Probability Simulator
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Investigation questions

64 questions. Work through these with the simulator above. Set it up loads the right experiment for you, and Show on board puts the question on screen, full size, with the tool ready to go.

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First experiments
5 questions

  1. Press ▶ Run 1 ten times (or press Space). Before each flip, everyone predicts Heads or Tails. How many predictions were right? Could anyone get good at predicting a fair coin?

    Teacher notes

    Nobody can get good at it. Each flip is independent, so over many predictions everyone scores about half. Ten predictions are too few to separate luck from skill: someone gets 8 or more out of 10 by pure guessing about 1 time in 18.

  2. Open the Table tab. The Frequency column shows your counts. Click the Relative frequency heading to reveal it. What do the numbers mean, and how could you work them out from the frequencies? Switch to ¼ to see them as fractions.

    Teacher notes

    Relative frequency = frequency ÷ number of trials, so the column always adds up to 1. The ¼ view shows it as frequency over total, unsimplified, which makes the link explicit.

  3. Press ×1000 and look at the relative frequencies again. Are they closer to 0.5 than after ten flips? Press ×1000 twice more. What is happening to the relative frequencies as the number of trials grows?

    Teacher notes

    The relative frequencies drift towards 0.5, but not smoothly. After 10 flips, 0.3 or 0.7 is common. After 3000, they are almost always within 0.02 of 0.5.

  4. Open Convergence and press 👁 Theoretical. Now press ⏩ Auto and watch. Describe what happens to the two lines. Why do they jump about so much at the start and hardly at all later?

    Teacher notes

    Early on, one flip moves the relative frequency a lot (one flip in five is 0.2). After 1000 flips, one flip moves it by only 0.001. This is the law of large numbers. The two lines mirror each other because they always add up to 1.

  5. On the Bar chart the two bars are almost never exactly the same height. A student says: ‘This coin must be unfair.’ Do you agree? How different would the bars need to be before you became suspicious?

    Teacher notes

    Disagree. Random variation makes exactly equal bars unlikely: exactly 500 Heads in 1000 flips happens only about 2.5% of the time. For a fair coin, Heads usually lands within about 30 of 500 in 1000 flips, so that is a sensible ‘suspicious’ threshold to discuss.

Coins
7 questions

  1. Set 2 coins and Count heads. The outcomes are 0, 1 and 2 Heads. Before you run anything, predict: are they equally likely? Run 300 trials. Were you right?

    Teacher notes

    Not equally likely: P(0 Heads) = ¼, P(1 Head) = ½, P(2 Heads) = ¼. The classic misconception is ‘three outcomes, so ⅓ each’. Collect predictions first so the surprise lands.

  2. Now choose Show order. The outcomes become HH, HT, TH and TT. Run 400 trials. Are these four equally likely? Use them to explain why 1 Head happens about twice as often as 2 Heads.

    Teacher notes

    HH, HT, TH and TT are each ¼. ‘1 Head’ is made of two of them (HT and TH), so it is 2/4 = ½. If students think HT and TH are the same, imagine one coin is a 1p and the other a 2p.

  3. Open the Tree diagram for 2 coins and run 200 trials. The boxes show how often each branch happened. Press 👁 Probabilities. Why do you multiply along the branches? Check that the four end probabilities add up to 1.

    Teacher notes

    Half the trials go each way at the first branch, then half of those split again, so ½ × ½ = ¼. Multiplying ‘takes a fraction of a fraction’. The four ends add to 1 because every trial ends up somewhere.

  4. Change to 3 coins. Predict which number of heads is most common. Run 1000 trials and check. Then use the tree diagram: how many of the 8 paths give exactly 2 Heads? Why is 0 Heads less likely than 1 Head?

    Teacher notes

    1 Head and 2 Heads tie as most common (3/8 each). Three paths give exactly 2 Heads: HHT, HTH and THH. 0 Heads has only one path (TTT, 1/8), while 1 Head has three.

  5. Set 5 coins, run 2000 trials and look at the Bar chart. Describe the shape. Where is the peak? Is it symmetrical? Press 👁 Theoretical. How closely do your bars match?

    Teacher notes

    Symmetrical, with peaks at 2 and 3 Heads (10/32 each) and only 1/32 for 0 or 5 Heads. The numbers of paths, 1 5 10 10 5 1, are a row of Pascal’s triangle: a lovely link for able groups.

  6. With 1 coin, set P(Heads) to 0.80 and run 500 trials. Take a 📌 Snapshot. Press Fair and run another 500. The dashed purple bars are the biased coin. What differences can you see?

