Question

At a carnival, there is a game where you can draw one of 20 balls from a bucket if you pay $7. The balls are numbered from 1 to 20. If the number on the ball is odd, you win $13. If the number on the ball is even, you win nothing. If you play the game, what is the expected profit?

Answer 1

We have a game where we win if we get an odd ball from the bucket.

As the balls are numbered from 1 to 20, we have 20 balls where 10 of them are odd.

Then, we have a probability of 50% of getting an odd ball and 50% of not getting it.

Then, we can calculate the expected prize as the sum of the outcomes weighted by the probability of having that outcome.

In this case we have two outcomes:

- Odd ball: The outcome prize is x1=13 and the probability of this outcome is p1=0.5.

- Even ball: The outcome prize is x2=0 and the probability of this outcome is p2=0.5.

Then, the expected prize is:

`E(x)=\sum ^2_(i=1)p_ix_i=0.5\cdot13+0.5\cdot0=6.5+0=6.5`

As the game cost $7 and we expect to get $6.5 from playing, the expected profit is the difference between the expected prize and the cost:

`E(p)=E(x)-c=6.5-7=-0.5`

Answer: The expected profit is $ -0.50 (a loss of 50 cents per game).