Teaching objectives
In this lab the student does not read about probability: they play it. They toss real coins and dice (with physics and all), bet before each experiment and then check whether their decision was a good one. Each student starts with 10 Ᵽ; a won bet adds 5 Ᵽ and a lost bet takes away 3 Ᵽ. The bet is settled with a long series (normally 100 rounds), so a good decision pays off almost every time; in one room, the Grand Duke’s, the series is deliberately short and misleading, and the lab says so at the end. The Ᵽ points make choosing with care matter: betting by gut feeling and betting after counting have different consequences, and the student sees it in their purse.
There are no explanation screens: the theory comes as challenges. In front of the student there is a board (the 6×6 square, the 6×6×6 cube, a grid of paths, a tree) and they are asked to mark the winning cells and count them. Many rooms set 📐 calculation against 💻 simulation: first you bet or calculate, then you simulate ×10, ×100 and ×1000 with the calculated value as a dashed line over the histogram, and it ends with “do they agree? How many repetitions are needed?”.
What you'll learn
The lab takes the student from tossing a coin to binomial coefficients and the formula for P(X = k out of n), by way of counting properly before dividing.
- Describing an experiment with all its possible outcomes and telling apart outcomes that look the same (D’Alembert) but are not (rooms 1-4).
- Computing two-dice probabilities with the 6×6 square: marking and counting cells (rooms 5-8).
- Extending the idea to three dice (the 6×6×6 cube) and seeing why counting ordered outcomes changes the answer (rooms 9-10).
- Counting paths on a grid, arriving at Pascal’s triangle and reading the binomial coefficients C(n,k) from it (rooms 11-14).
- Computing the probability of exactly k successes in n attempts, with coins, with dice and with a loaded coin (rooms 15-17).
- Comparing with simulations and seeing when and why they approach the calculation (rooms 1-20).
- Analysing turn-taking games and races between coin patterns, where intuition fails (rooms 18-20).
Key mathematical ideas
- Probability as favourable cases over possible cases, when all cases are equally likely. The key is listing the cases correctly: that is why the 6×6 square has 36 cells and not 21 “different sums”.
- The 6×6 square as a map of the outcomes of two dice: P(sum 7) = 6/36 = 1/6, P(sum ≥ 10) = 6/36, P(sum < 4) = 3/36; an even sum has probability 1/2 and an even product 27/36 = 3/4.
- Counting paths: the grid in which each crossing adds up the paths
arriving from above is Pascal’s triangle; row 6 (1, 6, 15, 20, 15, 6,
- is how the balls spread out in the Galton board.
- Combinations C(n,k): choosing in which k of the n tosses the event happens. That is why there are as many 5-toss sequences with 2 heads as with 3, and why C(10,3) = 120.
- P(X = k out of n) = C(n,k)·p^k·(1−p)^(n−k): with p = 1/2 it is C(n,k)/2ⁿ; 5 heads out of 10 has probability 252/1024 ≈ 0.246 and 50 out of 100 only ≈ 0.080.
- The law of large numbers, intuitively: with few repetitions the simulated result strays from the calculated one; with hundreds or thousands, the histogram settles on the theoretical line.
- Intuition misleads: D’Alembert’s mistake (two heads is 1/4, not 1/3), the Chevalier de Méré’s bet (1 − (5/6)⁴ ≈ 0.518 versus ≈ 0.491), the Grand Duke’s problem (9 comes up in 25 ways, 10 in 27) and Penney’s game (the second player to choose a three-toss pattern has the edge and can win 7:1).
Level: lower secondary and early upper secondary (ESO, ages 12-16). It is the companion lab of “Expectation”: this one asks with what probability, that one how much you win. In class it works well to project the 6×6 square and the Arthur, Bertha and John room, asking the class to bet before counting; the Galton board and Penney’s game also work well and invite discussion.
Room-by-room contents
Room 1 · The spinning coin
A coin spins on its edge, slows down, wobbles and lands (X = tails, O = heads). Bet: in 10 tosses, which is more likely, a 5-5 tie or more heads than tails? It is settled with 100 batches. Then it is calculated: a tie 63/256 ≈ 0.25 versus more heads 193/512 ≈ 0.38.
Student tasks
- List the 8 sequences of 3 tosses.
- Probability of 3 heads in a row (1/8).
- Compare the tie and “more heads” with the simulation and with the calculation.
Room 2 · D’Alembert’s mistake
Two coins, one gold and one silver so they can be told apart. Bet: “one of each” or “two heads”? Marking the 4 outcomes shows that one of each has 1/2 and two heads 1/4. D’Alembert thought there were only three cases (two heads, two tails, one of each) and that each was worth 1/3.
Student tasks
- Mark the 4 outcomes of two distinguishable coins.
- Probability of “one of each” (1/2) and of “two heads” (1/4).
- Explain why there are not 3 equally likely cases.