    Teacher notes

    Biased: about 400 Heads and 100 Tails. Fair: about 250 of each. The snapshot bars are rescaled to the current number of trials, so the two charts compare fairly even if you ran different amounts.

  7. Set P(Heads) to 0.70 and run only 10 trials. Could you tell this coin is biased from 10 flips? Clear the results and run 1000. Could you tell now? What does this tell you about sample size and spotting bias?

    Teacher notes

    From 10 flips you can’t tell: a fair coin gives 7 or more Heads about 17% of the time. With 1000 flips you’d expect about 700 Heads, far outside the fair range of roughly 470 to 530. Bigger samples detect smaller biases.

Dice
6 questions

  1. Roll a fair d6 60 times. In the Table, reveal Expected frequency. Every row says 10.0. Why? Which face was furthest from 10? Does that mean the die is unfair?

    Teacher notes

    Expected = 60 × ⅙ = 10. A spread from about 5 to 15 is perfectly normal for a fair die over 60 rolls, so a face on 15 is not evidence of unfairness on its own.

  2. Run 600 trials and work out how far each frequency is from its expected value. Now run up to 6000 trials and do the same. Are the differences bigger or smaller? Are they a bigger or smaller proportion of the total?

    Teacher notes

    The differences get bigger in absolute terms (typically about 9 at 600 rolls, about 29 at 6000) but smaller as a proportion of the total. This absolute versus relative distinction is the heart of the law of large numbers.

  3. Try a d20 and run 1000 trials. What should each relative frequency be? Open Convergence and press 🔍 Zoom in. Why does this graph look so much messier than the coin’s?

    Teacher notes

    Each face should be 1/20 = 0.05. With 1000 rolls each face gets about 50, give or take 14, so relative frequencies between 0.036 and 0.064 are normal. Rarer outcomes have proportionally bigger wobbles, and 20 lines are crowded. Zoom helps.

  4. A student says: ‘A d4 is more predictable than a d20 because it has fewer outcomes.’ Run 200 trials of each and compare the Convergence graphs. What does ‘predictable’ mean here? Is the student right?

    Teacher notes

    It depends what ‘predictable’ means. Guessing the next roll: the d4 wins (¼ v 1/20). How close the relative frequencies are to their targets: the d20’s lines are closer in absolute terms, but much further away relative to their small targets. A rich discussion about defining terms.

  5. Choose Loaded and give face 6 a weight of 2. Before revealing anything, predict P(6). Then reveal the Theoretical column. Why isn’t the answer 2/6?

    Teacher notes

    The total weight is now 7, so P(6) = 2/7 ≈ 0.286. Students who say 2/6 increase the numerator but forget the denominator grows too.

  6. Load the die so that P(6) is exactly ½. What weights work? Is there more than one answer? Can you make P(6) = ⅓ as well?

    Teacher notes

    Give face 6 a weight equal to the total of the others, e.g. 6 → 5 with the rest on 1 (5/10). Doubling every weight gives the same probabilities, so there are many answers. For ⅓: others on 2 (total 10) and face 6 on 5, giving 5/15.

Spinners
5 questions

  1. Use a spinner with 4 equal sectors. Predict the theoretical probability of each sector, then run 400 trials and reveal the Theoretical column. How close are your relative frequencies?

    Teacher notes

    Each sector is ¼, so about 100 of 400 spins each. Ask how far from 100 would surprise them, then reveal.

  2. Use + to set the weights to 1 : 1 : 1 : 3. Before running, predict what fraction of spins land on Sector 4. Write it down, run 600 trials and check.

    Teacher notes

    Sector 4 should be 3/6 = ½. Watch for ‘¼, because it’s one of four sectors’. That is the equally-likely assumption again, when the weights make the sectors different sizes.

  3. Use a 2-sector spinner. Try the weights 1 : 1, 1 : 2, 1 : 3, 1 : 5 and 1 : 9, running 500 trials each time. Record P(Sector 1) each time. What ratio would make Sector 1 almost impossible?

    Teacher notes

    P(Sector 1) = ½, ⅓, ¼, ⅙ and 1/10. It never becomes impossible while its weight is positive, only smaller. On this spinner the smallest is 1/13 (1 : 12). Only a weight of 0 makes it impossible.

  4. Can you set up a sector with probability exactly ⅓? Exactly ¼? Exactly ⅗? Try different numbers of sectors and weights. Is every fraction possible? Which fractions can’t you make with this spinner?

