Topic 4.5 · teacher page

Running probability and sample spaces

The simulation that ends the "but it did not come out as a sixth" argument, and the expected value nobody can be.

The one thing to do with the simulation

Set it to 60 rolls first and ask whether the theory is wrong. At 60 the relative frequency can be a long way off a sixth, and that is exactly what happens when a class rolls dice for ten minutes and gets a disappointing answer.

Then push it to 4000 and watch the gap close, slowly and unevenly. The lesson is not that theory beats experiment; it is that a small experiment cannot distinguish them, which is why the counting argument exists.

The misconception this page is built around

Nearly every class produces 1⁄11 for a total of seven at some point, because there are eleven totals. The counting formula needs outcomes that are equally likely, and the eleven totals are not.

Draw the six by six grid once, properly, and count the diagonal. It takes two minutes and it prevents the error for good. Question 3 on the student page tests exactly this.

The answers

Sample space36 equally likely pairs. Six of them total 7, so the probability is 6/36 = 0.167.
Ways to make each total1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for totals 2 to 12. They sum to 36.
Expected absences200 × 0.15 = 30.
Expected sixes in 60 rolls60 × 1⁄6 = 10.
1, 2, 30.167; 30; B, the totals are not equally likely.

Where the marks go

1 markShowing the sample space, as a grid or a list. If the outcomes counted are not equally likely, everything after it is wrong.

1 markUsing the complement when the question says "at least". Shorter and far less error-prone than adding cases.

1 markLeaving an expected number as a decimal where it is one. 12.8 is the answer; 13 is a rounded version.

What each wrong answer tells you

They giveWhat it means
1/11 (Q1)The target error. Counted the totals rather than the pairs. Go to the grid, not to an explanation.
1/36 (Q1)Counted one specific pair. They have the right denominator and have not noticed there are six ways.
6 (Q1)Gave the count, not the probability. Usually a reading slip rather than a misunderstanding.
170 (Q2)Answered the complement. Worth praising the arithmetic and pointing at the question.
"The dice are not independent" (Q3)A reasonable-sounding wrong reason. They ARE independent, and that is precisely why the 36 pairs are equally likely.

Other things they will say

"We rolled 60 times and got 0.12, so the theory is wrong." The single most useful thing this page answers. Show them the same wobble on screen, then push the count up.

"You cannot have 12.8 students absent." Correct, and it does not matter. It is a long-run average. Ask how many you would expect over ten days: 128 is a perfectly sensible number.

"Is relative frequency just probability?" It is an estimate of it, and it gets better with more trials. Keeping the two words apart is worth insisting on, because 4.11 will rest on the distinction.

A possible order

 What is happening
1Roll real dice as a class. Collect the relative frequency and let it disappoint.
2The simulation. 60 rolls, then 4000. Why the small experiment could not have told you.
3The six by six grid, built on the board. Count the diagonal.
4The complement, and the "at least one" signal.
5Expected occurrences, including a non-integer one.
6Questions 1 to 3, with question 3 taken to the whole class.

Two things not to say

Do not say "probability is what happens in the long run" and leave it there. It makes the counting argument sound like a prediction about tomorrow, and then a class that rolls 0.12 thinks the mathematics failed.

Do not round an expected value to a whole number without saying why. If the question wants a sensible real-world statement, say both: 12.8 expected, so usually twelve or thirteen.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. At least oneA fair die is rolled 3 times. Find the probability of at least one six.
    1 − (5/6)³ = 91/216 ≈ 0.421.
  2. Without replacementA bag has 5 red counters and 7 blue. Two are drawn without replacement. Find P(both red).
    (5/12)(4/11) = 5/33 ≈ 0.152.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. Complement in contextA bus is late on 15% of mornings, independently. Find the probability it is late at least once in a 5 day week.
    1 − 0.85⁵ = 0.556, so more often than not.
  2. Two stage contextTwo of the 12 counters above are drawn with replacement instead. Find P(both red) and say why it differs.
    (5/12)² = 25/144 ≈ 0.174. Replacing restores the first counter, so the second draw has the same chances and the events are independent.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. Spot the errorA student computes P(at least one six in 3 rolls) as 3 × 1/6 = 1/2. Explain the mistake and why it gets worse.
    They added overlapping cases, double counting outcomes with more than one six. At 6 rolls the same method gives probability 1, which is visibly absurd. Use the complement.
  2. Reason about independenceExplain why drawing without replacement makes two draws dependent, using the counters above.
    The first draw changes what is left, so P(second red) is 4/11 after a red but 5/11 after a blue. Dependence means the second probability needs the first result to be stated.

Practicalities

The simulation uses one fixed sequence of rolls, so the picture does not reshuffle every time the slider moves and a class can discuss the same wobble. Nothing is loaded from any other site and nothing is stored.