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Roll a die: win $6 on a 6, lose $1 otherwise, expected value

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

Step by step

The question

Question: You roll a die.

You win $6 if you roll a 6.

Otherwise you lose $1.

What is the expected value?

1 Make a table

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

Win 6, with a chance of 1/6.

Lose: a loss is a negative value, minus 1.

The chance is 5/6.

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

2 Multiply each row

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

6 × 16 = 66

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 1 times 5/6 is minus 5/6.

(−1) × 56 = −56

3 Add them

Same bottoms: we add the tops.

A loss takes away.

E = 6 − 56

Careful!

The chances are not equal.

Each value counts as much as its probability.

Careful: not 6 + (−1)2 = 2.5 ✗

6 minus 5 is 1.

E = 16

1 divided by 6 is 0.167, and more digits.

1 ÷ 6 = 0.16667…

To the nearest cent, the expected value is 17 cents.

E ≈ $0.17

Positive: in the long run, you win about 17 cents a game.

you win $0.17 a game on average

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 up $1.

1 × $6 − 5 × $1 = $1

Shared over 6 games, that is about 17 cents a game.

The same.

So the answer is right!

$1 ÷ 6 ≈ $0.17 a game

Let's look back.

First: make a table.

Then: multiply each row.

Then: add them.

Last: check.

Answer
E ≈ $0.17

Your turn

Try this one yourself, then check:

Show the answer
E ≈ $0.33

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