Roll a die: win $3 on a 5 or 6, lose $2 otherwise, is it fair Step by Step

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The question

Question: You roll a die.

You win $3 if you roll a 5 or 6.

Otherwise you lose $2.

Is the game fair?

Find the expected value.

1 Make a table

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

Win 3, with a chance of 1/3.

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

The chance is 2/3.

The chances add up to 1: 1/3 and 2/3 make 3/3.

2 Multiply each row

Win: 3 times 1/3 is 3/3.

3 × 13 = 33

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 2/3 is minus 4/3.

(−2) × 23 = −43

3 Add them

Same bottoms: we add the tops.

A loss takes away.

E = 3 − 43

Careful!

The chances are not equal.

Each value counts as much as its probability.

Careful: not 3 + (−2)2 = 0.5 ✗

3 minus 4 is minus 1.

E = −13

1 divided by 3 is 0.333, and more digits.

1 ÷ 3 = 0.33333…

To the nearest cent, the expected value is minus 33 cents.

E ≈ −$0.33

Negative: in the long run, you lose about 33 cents a game.

A fair game has 0.

So this game is not fair.

you lose $0.33 a game on average

not fair

4 Check

Let's check.

Imagine 3 games.

The chances say: 1 win, 2 lose.

in 3 games: 1 win, 2 lose

Together: we are down $1.

1 × $3 − 2 × $2 = −$1

Shared over 3 games, that is about minus 33 cents a game.

The same.

So the answer is right!

−$1 ÷ 3 ≈ −$0.33 a game

Let's look back.

First: make a table.

Then: multiply each row.

Then: add them.

Last: check.

Answer