Pay $3 to play, roll a die, win $15 on a 6, is it fair Step by Step

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The question

Question: It costs $3 to play.

You roll a die and win $15 if you roll a 6.

Is the game fair?

Find the expected value.

1 Make a table

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

Win: the prize is $15, but we paid $3.

So we are up $12.

The chance is 1/6.

Lose: a loss is a negative value, minus 3.

The chance is 5/6.

The chances add up to 1: 1/6 and 5/6 make 6/6.

2 Multiply each row

Win: 12 times 1/6 is 12/6.

12 × 16 = 126

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 3 times 5/6 is minus 15/6.

(−3) × 56 = −156

3 Add them

Same bottoms: we add the tops.

A loss takes away.

E = 12 − 156

Careful!

The chances are not equal.

Each value counts as much as its probability.

Careful: not 12 + (−3)2 = 4.5 ✗

12 minus 15 is minus 3.

E = −36

Both 3 and 6 can be divided by 3.

That makes it simpler.

E = −3 ÷ 36 ÷ 3

So it is minus 1/2.

E = −12

1 divided by 2 is 0.5.

1 ÷ 2 = 0.5

So the expected value is minus 50 cents.

E = −$0.50

Negative: in the long run, you lose 50 cents a game.

A fair game has 0.

So this game is not fair.

you lose $0.50 a game on average

not fair

4 Check

Let's check.

Imagine 6 games.

The chances say: 1 win, 5 lose.

in 6 games: 1 win, 5 lose

Together: we are down $3.

1 × $12 − 5 × $3 = −$3

Shared over 6 games, that is minus 50 cents a game.

The same.

So the answer is right!

−$3 ÷ 6 = −$0.50 a game

Let's look back.

First: make a table.

Then: multiply each row.

Then: add them.

Last: check.

Answer