2}{5}\)

Run 1: add every category frequency to check \(104+210+291+395=1000\)

There are 395 successes, so observed relative frequency is \(\frac{395}{1000}=\frac{79}{200}\)

Expected successes: \(\text{expected successes}=1000\times\frac{2}{5}=400\)

Write \(\frac{79}{200}=\frac{395}{1000}\) and \(\frac{2}{5}=\frac{400}{1000}\); the distance is \(\frac{400}{1000}-\frac{395}{1000}=\frac{1}{200}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

Calculate the proportion from two thousand completed trials

Compare runs using proportions rather than raw counts

Runs with different trial counts can have different success counts even when their observed proportions are similar

Compute each success count divided by its own completed-trial count

Compare both observed proportions with the same model probability using non-negative distances

A larger raw count difference need not mean a larger proportional difference

Two independent runs are separate repetitions. A twenty-trial prefix and the same run extended to one hundred trials are not independent records

Compare two separate runs of equal size

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording red

These are supplied illustrative run records for analysis, not a claim that this learner collected them

Run 1: 50 completed trials

Run 2: 50 completed trials

The two displayed records represent separate runs of the same model, not a short prefix and its continuation

Calculate both observed proportions and their distances from the same model probability

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of red is \(\frac{1}{2}\)

Run 1: add every category frequency to check \(28+22=50\)

There are 28 successes, so observed relative frequency is \(\frac{28}{50}=\frac{14}{25}\)

Expected successes: \(\text{expected successes}=50\times\frac{1}{2}=25\)

Write \(\frac{14}{25}=\frac{28}{50}\) and \(\frac{1}{2}=\frac{25}{50}\); the distance is \(\frac{28}{50}-\frac{25}{50}=\frac{3}{50}\)

Run 2: add every category frequency to check \(23+27=50\)

There are 23 successes, so observed relative frequency is \(\frac{23}{50}\)

Expected successes: \(\text{expected successes}=50\times\frac{1}{2}=25\)

Write \(\frac{23}{50}=\frac{46}{100}\) and \(\frac{1}{2}=\frac{50}{100}\); the distance is \(\frac{50}{100}-\frac{46}{100}=\frac{1}{25}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

Which comparison is justified by these two exact proportional distances?

Run 1 has distance 3/50 from 1/2

Run 2 has distance 1/25 from 1/2

Run 2 is closer to the model probability

This comparison describes the two supplied runs. It does not prove that every larger run is closer or every future run has the same result

Find the closer result from exact distances

A bigger count deviation can have a smaller proportional distance

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording red

These are supplied illustrative run records for analysis, not a claim that this learner collected them

Run 1: 10 completed trials

Run 2: 100 completed trials

The two displayed records represent separate runs of the same model, not a short prefix and its continuation

Calculate both observed proportions and their distances from the same model probability

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of red is \(\frac{1}{2}\)

Run 1: add every category frequency to check \(4+6=10\)

There are 4 successes, so observed relative frequency is \(\frac{4}{10}=\frac{2}{5}\)

Expected successes: \(\text{expected successes}=10\times\frac{1}{2}=5\)

Write \(\frac{2}{5}=\frac{4}{10}\) and \(\frac{1}{2}=\frac{5}{10}\); the distance is \(\frac{5}{10}-\frac{4}{10}=\frac{1}{10}\)

Run 2: add every category frequency to check \(55+45=100\)

There are 55 successes, so observed relative frequency is \(\frac{55}{100}=\frac{11}{20}\)

Expected successes: \(\text{expected successes}=100\times\frac{1}{2}=50\)

Write \(\frac{11}{20}=\frac{22}{40}\) and \(\frac{1}{2}=\frac{20}{40}\); the distance is \(\frac{22}{40}-\frac{20}{40}=\frac{1}{20}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

Which comparison is justified by these two exact proportional distances?

Run 1 has distance 1/10 from 1/2

Run 2 has distance 1/20 from 1/2

Run 2 is closer to the model probability

This comparison describes the two supplied runs. It does not prove that every larger run is closer or every future run has the same result

Compare this pair without assuming a general guarantee

Larger runs reduce typical variation, not every individual discrepancy

Across many independent runs of an unchanged model, larger trial counts tend to give less variable observed proportions

This is a statement about typical proportional variation, not a guarantee that every larger run is closer

Within one accumulating run, the estimate can move towards or away from the model probability as new results arrive

The chance of success on the next trial remains the same under the unchanged model. It does not compensate for a run of earlier failures

Use several run numbers and trial sizes when investigating variability, rather than relying on one favourable pair

A longer run can be farther in a particular comparison

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording red

These are supplied illustrative run records for analysis, not a claim that this learner collected them

Run 1: 20 completed trials

Run 2: 200 completed trials

The two displayed records represent separate runs of the same model, not a short prefix and its continuation

Calculate both observed proportions and their distances from the same model probability

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of red is \(\frac{1}{2}\)

Run 1: add every category frequency to check \(10+10=20\)

There are 10 successes, so observed relative frequency is \(\frac{10}{20}=\frac{1}{2}\)

Expected successes: \(\text{expected successes}=20\times\frac{1}{2}=10\)

Write \(\frac{1}{2}=\frac{2}{4}\) and \(\frac{1}{2}=\frac{2}{4}\); the distance is \(\frac{2}{4}-\frac{2}{4}=0\)

Run 2: add every category frequency to check \(112+88=200\)

There are 112 successes, so observed relative frequency is \(\frac{112}{200}=\frac{14}{25}\)

Expected successes: \(\text{expected successes}=200\times\frac{1}{2}=100\)

Write \(\frac{14}{25}=\frac{28}{50}\) and \(\frac{1}{2}=\frac{25}{50}\); the distance is \(\frac{28}{50}-\frac{25}{50}=\frac{3}{50}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

Which comparison is justified by these two exact proportional distances?

