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 7 options:

$7.50


-$7.50


$0.75


-$0.75

Answer 1

To calculate the expected gain or loss from playing the game, we first need to understand the expected outcome of one game played.

1. **Cost to play the game**: $2
2. **Probability of winning**: \( \frac{1}{4} \)
3. **Probability of losing**: \( \frac{3}{4} \)
4. **Reward for winning**: $5
5. **Reward for losing**: $0 (loss of the money paid to play)

Next, we calculate the expected value (EV) for one game:

- If you win, your total outcome is the reward minus the cost:
\[
5 - 2 = 3
\]
- If you lose, your total outcome is:
\[
0 - 2 = -2
\]

Now, we can use the probabilities to calculate the expected value:

\[
EV = P(\text{win}) \times \text{outcome if win} + P(\text{lose}) \times \text{outcome if lose}
\]

Substituting the values we calculated:

\[
EV = \left(\frac{1}{4} \times 3\right) + \left(\frac{3}{4} \times -2\right)
\]

Calculating each part:

- Winning part:
\[
\frac{1}{4} \times 3 = \frac{3}{4}
\]

- Losing part:
\[
\frac{3}{4} \times -2 = -\frac{6}{4} = -1.5
\]

Now combine these:

\[
EV = \frac{3}{4} - 1.5 = \frac{3}{4} - \frac{6}{4} = -\frac{3}{4} = -0.75
\]

This means the expected loss per game is $0.75.

Since you play the game 10 times, you multiply this expected value by 10:

\[
\text{Total expected loss} = -0.75 \times 10 = -7.50
\]

Thus, the expected gain or loss after playing the game 10 times is:

**Answer**: -$7.50