E. Culminating Data Management Investigation · Culminating Data Management Investigation

Analysing the Data III: Probability Models and Expected Value

E1.4 E1.3About 75 minutes

Learning goals

I can investigate a game of chance as a culminating probability problem, from rules to data.
I can build a probability distribution for a game and calculate its expected value.
I can recognize when a binomial (or hypergeometric) model applies and use its mean and standard deviation.
I can collect experimental data by playing or simulating a game and compare it with theory.
I can use conditional probability and expected counts from my survey data.

Words to know

probability distribution
A list or table of all possible values of a random variable with their probabilities, which add to 1.

expected value E(X)
The long-run mean of a random variable: E(X) = Σ x · P(x).

fair game
A game in which the expected net gain of a player is 0.

binomial distribution
The distribution of the number of successes in n independent trials, each with the same probability p of success. Mean np, standard deviation √(np(1 − p)).

experimental probability
The relative frequency of an event in actual trials: number of times it happened ÷ number of trials.

simulation
Using random tools (dice, cards, random numbers) to imitate a random process many times.

When the question is about chance

The curriculum's footnote for the culminating investigation allows two smaller problems, one on probability and the other on statistics, and one of its examples is primary data collected from a student-designed game of chance. The analysis expectation names two probability tools: recognizing and applying a probability distribution model and calculating the expected value of a probability distribution.

Amira's statistics problem is about sleep. For her probability problem she designs a dice game for the school's winter carnival, where turns are paid for with carnival tickets (no money changes hands). Her question: Does my game make a profit of tickets for the carnival, how big should that profit be per play, and do 300 real plays match the theory?

The game

A player pays 2 tickets and rolls two standard dice.

Double six: win 20 tickets.
Any other double (1-1, 2-2, 3-3, 4-4, 5-5): win 5 tickets.
Not a double: win nothing.

The tools come from Probability (sample spaces), Discrete Probability Distributions (expected value, binomial distribution) and Continuous Distributions and the Normal Distribution (normal approximation).

The 36 equally likely outcomes of rolling two dice. The diagonal holds the 6 doubles; the corner cell is double six.

Worked example

The model investigation, Stage 15: theory and expected value

Find the probability distribution of the prize X (in tickets), the expected prize, and the carnival's expected profit per play.

Sample space: 6 × 6 = 36 equally likely outcomes.
P(X = 20) = 1/36 (double six). P(X = 5) = 5/36 (the other five doubles). P(X = 0) = 30/36. Check: (1 + 5 + 30)/36 = 1 ✓.
E(X) = 20(1/36) + 5(5/36) + 0(30/36) = 20/36 + 25/36 = 45/36 = 1.25 tickets.
A player pays 2 tickets, so the player's expected net gain is 1.25 − 2 = −0.75 tickets per play, and the carnival's expected profit is +0.75 tickets per play.
Over 300 plays, the expected profit is 300 × 0.75 = 225 tickets (600 collected, 375 expected to be paid out).
The game is not fair: a fair price would be 1.25 tickets per play. The carnival wants a modest profit, so the game is designed this way on purpose, and the rules and odds will be posted for players.

E(X) = 1.25 tickets; expected profit 0.75 tickets per play, or 225 tickets in 300 plays.

Worked example

The model investigation, Stage 16: a binomial model

Let D be the number of double sixes in 300 plays. Explain why D is binomial, and find its mean and standard deviation. Do the same for the number of doubles of any kind.

Each play is a trial with two outcomes for this question (double six or not), the plays are independent, the probability p = 1/36 is constant, and n = 300 is fixed. So D ~ B(300, 1/36).
Mean: np = 300 × 1/36 ≈ 8.33. Standard deviation: √(np(1 − p)) = √(300 × 1/36 × 35/36) ≈ √8.10 ≈ 2.85.
Number of doubles of any kind: B(300, 1/6), mean 300 × 1/6 = 50, standard deviation √(300 × 1/6 × 5/6) ≈ 6.45.
So in 300 plays we should expect about 8 double sixes, and, by the normal approximation, roughly 95% of the time between about 8.33 − 2(2.85) ≈ 2.6 and 8.33 + 2(2.85) ≈ 14.0, that is, from 3 to 14 double sixes.

D ~ B(300, 1/36): mean ≈ 8.33, SD ≈ 2.85. Doubles of any kind: mean 50, SD ≈ 6.45.

Worked example

The model investigation, Stage 17: the experimental data

Example data (made up for teaching; generated by a computer simulation of 300 plays): 9 double sixes, 38 other doubles, 253 non-doubles. Compare with the theory and find the actual profit.

Expected counts: double six 300/36 ≈ 8.3; other doubles 300 × 5/36 ≈ 41.7; not doubles 300 × 30/36 = 250. Observed: 9, 38, 253. Every count is close to its expected value.
Doubles of any kind: 9 + 38 = 47, versus an expected 50. z = (47 − 50) ÷ 6.45 ≈ −0.47: well within normal variation.
Double sixes: 9 versus 8.33. z = (9 − 8.33) ÷ 2.85 ≈ 0.23. A spreadsheet's binomial function gives P(D ≥ 9) ≈ 0.45, so 9 or more happens almost half the time: nothing unusual.
Payout: 9 × 20 + 38 × 5 = 180 + 190 = 370 tickets. Collected: 300 × 2 = 600. Actual profit: 600 − 370 = 230 tickets, or 230 ÷ 300 ≈ 0.77 tickets per play, close to the expected 0.75.
Experimental probability of a double: 47/300 ≈ 0.157, close to the theoretical 1/6 ≈ 0.167.

