Step by step
The question
Question: In a game, you win $12 with a 1 in 6 chance.
Otherwise you lose $3.
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 12, with a chance of 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.
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 3 times 5/6 is minus 15/6.
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.
12 minus 15 is minus 3.
Both 3 and 6 can be divided by 3.
That makes it simpler.
So it is minus 1/2.
1 divided by 2 is 0.5.
So the expected value is minus 50 cents.
Negative: in the long run, you lose 50 cents a game.
4 Check
Let's check.
Imagine 6 games.
The chances say: 1 win, 5 lose.
Together: we are down $3.
Shared over 6 games, that is minus 50 cents a game.
The same.
So the answer is right!
5 Pick the choice
Our answer is E = −$0.50.
That is choice A.
6 Why not the others
B is $0.50.
Wrong: lost the minus: you lose more than you win.
C is $2.
Wrong: forgot the losing part.
D is $9.
Wrong: 12 − 3: left out the probabilities.
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.