    Teacher notes

    ⅓: three equal sectors, or 1 : 2. ¼: 1 : 3. ⅗: 3 : 2. On this spinner each weight is at most 12 and the total at most 96, so fractions like 1/100 or 13/14 can’t be made. On a real spinner, any fraction is possible.

  5. Set one sector’s weight to 0. What happens to the spinner picture? What is its theoretical probability? Now can you make a spinner where one event is certain?

    Teacher notes

    The sector disappears from the picture and its probability is 0: impossible. Set all the other sectors to 0 and the remaining sector has probability 1: certain. A good moment to sketch the 0 to 1 probability scale.

Two dice and sample spaces
6 questions

  1. Two d6, Add. Open the Sample space tab before rolling. It is blank. Press ▶ Run 1 twenty times and watch it fill. How many of the 36 cells have you found? Which cells do you think will be hardest to find?

    Teacher notes

    After 20 rolls you’ll usually have found about 15 or 16 cells. The twist: every cell is equally likely (1/36), so no single cell is harder to find. It’s the totals 2 and 12 that are rare, because each has only one cell.

  2. Run 200 trials, then press 👁 Reveal all. Click 7 under the grid to highlight its cells. Why do the 7s make a diagonal line? How many cells show 7? How many show 2, and 12?

    Teacher notes

    Moving one step along the diagonal adds 1 to one die and takes 1 off the other, so the total stays the same. 7 has 6 cells, 2 and 12 have 1 each. That’s why 7 is six times as likely as 2.

  3. With the grid revealed, the key shows things like ‘7: 6/36’. Simplify that fraction. Do the same for totals of 2, 6 and 11. What pattern links the total to the number of ways of making it?

    Teacher notes

    6/36 = ⅙. Totals of 2 = 1/36, 6 = 5/36, 11 = 2/36 = 1/18. Number of ways = 6 − (distance from 7).

  4. Switch to Difference (larger − smaller). Run some trials, then reveal the grid. Where are the 0s, and why? Which difference is most likely? Which is least likely?

    Teacher notes

    The 0s run down the leading diagonal: doubles. Most likely difference is 1 (10/36), least likely 5 (2/36). 0 is 6/36. Many students expect 0 to be most likely.

  5. Run 1000 trials with Add and take a 📌 Snapshot. Switch to Difference and run 1000 more. Compare the shapes on the Bar chart. Describe the differences. Which distribution is more spread out?

    Teacher notes

    Add: a symmetrical triangle from 2 to 12, peaking at 7. Difference: from 0 to 5, lopsided, peaking at 1. Add is more spread out (range 10 v 5).

  6. Try two d4: a 4 × 4 grid. What is the most common total? Now two d8 and two d12. Can you write a rule for the most common total when two dN dice are added?

    Teacher notes

    Two d4: 5. Two d8: 9. Two d12: 13. Rule: for two dN, the most common total is N + 1. The longest diagonal of the grid explains why.

Bags of counters
5 questions

  1. The bag has 3 Red, 3 Blue and 3 Green. Run 90 trials and reveal Expected frequency: each colour says 30.0. Why? Add 3 more Red counters and run 90 trials. How have the expected frequencies changed?

    Teacher notes

    90 × 3/9 = 30. With 6 Red, 3 Blue and 3 Green: Red 90 × 6/12 = 45, Blue and Green 22.5 each. Expected frequencies don’t have to be whole numbers, which surprises some students.

  2. Set up 5 Red and 1 Blue. Predict what fraction of draws will be Red, then run 600 trials. Now add Blue counters one at a time, running 600 trials each time. When do Red and Blue become equally likely?

    Teacher notes

    P(Red) = 5/6, then 5/7, 5/8 and so on. Red and Blue are equally likely when there are 5 Blue (5/10 each).

  3. A bag has 3 Red and 1 Blue. A student says: ‘There are only two colours, so P(Red) = ½.’ Run 400 trials. What is wrong with the student’s reasoning? What is P(Red) really?

    Teacher notes

    P(Red) = 3/4. ‘Two outcomes, so ½’ is the equally-likely misconception (sometimes called equiprobability bias), one of the most common in probability. Ask what would happen with 99 Red and 1 Blue.

  4. Choose Keep it out. Press ▶ Run 1 and watch the counters disappear from the bag. What happens when the bag is empty? Look at the notes in the Log.