Room 3 · The bouncing die
A 3D die is thrown, bounces and settles. Bet: in 6 rolls, will all six numbers appear or will one repeat? Usually one repeats. Probability of no 6 at all in 6 rolls: (5/6)⁶ ≈ 0.335; of all six different: 720/46 656 ≈ 0.015.
Student tasks
- Probability of no 6 in 6 rolls (≈ 0.335).
- How many sequences of 6 rolls there are (46 656) and how many with all different (720).
- Compare with a simulation of 1000 series.
Room 4 · The Chevalier de Méré’s bet
Two gaming tables in parallel: at least one six in 4 rolls of one die, or at least one double six in 24 rolls of two dice (series of 2000). It is settled with the opposite event, “no six”: 1 − (5/6)⁴ = 671/1296 ≈ 0.518 versus 1 − (35/36)²⁴ ≈ 0.491. A small difference that Méré noticed by playing.
Student tasks
- Calculate the probability of “no six” in 4 rolls.
- Calculate P(at least one six) = 1 − (5/6)⁴ ≈ 0.518.
- Compare with a double six in 24 rolls (≈ 0.491) and with the simulation.
Room 5 · Which sum comes up most?
Two physical dice and a live histogram. You bet on a sum from 2 to 12. After 10, 100 and 1000 pairs you see the histogram settle: the most frequent sum is 7 and the least frequent are 2 and 12. With 10 pairs anything seems possible; with 1000 the roof shape appears.
Student tasks
- Choose a sum before rolling.
- See which sum comes up most after 10, 100 and 1000 pairs (the 7).
- Say which come up least (2 and 12).
Room 6 · The 6×6 square
A room of challenges, with no bet: the board of 36 cells, one for each pair of dice, lights up as the winning cells are marked. You count those with sum 7 (6), sum ≥ 10 (6) and sum < 4 (3). So P(7) = 6/36 = 1/6.
Student tasks
- Mark and count the cells with sum 7 (6).
- Mark and count those with sum ≥ 10 (6).
- Mark and count those with sum < 4 (3).
- Give the probability of rolling a 7 (1/6).
Room 7 · Arthur, Bertha and John
Three characters with their scoreboard: Arthur wins if the sum is ≥ 10, Bertha if it is 6 or 7, and John if it is < 4. You bet on who will win the most rounds. On the square: 6, 11 and 3 cells, that is 6/36, 11/36 and 3/36. Bertha beats Arthur in 97.6 % of 100-round series.
Student tasks
- Mark each character’s cells and count them (6, 11 and 3).
- Give the three probabilities (6/36, 11/36, 3/36).
- Say who wins most often, and why in a single roll anyone can win.
Room 8 · Even or odd?
A board with two colours. Bet: is it more likely that the sum is even or that the product is even? An even sum has probability 1/2; the product is odd in only 9 cells, so it is even in 27/36 = 3/4.
Student tasks
- Mark the cells with an even sum (18).
- Mark the cells with an odd product (9).
- Give P(even product) = 27/36 = 3/4.
Room 9 · Three dice
A 6×6×6 cube of little cubes that light up layer by layer: 216 possible outcomes. You bet on a sum from 3 to 18. The most likely are 10 and 11 (27/216 each) and P(sum ≤ 5) = 10/216.
Student tasks
- How many outcomes there are with three dice (216).
- Which sum is most likely and with what probability (10 and 11, 27/216).
- Calculate P(sum ≤ 5) (10/216).
Room 10 · The Grand Duke’s problem
Scoreboard 9 versus 10: which sum comes up more with three dice? This room deliberately uses a series of 100 rolls, which often misleads, and then 100 000 (with 20 000, 10 still lost about 1 time in 285). The trick: {3,3,4} comes up in 3 ordered ways and {2,3,4} in 6; in total 9 comes up in 25 ways and 10 in 27.
Student tasks
- Bet on 9 or on 10 and play the series of 100.
- Repeat with 100 000 rolls and compare.
- Count the ordered ways of {3,3,4} (3) and of {2,3,4} (6) and reach 25 versus 27.
Room 11 · The coin ritual
Each person tosses 4 coins: tails, one step forward and to the right; heads, forward and to the left. The little figures walk and pile up under their column. You bet on the column with the most people. P(ending in column 0) = 6/16 and the spread is 1-4-6-4-1.
Student tasks
- Bet on the column with the most people.
- Find P(ending in the middle column) = 6/16.
- Write the spread over the 5 columns (1, 4, 6, 4, 1).
Room 12 · How many paths?
A room of challenges: at each crossing of a grid you write how many paths arrive, adding the two above. The grid turns into Pascal’s triangle. Row 6 is 1, 6, 15, 20, 15, 6, 1.
Student tasks
- Fill in the number of paths at each crossing.
- Write row 6 (1, 6, 15, 20, 15, 6, 1).
- Say which rule takes you from one row to the next.
Room 13 · The Galton board
Balls bounce off the pegs with real physics. You bet on the slot where most balls will land. With 6 rows of pegs, P(middle slot) = 20/64, and the spread of 1000 balls draws the bell that was counted earlier with paths; at the end the curve is drawn and the Gaussian bell is named.
