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A game: win $8 with probability 1/4, lose $2 otherwise, expected value

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

Step by step

The question

Question: In a game, you win $8 with a 1 in 4 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 8, with a chance of 1/4.

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

The chance is 3/4.

The chances add up to 1: 1/4 and 3/4 make 4/4.

2 Multiply each row

Win: 8 times 1/4 is 8/4.

8 × 14 = 84

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 3/4 is minus 6/4.

(−2) × 34 = −64

3 Add them

Same bottoms: we add the tops.

A loss takes away.

E = 8 − 64

Careful!

The chances are not equal.

Each value counts as much as its probability.

Careful: not 8 + (−2)2 = 3 ✗

8 minus 6 is 2.

E = 24

Both 2 and 4 can be divided by 2.

That makes it simpler.

E = 2 ÷ 24 ÷ 2

So it is 1/2.

E = 12

1 divided by 2 is 0.5.

1 ÷ 2 = 0.5

So the expected value is 50 cents.

E = $0.50

Positive: in the long run, you win 50 cents a game.

you win $0.50 a game on average

4 Check

Let's check.

Imagine 4 games.

The chances say: 1 win, 3 lose.

in 4 games: 1 win, 3 lose

Together: we are up $2.

1 × $8 − 3 × $2 = $2

Shared over 4 games, that is 50 cents a game.

The same.

So the answer is right!

$2 ÷ 4 = $0.50 a game

Let's look back.

First: make a table.

Then: multiply each row.

Then: add them.

Last: check.

Answer
E = $0.50

Your turn

Try this one yourself, then check:

Show the answer
E = $1

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