    Teacher notes

    When the bag is empty it is refilled with the original counters (the Log says ‘bag refilled’). A nice question: when one counter is left, what is the probability of its colour? (1: certain.)

  5. Set up 1 Red and 9 Blue with Keep it out. Run 100 trials. Why do you get exactly 10 Reds every time? Clear the results and choose Put it back. Run 100 trials again. Do you still get exactly 10?

    Teacher notes

    Each bagful of 10 holds exactly one Red, and 100 trials are exactly 10 bagfuls, so it’s 10 every time. With replacement, the Red count varies, usually somewhere between about 4 and 16. Without replacement, the long run is more ‘even’ than chance.

Two counters and tree diagrams
6 questions

  1. A bag has 3 Red and 2 Blue. Take 2 counters and Put it back in between. Run 500 trials and open the Tree diagram. Compare the second branches after Red with the second branches after Blue. Why are they split in about the same ratio?

    Teacher notes

    With replacement, the bag is identical for the second counter, so after a Red or after a Blue the split is about 3 : 2 both times. The second counter doesn’t depend on the first: the events are independent.

  2. Now switch to Keep it out and press 👁 Probabilities. Why does the Red branch after a Red say 2/4 and not 3/5? Why do all the second branches have 4 on the bottom?

    Teacher notes

    After a Red is taken out, 4 counters remain and 2 of them are Red. Every second-counter branch is out of 4 because one counter is missing, whichever it was. The events are now dependent.

  3. Still with Keep it out, use the tree to work out P(both Red). Then run 1000 trials and compare with the relative frequency of ‘Red, Red’ in the Table. How close were you?

    Teacher notes

    P(both Red) = 3/5 × 2/4 = 6/20 = 3/10 = 0.3. With 1000 trials, expect about 300.

  4. What is P(one of each colour)? Which two ends of the tree do you need? Why do you multiply along the branches but add the two ends together?

    Teacher notes

    Red then Blue (6/20) plus Blue then Red (6/20) = 12/20 = ⅗. Multiply along a path (‘and then’), add different paths (‘or’). The common error is to count only Red then Blue.

  5. A bag has 2 Red and 1 Blue. Take 2 counters with Keep it out. One end of the tree never gets any trials. Which one, and why? What is its probability?

    Teacher notes

    Blue, Blue: there is only one Blue, so once it’s taken the second counter can’t be Blue. P = ⅓ × 0/2 = 0. The tree shows an impossible event naturally.

  6. A student says replacement makes no difference to P(Red, Red). With 4 Red and 4 Blue, run 1000 trials with Put it back and take a 📌 Snapshot. Switch to Keep it out and compare on the Bar chart. Then try 1 Red and 1 Blue. When does replacement make the biggest difference?

    Teacher notes

    With 4 Red and 4 Blue: P(Red, Red) is 16/64 = ¼ with replacement and 4/8 × 3/7 = 3/14 ≈ 0.21 without, a small difference. With 1 Red and 1 Blue: ¼ v 0. Replacement matters most when the bag is small.

Mystery experiments
5 questions

  1. Press Set it up for a mystery coin. Run 20 trials. Is it fair? Now run 500 more. Estimate P(Heads) to 2 decimal places, then press 👁 Reveal. How close were you?

    Teacher notes

    20 flips can’t settle it: even a fair coin’s relative frequency often lands anywhere from about 0.3 to 0.7. After 500 more, estimates are usually within about 0.04 of the truth. Ask: what evidence would convince you it’s fair? (You can never be certain.)

  2. A mystery bag: you know how many counters there are, but not their colours. Run 50 trials and estimate how many of each colour are inside. Run 500 more and improve your estimate. Reveal. Which estimate was better, and why?

    Teacher notes

    Estimate = relative frequency × number of counters, then round. Remind them that counts must be whole numbers and must add up to the total, a useful check. The 500-trial estimate is usually better because relative frequencies settle with more trials.

  3. Is this mystery die fair? Before you start, agree a rule, for example ‘run 600 trials; if any face is more than 20 away from 100, call it loaded’. Apply your rule, then reveal. Did it work? Try three more mysteries with 🔀 New mystery.

    Teacher notes

    Agreeing a rule first is a significance test in disguise. With this rule a fair die gets wrongly called ‘loaded’ about 1 time in 7. A face with weight 2 is about 171 in 600 and easy to spot. Discuss false alarms versus misses.

  4. A mystery spinner. Run enough trials to estimate what fraction of the spinner each sector takes up, then sketch it. Reveal and compare. Which sectors were hardest to estimate?