Student tasks
- Bet on a slot.
- Calculate P(middle slot) = 20/64.
- Compare the histogram of balls with row 6 of the triangle.
Room 14 · The Galton board and the Gaussian bell curve
The Galton board with 10 rows of pegs and a control p, the probability of going right at each peg. Bet: if p goes up from 0.5 to 0.8, how does the shape of the bell change? 400 balls are dropped with each p and the bet is settled with the calculated curve: narrower and taller (the fullest bin goes from 25% to 30%). Then the control is free: the peak is at bin 10·p, p = 0.5 gives the widest bell, p = 0.2 and p = 0.8 are mirror images and with p = 1 every ball lands in bin 10.
Student tasks
- Bet on the shape of the bell with p = 0.8.
- Find the peak with p = 0.3 (bin 3).
- Say which p gives the widest bell (0.5) and compare p = 0.2 with p = 0.8.
- Explain what happens with p = 1.
Room 15 · Choosing which
Bet: are there more sequences of 5 tosses with 2 heads or with 3? They are equal. You build the 10 sequences with 2 heads by flipping tokens and read C(10,3) = 120 in Pascal’s triangle: choosing in which tosses heads comes up.
Student tasks
- Build the 10 sequences of 5 tosses with 2 heads.
- Explain why there are as many with 2 as with 3.
- Read C(10,3) = 120 in the triangle.
Room 16 · Exactly k out of n
Two overlaid histograms that flatten out. Bet: which is more likely, 5 heads out of 10 tosses or 50 heads out of 100? With P = C(n,k)/2ⁿ: 252/1024 ≈ 0.246 versus ≈ 0.080. Landing exactly on half gets rarer, even though half is still the most likely value.
Student tasks
- Calculate P(5 heads out of 10) = 252/1024 ≈ 0.246.
- Compare with P(50 out of 100) ≈ 0.080.
- Explain why it drops even though 50 is the most likely value.
Room 17 · Sixes in n rolls
Bet: in 6 rolls, no six or exactly one? First a tree of 2 rolls whose branch thickness is proportional to probability: a path is worth the product of its branches, and exactly one six means 2 paths of (1/6)·(5/6), 10/36. With 6 rolls there is no need to draw it: the single six can be in 6 places and each path is worth (1/6)·(5/6)⁵, so 6·(1/6)·(5/6)⁵ ≈ 0.402, more than no six (≈ 0.335).
Student tasks
- Use the tree to find P(exactly one six in 2 rolls) = 10/36.
- Count the 6 possible places of the single six in 6 rolls.
- Write what one path is worth: (1/6)·(5/6)⁵.
- Put it together: 6·(1/6)·(5/6)⁵ ≈ 0.402.
Room 18 · The loaded coin
A bulging coin that spins with a limp: it goes right with probability p = 0.4. Bet: in 2, 3 or 4 steps, when is it easiest to be at position 2? With 2 steps 0.16; with 3 steps 0, by parity; with 4 steps 96/625 = 0.1536. The general formula C(n,k)·p^k·(1−p)^(n−k) appears.
Student tasks
- Calculate P(being at 2 after 2 steps) = 0.16.
- Explain why after 3 steps it is impossible (parity).
- Calculate P(being at 2 after 4 steps) = 96/625.
Room 19 · Who gets tails first?
Two players alternate tosses and the first to get tails wins. Bet: the player who tosses on the odd turns or the one on the even turns? If A does not win on the first toss, the same game starts with B first: B wins with the same probability as A, but only in half of the games, P(B wins) = ½·P(A wins). Since P(A wins) + ½·P(A wins) = 1, A wins with 2/3. With a die and a six, P + 5/6·P = 1 gives 6/11.
Student tasks
- Bet on the odd-turn or even-turn player.
- See that, if A misses the first toss, the game goes on as a new game started by B.
- Solve P(A wins) + ½·P(A wins) = 1 (2/3).
- Calculate with a die and a six (6/11).
Room 20 · XO or XX?
A strip of coins that moves along. Bet: which pair shows up first? XO before XX happens half of the time (1/2). XO before OO wins 3/4: OO can only win at the start (OO, 1/4) and, as soon as an X comes up, only XO can win. One pair can be better even if it seems equally likely.
Student tasks
- Bet between XO and XX and check it.
- Reason why XO before XX is 1/2.
- Find the probability of XO before OO (3/4).
Room 21 · Penney’s game
A race between two strips of coins. You pick a pattern of 3; the computer then picks another. The computer wins 7 to 1 against XXX/OOO, 3 to 1 against XXO/OOX and 2 to 1 against the rest. The trick: if you pick XOX, it answers XXO (the opposite of your 2nd + your first two).
Student tasks
- Pick a pattern of 3 and play against the computer.
- Find out how the computer chooses its answer.
- Say what it picks if the student picks XOX (XXO).
Rooms to project
The most striking ones to show and discuss in class.