To find the Expected Value (EV) for Game 2 (the dice game), we need to compute the average outcome of a single roll, taking into account the possible winnings and their probabilities.

Given:

• Each number (1 to 6) has an equal probability of being rolled: $\frac{1}{6}$.
• Winnings for each outcome:
  • Roll a 1: Win $0.3
  • Roll a 2: Win $0.5
  • Roll a 3: Win $0.7
  • Roll a 4: Win $0.9
  • Roll a 5: Win $1.10
  • Roll a 6: Win $1.3

The formula for expected value is:

$\text{EV} = \sum (\text{probability of outcome}) \times (\text{value of outcome})$

Since each outcome is equally likely, the probability of each outcome is $\frac{1}{6}$. The expected value is:

$\text{EV} = \frac{1}{6} \times (0.3 + 0.5 + 0.7 + 0.9 + 1.1 + 1.3)$

Let's calculate that:

$0.3 + 0.5 + 0.7 + 0.9 + 1.1 + 1.3 = 4.8$

Now, dividing by 6:

$\text{EV} = \frac{1}{6} \times 4.8 = 0.8$

Thus, the expected value per roll in Game 2 is $0.80.

Expected Value for 100 plays:

If you play 100 times, spending $100 (since each play costs $1), the total expected return is:

$100 \times 0.8 = 80$

So, you expect to win $80 in total if you play all $100 on Game 2.

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Would you like a more detailed breakdown or have any questions?

5 Related Questions:

1. How would the expected value change if the payouts for each number were adjusted?
2. What is the probability of winning more than $1 in a single roll?
3. What is the variance and standard deviation of the winnings in this game?
4. How would the expected value be affected if the cost of playing the game changed?
5. What is the breakeven point for this game in terms of payout?

Tip:

Expected value gives a good sense of the long-term average, but it doesn't guarantee individual outcomes—some short-term results may vary significantly!