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A game: win $10 with probability 1/5, lose $2 otherwise, expected…

Step by step, with the reason for every step · Expected value · Probability

Step by step

The question

Question: In a game, you win $10 with a 1 in 5 chance.

Otherwise you lose $2.

What is the expected value of the game?

1 Make a table

One row for each outcome: its value and its probability.

Win 10, with a chance of 1/5.

Lose: a loss is a negative value, minus 2.

The chance is 4/5.

The chances add up to 1: 1/5 and 4/5 make 5/5.

2 Multiply each row

Win: 10 times 1/5 is 10/5.

10 × 15 = 105

The expected value: each value times its probability, all added up.

It is what we win or lose per game, on average, if we play many times.

E = expected value:
each value × its probability,
all added

Lose: minus 2 times 4/5 is minus 8/5.

(−2) × 45 = −85

3 Add them

Same bottoms: we add the tops.

A loss takes away.

E = 10 − 85

Careful!

The chances are not equal.

Each value counts as much as its probability.

Careful: not 10 + (−2)2 = 4 ✗

10 minus 8 is 2.

E = 25

2 divided by 5 is 0.4.

2 ÷ 5 = 0.4

So the expected value is 40 cents.

E = $0.40

Positive: in the long run, you win 40 cents a game.

you win $0.40 a game on average

4 Check

Let's check.

Imagine 5 games.

The chances say: 1 win, 4 lose.

in 5 games: 1 win, 4 lose

Together: we are up $2.

1 × $10 − 4 × $2 = $2

Shared over 5 games, that is 40 cents a game.

The same.

So the answer is right!

$2 ÷ 5 = $0.40 a game

Let's look back.

First: make a table.

Then: multiply each row.

Then: add them.

Last: check.

Answer
E = $0.40

Your turn

Try this one yourself, then check:

Show the answer
E = $1.40

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