Run 1 has distance 0 from 1/2

Run 2 has distance 3/50 from 1/2

Run 1 is closer to the model probability

This comparison describes the two supplied runs. It does not prove that every larger run is closer or every future run has the same result

Check the proportions before choosing the longer run

Different observed proportions can be equally close

Model: a bag contains the labelled tokens shown. Tokens are identical except for their labels; each individual token is equally likely after mixing

One trial selects one token, records its label, then replaces and remixes it. Success means recording red

These are supplied illustrative run records for analysis, not a claim that this learner collected them

Run 1: 20 completed trials

Run 2: 40 completed trials

The two displayed records represent separate runs of the same model, not a short prefix and its continuation

Calculate both observed proportions and their distances from the same model probability

Keep exact integers or simplified fractions for probabilities and distances; do not round

The model probability of red is \(\frac{1}{2}\)

Run 1: add every category frequency to check \(8+12=20\)

There are 8 successes, so observed relative frequency is \(\frac{8}{20}=\frac{2}{5}\)

Expected successes: \(\text{expected successes}=20\times\frac{1}{2}=10\)

Write \(\frac{2}{5}=\frac{4}{10}\) and \(\frac{1}{2}=\frac{5}{10}\); the distance is \(\frac{5}{10}-\frac{4}{10}=\frac{1}{10}\)

Run 2: add every category frequency to check \(24+16=40\)

There are 24 successes, so observed relative frequency is \(\frac{24}{40}=\frac{3}{5}\)

Expected successes: \(\text{expected successes}=40\times\frac{1}{2}=20\)

Write \(\frac{3}{5}=\frac{6}{10}\) and \(\frac{1}{2}=\frac{5}{10}\); the distance is \(\frac{6}{10}-\frac{5}{10}=\frac{1}{10}\)

A finite observed proportion can differ from the model probability without proving that the model is wrong

More trials tend to reduce typical proportional variation across repeated runs, but one longer run need not be closer and a cumulative estimate can move away temporarily

Which comparison is justified by these two exact proportional distances?

Run 1 has distance 1/10 from 1/2

Run 2 has distance 1/10 from 1/2

The two observed proportions are equally far from the model probability

This comparison describes the two supplied runs. It does not prove that every larger run is closer or every future run has the same result

Retain the exact tie instead of choosing by sample size

The bag has three red and two blue tokens, so the probability of red is three fifths. One trial is one draw, followed by replacement and remixing. Choose several trial totals up to two thousand, then compare several run numbers. Increasing the total at the same run number continues its existing prefix

Which statement correctly describes what these controls and records show?

A fair-coin simulation records six Tails in a row. It will continue with the same fair-coin model. What is the probability of Heads on its next toss?

Report the model, procedure and observed evidence

State the experiment, success definition, mapping or device, reset rule and trial totals

Report observed counts and proportions alongside the model probability and expected counts

Describe differences without claiming that a finite match proves fairness or a finite mismatch proves bias

For simulated runs, keep the settings and run number so the displayed result can be reproduced

When pooling compatible records, add success counts and trial counts before dividing. Do not average proportions from different run sizes without weighting

Two runs used the same unchanged model and success rule. One recorded eight successes in ten trials; another recorded forty in one hundred. What is their combined observed relative frequency?

Conduct, record and compare your own experiments

Practical and written investigation task

Use a fair coin or a bag containing three identical red and two identical blue counters. Define one-stage success and keep the trial conditions fixed

Conduct twenty physical trials, recording every outcome. For the bag, replace and remix after each draw. Extend the same record to one hundred trials and retain its twenty-trial prefix

Calculate each cumulative observed proportion and compare it with the appropriate model probability

Use the simulator to compare several independent run numbers at twenty, two hundred and two thousand trials. Record the settings and results without choosing only runs that fit a preferred conclusion

Create an exact uniform-integer mapping for the same model and explain why it preserves the probabilities. If using another digital tool, verify its integer bounds, mapping and reset rule before collecting results

Report what varies, what stays fixed and why a larger run is not guaranteed to be closer in each case. Distinguish cumulative prefixes from independent runs

Actual physical collection, independent software actions, unrestricted mapping design and the written report are not automatically checked. The supplied numerical records and selected decisions are checked

Connect the chance model with repeated evidence

Define one complete trial and preserve its intended probabilities in the random device and reset rule

Record every result, including failures, and distinguish no observations from zero successes

Observed relative frequency is successes divided by trials; expected count is model probability times trials

Use exact non-negative proportional distances when comparing estimates

Larger runs tend to reduce typical proportional variation across repeated runs, but no individual finite record is guaranteed to match the model or improve monotonically

Keep provenance, settings and limitations with every comparison