The data agree with the model: profit 230 tickets (expected 225), and the counts of double sixes and doubles are within one standard deviation of their expected values.

Expected and observed counts in 300 plays of the carnival game (observed counts are example data from a simulation). Each observed bar is close to its expected bar.

The binomial distribution B(300, 1/36) for the number of double sixes. The observed value, 9, sits right in the middle of the likely values.

Probability inside a statistics project

Even a pure statistics project uses probability. From Amira's contingency table (Analysing the Data II), if one of the 40 students is chosen at random:

P(meets guideline) = 22/40 = 0.55;
P(meets | job) = 6/22 ≈ 0.273 and P(meets | no job) = 16/18 ≈ 0.889 (conditional probability);
if the job status of the 240 Grade 12 students were split in the same proportions as the sample (22/40 = 55% with jobs), the expected number of Grade 12 students at Pinecrest with jobs would be about 0.55 × 240 = 132, and the expected number of those meeting the guideline about 132 × 0.273 ≈ 36. These are estimates with large margins of error, because the sample has only 40 students.

The next lesson, Drawing Conclusions, uses a simulation to ask a probability question about the survey: could a difference as big as 1.14 hours appear just by chance?

Make the probability problem your own

Good game-of-chance investigations change one rule and study the effect ("What prize for double six makes the game fair?"), compare with and without replacement (for card games, a hypergeometric model), or compare two designs. Always do the theory first, then collect at least 100 plays or simulated trials, then compare. A spreadsheet can simulate thousands of plays with =RANDBETWEEN(1,6).

Just for fun

Dice have no memory. That's why, after three double sixes in a row, the dice feel nothing at all, while the carnival treasurer feels quite a lot.

Hands-on activity

Your investigation, step 8: a game of chance

You need: two dice (or a deck of cards or a spinner made from paper and a pencil with a paper clip), paper, and a spreadsheet (optional).

Design a simple game with ticket or point prizes (no money). Write the rules.
Theory: list the sample space, build the probability distribution of the prize, and calculate the expected value and the organizer's expected profit per play.
Identify one binomial (or, for cards without replacement, hypergeometric) count in your game and find its mean and standard deviation for your number of plays.
Data: play at least 100 rounds with family members (or simulate in a spreadsheet: =RANDBETWEEN(1,6) in two columns, then =IF(A2=B2, ...)). Record every play in its own row.
Compare: expected vs observed counts (a side-by-side bar graph), actual vs expected profit, and z-scores for your binomial counts.
Redesign: find the cost per play that would make the game fair.

What to look for: correct probabilities that add to 1, E(X) calculated with units, a clear comparison, and a sentence explaining why observed and expected differ a little.

Your turn

What cost per play would make Amira's game fair?

If the double-six prize were raised to 30 tickets, find the new E(X) and the carnival's expected profit per play (cost still 2).

Find the expected number of non-doubles in 300 plays.

Why is the number of double sixes in 300 plays a binomial random variable?

In the simulation, what was the experimental probability of a double six? Compare with the theory.

Using the contingency table, find P(no job | meets guideline).

More practice: warm-up

How many outcomes are there when rolling two dice?

What is the probability of rolling a double with two dice?

State the formula for expected value.

What is the mean of B(300, 1/6)?

Is Amira's game fair? Explain briefly.

More practice: on your own

Find the standard deviation of the prize X in Amira's game. (E(X²) = 400/36 + 25 × 5/36.)

A different game: draw 2 cards without replacement from a standard deck; win 10 tickets if both are hearts. Find P(win) and E(prize).

In 200 plays of the card game, which model fits the number of wins, and what is its mean?

A student observed 70 doubles in 300 plays. Find the z-score and comment.

Explain why one evening of 20 plays could easily lose tickets for the carnival even though the expected profit is positive.

Challenge

What prize for double six (other prizes unchanged, cost 2 tickets) makes the game exactly fair?

Use the normal approximation (with continuity correction) to estimate P(D ≥ 9) for D ~ B(300, 1/36), and compare with the exact value 0.45.

The carnival wants an expected profit of exactly 0.5 tickets per play and will keep the 2-ticket cost and the 20-ticket double-six prize. What should the prize for other doubles be?

Check your understanding

Quiz

In Amira's game, E(X) for the prize is:

The carnival's expected profit over 300 plays is:

Which conditions make the number of double sixes in 300 plays binomial?

Observed: 47 doubles in 300 plays (expected 50, SD ≈ 6.45). This result is:

A game is fair when:

Why does Amira use tickets instead of money?

For parents and teachers

Big idea (E1.4, E1.3): a probability investigation predicts with theory (sample space, distribution, expected value, binomial model), then tests the prediction with real or simulated plays. Expected value describes the long run, not any single evening.

How to help: Be a player! Play the game your teen designs 50 to 100 times while they record each round. Ask them to explain before you start how many prizes they expect to give away, and compare afterwards.

Safety and values: keep games to points or tickets with no money involved. This is a good moment to talk about why real gambling games are designed so that the house wins in the long run.

Common difficulties:

Probabilities that do not add to 1.
Confusing the expected prize with the expected net gain (forgetting the cost to play).
Expecting observed counts to equal expected counts exactly.
Using a binomial model for draws without replacement from a small group (that is hypergeometric).