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Roll a die: win $10 on a 1 or 2, lose $5 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 $10 if you roll a 1 or 2.

Otherwise you lose $5.

What is the expected value?

1 Make a table

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

Win 10, with a chance of 1/3.

Lose: a loss is a negative value, minus 5.

The chance is 2/3.

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

2 Multiply each row

Win: 10 times 1/3 is 10/3.

10 × 13 = 103

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 5 times 2/3 is minus 10/3.

(−5) × 23 = −103

3 Add them

Same bottoms: we add the tops.

A loss takes away.

E = 10 − 103

Careful!

The chances are not equal.

Each value counts as much as its probability.

Careful: not 10 + (−5)2 = 2.5 ✗

10 minus 10 is 0.

E = 03

So the expected value is $0.

E = $0

The expected value is 0: in the long run, nobody wins or loses.

That is called a fair game.

0: a fair game

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 up $0.

1 × $10 − 2 × $5 = $0

Shared over 3 games, that is $0 a game.

The same.

So the answer is right!

$0 ÷ 3 = $0 a game

Let's look back.

First: make a table.

Then: multiply each row.

Then: add them.

Last: check.

Answer
E = $0

Your turn

Try this one yourself, then check:

Show the answer
E = $1

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