Step by step
The question
Question: In a game, you win $10 with a 1 in 5 chance.
Otherwise you lose $2.
What is the expected value of the game?
There are four choices.
First we solve it, the normal way.
Then we pick.
1 Make a table
One row for each outcome: its value and its probability.
Win 10, with a chance of 1/5.
Lose: a loss is a negative value, minus 2.
The chance is 4/5.
The chances add up to 1: 1/5 and 4/5 make 5/5.
2 Multiply each row
Win: 10 times 1/5 is 10/5.
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.
Lose: minus 2 times 4/5 is minus 8/5.
3 Add them
Same bottoms: we add the tops.
A loss takes away.
Careful!
We cannot just take the middle of the values: they do not happen equally often.
Each value counts as much as its chance.
10 minus 8 is 2.
2 divided by 5 is 0.4.
So the expected value is 40 cents.
Positive: in the long run, you win 40 cents a game.
4 Check
Let's check.
Imagine 5 games.
The chances say: 1 win, 4 lose.
Together: we are up $2.
Shared over 5 games, that is 40 cents a game.
The same.
So the answer is right!
5 Pick the choice
Our answer is E = $0.40.
That is choice C.
6 Why not the others
A is 8.
Wrong: used 10 − 2: weight each by its probability.
B is 2.
Wrong: 10 × 1/5 only: the losses count too.
D is −1.6.
Wrong: 2 × 4/5 with a minus only.
Let's look back.
First: make a table.
Then: multiply each row.
Then: add them.
Then: check.
Then: pick the choice.
Last: why not the others.