    Teacher notes

    Each sector’s fraction ≈ its relative frequency. For the sketch, angle = 360° × relative frequency, a neat link to pie charts. Small sectors are hardest to estimate in relative terms.

  5. How many trials are enough? Start a new mystery bag. After every 50 trials, write down your estimate of the contents. When do your estimates stop changing? Open Convergence to see why.

    Teacher notes

    Estimates jump about at first and then settle. To halve the typical error you need about four times as many trials, which is why going from 50 to 200 helps far more than going from 500 to 650.

The law of large numbers
4 questions

  1. Choose any experiment. Run 5 trials and note the relative frequency of one outcome. Then 50, 500 and 5000. Describe what happens on the Convergence graph. Write a sentence explaining the law of large numbers in your own words.

    Teacher notes

    As the number of trials grows, the relative frequency tends to get closer to the theoretical probability. Push back on ‘the results even out’: the frequencies don’t have to balance, only the proportions settle.

  2. Roll a fair d6 exactly 6 times. Did you get one of each number? Clear and try again, ten times. How often do you get a ‘perfect’ set? Now run 6000 trials. Are any relative frequencies exactly ⅙? What does this tell you?

    Teacher notes

    One of each in six rolls happens only 720/46656 ≈ 1.5% of the time, so almost never. After 6000 rolls a relative frequency of exactly ⅙ needs exactly 1000 of that face, which happens only about 1.4% of the time. Theory describes the long run, not any exact result.

  3. Fair coin, Convergence, 🔍 Zoom in, then ⏩ Auto. Roughly how many trials until the Heads line stays between 0.45 and 0.55? Clear and repeat. Is it the same every time?

    Teacher notes

    Usually it’s a few hundred trials, but it varies a lot from run to run, and that variation is the point. Repeat it with the class and compare everyone’s answers.

  4. After 100 flips, the gap between Heads and Tails might be about 6. After 10,000 flips it might be about 60. Is the coin becoming less fair? Run both and compare the gap in frequency with the gap in relative frequency.

    Teacher notes

    The gap in frequency tends to grow (roughly with the square root of the number of flips), but the gap in relative frequency shrinks: 6/100 = 0.06 versus 60/10,000 = 0.006. The coin isn’t ‘catching up’. The early imbalance is just swamped.

Comparing with snapshots
3 questions

  1. Run 500 trials on a fair coin and take a 📌 Snapshot. Set P(Heads) to 0.60 and run 500. Can you see the difference on the Bar chart? Now try 0.55. What is the smallest bias you can spot with 500 trials?

    Teacher notes

    With 500 trials, a fair coin’s relative frequency varies by about ±0.045. So 0.60 is clearly visible, while 0.55 is borderline. Spotting a bias half the size needs about four times as many trials.

  2. Run 1000 trials on a d6 and take a 📌 Snapshot. Switch to a d8 and run 1000. The Table now has a Snapshot column. For faces 1 to 6, why are the d8’s relative frequencies smaller?

    Teacher notes

    Each d8 face is 1/8 = 0.125, against ⅙ ≈ 0.167 for a d6. The total probability is still 1, but it is shared between more outcomes.

  3. Run 500 trials on a 4-sector equal spinner and take a Snapshot. Change the weights to 1 : 1 : 1 : 3 and run 500. On Convergence, show the theoretical lines too. You now have three layers on one graph. What does each layer tell you?

    Teacher notes

    Solid lines: what this spinner actually did. Coloured dashed lines: the theory for 1 : 1 : 1 : 3 (⅙, ⅙, ⅙, ½). Purple dashed lines: the snapshot of the equal spinner, settling at ¼. Theory, this experiment and the earlier experiment, all on one graph.

Predictions and misconceptions
6 questions

  1. Leave the hidden columns hidden. Run 100 trials on a d6 and look only at the frequencies. Write down your predictions for every hidden cell, then reveal one column at a time. How accurate were you?

    Teacher notes

    Relative frequency = frequency ÷ 100. Expected = 100 × ⅙ ≈ 16.7 for every face. Theoretical = ⅙ ≈ 0.167. Students who get the first column right but expect the frequencies to match the expected column have found the key idea.

  2. One student says: ‘If I flip a fair coin 100 times I will get exactly 50 Heads.’ Another says ‘about 50’. Run 100 flips, clear, and repeat ten times. Record the number of Heads each time. What were the smallest and largest? Who is right?

