Expected value in context

Fair games, totals and the least number

In 7 of 20 recent exams
Paper 2, calculator
1 to 6 marks

How it is set in the exam

Paper 2 (calculator) mostly (5 of its 7 exam versions), 1–6 marks per part: usually the last parts of a Section B question about a game, prize tickets or sales. Common (7 of 20 exam versions).

Spot it in the exam

Exam phrases to spot (5)

“is the game fair?”
“expected gain per game”
“the expected total after $n$ games”
“how many … must be sold so that the expected profit is at least …”
“the smallest prize for which …”

In disguise (5)

the cost to play is the unknown: set the expected prize equal to the cost
an average per customer, scaled up to a whole day, a population, or a number of boxes to order
the organiser's point of view: the organiser's expected profit is the players' expected loss
a prize that grows each week: the expected prize becomes an expression in the week number (Where it hides)
nothing is random: a stock of items in several grades, each with its own price, is assumed to be sold in the same proportions as it was bought. The proportions play the part of the probabilities, so the mean income from one item is $\sum \text{price} \times \text{proportion}$ , then the usual total or least-number finish

Variants (2)

$X$ is a gain, a score or a revenue; filling in one gain entry is often a 1-mark first part
asked: fair or not, the prize that makes it fair, an expected total, the number of games for a given expected total, the least number sold, the least prize

Looks like it, but isn't (3)

the least number of games so that the probability of at least one prize passes a target: Find n, with $1 - q^n$
the expected number of times an event happens, $n \times \mathrm{P}(A)$ : the expected number of occurrences, a concept (or $np$ in Binomial probabilities)
$\mathrm{E}(X)$ from a table, with no cost, total or target: Probability distribution tables

Worked example

Is the game fair? Then make it fair

Paper 2 (calculator)

Question

It costs $\text{\textdollar}3$ to play a game at a school fair. The prizes and their probabilities are shown below.

Prize ( $\text{\textdollar}$ )	Probability
$30$	$0.02$
$15$	$0.04$
$3$	$0.3$
$0$	$c$

Prize ( $\text{\textdollar}$ )	$30$	$15$	$3$	$0$
Probability	$0.02$	$0.04$	$0.3$	$c$

(a) Find $c$ . [1]
(b) Find the player's expected gain per game, and determine whether the game is fair. [4]
(c) The organiser changes the top prize, keeping everything else the same, so that the game is fair. Find the new top prize. [3]
(d) With the original prizes, $400$ games are played. Find the organiser's expected profit. [2]

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Steps

Define $X$
for the person the question is about: the player's gain is prize $-$ cost (a prize that returns the cost is a gain of $0$ ); the organiser's profit is cost $-$ prize; a shop's profit on an item is price $-$ cost of the item. Write its table; a missing probability comes from the total of $1$ .
$\mathrm{E}(X) = \sum x\,\mathrm{P}(X = x)$ ,
written out with its terms.
Fair game:
fair exactly when $\mathrm{E}(\text{gain}) = 0$ , i.e. when the expected prize equals the cost. Conclude with the value as the reason: " $\mathrm{E}(G) = -0.9 \ne 0$ , so not fair". To make a game fair, set $\mathrm{E}(\text{gain}) = 0$ with the unknown prize or cost, and solve.
Totals:
over $n$ repeats, the expected total is $n\,\mathrm{E}(X)$ . Subtract a fixed cost once, not once per item.
Least number, least prize:
write the inequality (such as $n\,\mathrm{E}(X) - \text{fixed cost} \ge \text{target}$ , or $\mathrm{E}(\text{gain}) > 0$ ), solve it, then take the least whole number that satisfies it (for a price in dollars and cents, the least whole number of cents): round the boundary up, never to the nearest. If the boundary is itself a whole number, it is the answer for $\ge$ ( $n \ge 240$ gives $240$ ) and the next one up for $>$ ( $n > 240$ gives $241$ ). Show the values either side, and never give a part of a game, a ticket or a box.

Common mistakes

Using the prize instead of the gain
forgetting the cost to play, or subtracting it twice (once in the table and again at the end).
"Fair" with no reason, or the wrong comparison
the reason is $\mathrm{E}(\text{gain}) = 0$ (or expected prize $=$ cost), with the value. A correct "not fair" with no reason loses the .
Rounding down, or giving a decimal,
for a number of games, boxes or tickets: $417.91\ldots$ boxes means $418$ . A non-whole number of games loses the final mark.

More mistakes (4)

A fixed cost counted once per item,
or forgotten: it is subtracted once from the expected total.
Mixing up whose gain it is
the organiser's expected profit is the negative of the player's expected gain.
Confusing an expected total with a probability
"the expected total is at least $40$ " is $n\,\mathrm{E}(X) \ge 40$ ; "the probability of at least one prize is more than $0.9$ " is Find n.
Follow-through needs a valid table
markschemes give follow-through marks only when your probabilities add up to $1$ .

Easy to confuse with

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