Topic 4.7 · teacher page · Standard Level

Expected does not mean likely

E(X) as a balance point, and why the expected value is often a number that cannot happen.

The one thing to do with the animation

Ask where the balance point will be before you reveal it.

Students guess the tallest bar, because “expected” sounds like “most likely”. The beam tips until it settles somewhere that is frequently not a bar at all.

Then give them the 2.5 children example. Nobody has 2.5 children, and the mean is still 2.5. The word is a technical term for a balance point and not a prediction, and the balance beam is the only explanation of that which survives the week.

Why the probabilities must sum to 1 first

A table with probabilities summing to 0.95 is not a distribution, and every number computed from it is wrong. Make the sum the first line of working, always, before any multiplying.

Questions that give an unknown k in the table are testing exactly this: the sum to 1 is the equation, and finding k is the first mark.

The answers

1. E(X) from the table0.1 + 0.4 + 0.9 + 1.6 = 3.
2. E(X) for (4 + x)/18(5 + 12 + 21) over 18 = 38/18 = 2.111.
3. The fair stakeExpected winnings are 0.3 × 5 = 1.50, so a stake of 1.50 makes the expected gain zero.

Where the marks go

1 markMultiplying each value by its own probability and summing, with the products visible.

1 markChecking or imposing that the probabilities sum to 1.

1 markInterpreting a fair game as expected gain zero, rather than as equal chances.

What each wrong answer tells you

They giveWhat it means
2.5 (Q1)Averaged the x values and ignored the probabilities. Very common, and it looks plausible on a symmetric-ish table, which is why a deliberately lopsided table is the better teaching example.
4 (Q1)Gave the most likely value. This is the misreading the balance beam exists to break.
1 (Q1 or Q2)Summed the probabilities instead. They have done the check and reported it as the answer.
2 (Q2)Gave the middle value of 1, 2, 3 by symmetry, but the distribution is not symmetric.
6.333 (Q2)Forgot to divide by 18, or divided only one term.
5 (Q3)Gave the payout. A fair stake is the expected payout, not the payout.

Other things they will say

"Can E(X) be a value X never takes?" Yes, and that is the point. A fair die has E(X) = 3.5. If this surprises them, it means they were still reading it as a prediction.

"Is a fair game one where I win half the time?" No. Fair means expected gain zero. A game you win one time in ten paying ten times your stake is fair, and feels nothing like it.

"Why do casinos exist?" Because every game has expected gain negative for the player, by design and by a known margin. A two minute digression here does more for retention than another table.

A possible order

 What is happening
1Guess the balance point. Reveal. Let the gap between guess and answer do the work.
2The 2.5 children example and the definition.
3The three questions, with the sum-to-1 check written first every time.
4A table with an unknown k, then a fair game question.
5Where this reappears: the binomial mean next lesson is a special case of exactly this.

Two things not to say

Do not say “the expected value is what you expect to get”. It is circular and it reinforces the misreading.

Do not only use symmetric distributions. They let the wrong method give the right answer, which is the worst possible feedback.

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. Expected valueX takes 0, 1, 2, 3 with probabilities 0.1, 0.3, 0.4, 0.2. Find E(X).
    1.7
  2. VarianceFor the same X find Var(X) and the standard deviation.
    E(X²) = 3.7, so Var(X) = 3.7 − 1.7² = 0.81 and the standard deviation is 0.9.

2In context

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

  1. Linear transformationA stall's profit is P = 3X − 2 hundred baht, with X as above. Find E(P) and Var(P).
    E(P) = 3(1.7) − 2 = 3.1; Var(P) = 3²(0.81) = 7.29. The −2 shifts the mean and leaves the variance alone.
  2. Use E(X) to decideA game costs 2 counters to play and pays X counters. Using E(X) = 1.7, say whether to play.
    No. The expected return is 1.7 against a cost of 2, so the average loss is 0.3 a go, however it feels over a few turns.

3Challenge

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

  1. Why the mean need not occurE(X) = 1.7 but X is only ever a whole number. Explain what E(X) means given that.
    It is the long-run average, the balance point of the distribution, not a value X can take. A mean family size of 1.7 children is the same idea.
  2. Find a missing probabilityY takes 1, 2, 5 with probabilities 0.5, p, q and E(Y) = 2.3. Find p and q.
    p + q = 0.5 and 0.5 + 2p + 5q = 2.3. So 2p + 5q = 1.8 with p = 0.5 − q gives 1 + 3q = 1.8, q = 0.2666... Checking: q = 4/15, p = 7/30. Students should notice the total must be 1 before starting.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.