    Teacher notes

    Exactly 50 Heads happens only about 8% of the time. Over ten runs, results anywhere from about 40 to 60 are normal. ‘About 50’ is right, and the activity makes ‘about’ precise.

  3. Two d6 added. ‘I need a 6 to win. That’s easy, 6 is in the middle.’ ‘No, hope for a 7, it’s the most common.’ Use the Sample space to count the ways of making 6 and 7. Who is right? What is the probability of each?

    Teacher notes

    6 has 5 ways (5/36) and 7 has 6 ways (6/36). The second student is right, but only just: 6 is still a good bet. Also, 6 isn’t ‘the middle’. The middle of 2 to 12 is 7.

  4. A bag has 3 Red and 2 Blue. Write P(Red) as a fraction, a decimal and a percentage, then check with the ¼, 0.25 and 25% buttons in the Table. Run 1000 trials. Multiply your relative frequency by 1000. What do you get, and why?

    Teacher notes

    ⅗ = 0.6 = 60%. The relative frequency × 1000 gives exactly the number of Reds, because relative frequency = Reds ÷ 1000. Expected ≈ 600.

  5. The gambler’s fallacy. Run 1000 flips of a fair coin. The box at the top of the Log shows the longest run. Is it longer than you expected? A student says that after 5 Heads in a row, Tails is ‘due’. Find a long run in the Log. What came next? Does the coin remember?

    Teacher notes

    In 1000 fair flips the longest run is typically about 10, so long runs are normal. After 5 Heads, the next flip is still ½ Heads. The coin has no memory. Download the CSV to collect more evidence across the class.

  6. Is it possible to set up each experiment so that every outcome has the same theoretical probability? Try coins, dice, the spinner, two dice and the bag. Which can and which can’t? Why?

    Teacher notes

    Coins: yes for 1 fair coin, or in Show order. Counting heads with 2 or more coins: no. Dice and spinner: yes with equal weights. Bag: yes with equal counts (and with replacement if taking 2). Two dice: never, because totals and differences have different numbers of ways.

Open investigations
6 questions

  1. Two d6 added: which total takes longest to appear? Run single trials until every total from 2 to 12 has appeared. How many trials did it take? Clear and repeat. Why do some totals take longer?

    Teacher notes

    2 and 12 (1/36 each) are usually the last to appear. On average it takes about 61 rolls to see all eleven totals, and most of that wait is for the rarer totals.

  2. Design a spinner game where Sector 1 wins a prize with a chance of 1 in 10. Run 1000 trials to check. Now change it to exactly 1 in 4. Is there more than one way to set the weights?

    Teacher notes

    1 in 10: Sector 1 on 1 and the rest totalling 9, e.g. 1 : 3 : 3 : 3. 1 in 4: 1 : 3, 2 : 6, 1 : 1 : 1 : 1, and so on. Scaling all the weights by the same amount never changes the probabilities.

  3. Run 1000 rolls of a d6 and press 📊 to download the data. In a spreadsheet, plot one face’s relative frequency against trial number. Describe the graph for the first 10 trials, the first 100, and all 1000.

    Teacher notes

    For the first 10 trials the line jumps around wildly. For the first 100 it is settling. Over 1000 it hugs ⅙. A good chance to practise =COUNTIF or a running total in a spreadsheet.

  4. Put one counter of each of the 8 colours in the bag, with Put it back. Draw single counters until you get a colour you’ve already seen. How many draws did it take? Repeat ten times and find the average. This is the idea behind the birthday problem: in a room of 23 people, two sharing a birthday is more likely than not.

    Teacher notes

    With 8 colours, the first repeat takes about 4.2 draws on average, sooner than most people guess. Use it to introduce the birthday problem: with 365 days you need only 23 people for a better-than-even chance of a shared birthday.

  5. Using any experiments and the 📌 Snapshot, design a demonstration that convinces a sceptical student that (a) results are unpredictable in the short run and (b) they become predictable in the long run. Write instructions so another student can repeat it, then press 🔗 to share your set-up.

    Teacher notes

    Strong answers use a snapshot of a short run beside a long run, or repeat 10-trial runs to show how much they vary. Look for clear instructions someone else could follow without help.

  6. Write your own investigation. Choose a focus (fairness, sample size, convergence, comparing distributions, or something else), decide which experiments to use, and write three questions for a partner. Test them yourself first. Which question led to the most interesting discovery?

    Teacher notes

    Good questions start with a prediction, have a clear focus, and can actually be answered with the tool. Encourage partners to try a question before reading the author’s answer.