Question

In a game there is a 1 in 4 chance of winning the game and a 3 in 4 chance of losing. The game costs $2 to play. If the game is won, the winner receives $5. If the game is lost, the player receives nothing. If you play the game 10 times, what is the expected gain or loss?

Question 9 options:

-$0.75


$7.50


-$7.50


$0.75

Answer 1

To calculate the expected gain or loss when playing the game 10 times, we need to analyze the expected outcome for a single game and then scale it up for 10 games.

1. **Determine the expected value (EV) for one game.**
- Probability of winning = \( \frac{1}{4} \)
- Probability of losing = \( \frac{3}{4} \)

- If a player wins, they gain $5, but they also pay $2 to play, resulting in a net gain of \( 5 - 2 = 3 \) dollars.
- If a player loses, they lose the $2 they paid to play, resulting in a net loss of \( -2 \) dollars.

Now, we can calculate the expected value:
\[
\text{EV} = \left( \frac{1}{4} \times 3 \right) + \left( \frac{3}{4} \times (-2) \right)
\]
\[
= \left( \frac{3}{4} \right) + \left( -\frac{6}{4} \right)
\]
\[
= \frac{3}{4} - \frac{6}{4} = -\frac{3}{4} = -0.75
\]

2. **Expected value for 10 games:**
The expected value for 10 games is simply 10 times the expected value of a single game:
\[
\text{Total EV} = 10 \times (-0.75) = -7.50
\]

The expected gain or loss after playing the game 10 times is \(-7.50\).

Thus, the answer is:

**-$7.50**