Question

4
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a dice game involves rolling two dice. a player who rolls a 3, 4, 10, 11, or 12 wins 5 points. a player who rolls a 5, 6, 7, 8, or 9 loses 5 points.
if the expected value of playing this game is to lose \\(\frac{5}{3}\\), a player will \\(\square\\) \\(\square\\) points by rolling a 2.

Explanation:

Step1: Determine total outcomes

When rolling two dice, the total number of possible outcomes is \(6\times6 = 36\).

Step2: Count outcomes for each case

• Outcomes for rolling 3: (1,2), (2,1) → 2 outcomes.
• Outcomes for rolling 4: (1,3), (2,2), (3,1) → 3 outcomes.
• Outcomes for rolling 10: (4,6), (5,5), (6,4) → 3 outcomes.
• Outcomes for rolling 11: (5,6), (6,5) → 2 outcomes.
• Outcomes for rolling 12: (6,6) → 1 outcome.
• Total for winning (3,4,10,11,12): \(2 + 3+3 + 2+1=11\) outcomes.
• Outcomes for rolling 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes.
• Outcomes for rolling 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes.
• Outcomes for rolling 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes.
• Outcomes for rolling 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes.
• Outcomes for rolling 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes.
• Total for losing (5,6,7,8,9): \(4 + 5+6 + 5+4=24\) outcomes.
• Let the number of outcomes for rolling 2 be 1 (since (1,1) is the only way to roll a 2) and let the points for rolling 2 be \(x\).

Step3: Calculate expected value

The expected value \(E\) is given by the formula:

$$ E=\frac{\text{Number of winning outcomes}\times\text{Winning points}+\text{Number of losing outcomes}\times\text{Losing points}+\text{Number of rolling 2 outcomes}\times\text{Points for 2}}{\text{Total outcomes}} $$

We know \(E =-\frac{5}{3}\) (negative because it's a loss), winning points = 5, losing points = - 5 (since losing 5 points), number of winning outcomes = 11, number of losing outcomes = 24, number of rolling 2 outcomes = 1, total outcomes = 36.
Substitute into the formula:

$$ -\frac{5}{3}=\frac{11\times5 + 24\times(- 5)+1\times x}{36} $$

Step4: Solve for \(x\)

First, simplify the numerator:

$$ 11\times5+24\times(-5)+x = 55-120 + x=x - 65 $$

So the equation becomes:

$$ -\frac{5}{3}=\frac{x - 65}{36} $$

Multiply both sides by 36:

$$ -\frac{5}{3}\times36=x - 65 $$

$$ - 60=x - 65 $$

Add 65 to both sides:

$$ x=65 - 60 = 5 $$

Wait, no, wait. Wait, if we get \(x = 5\), but let's check again. Wait, maybe I made a mistake in sign. Let's re - define: Let the points for winning be \(+ 5\), points for losing be \(-5\), and points for rolling 2 be \(y\).
Expected value \(E=\frac{11\times5+24\times(-5)+1\times y}{36}\)
We know \(E =-\frac{5}{3}\)
So:

$$ \frac{55-120 + y}{36}=-\frac{5}{3} $$

$$ \frac{y - 65}{36}=-\frac{5}{3} $$

Cross - multiply: \(3(y - 65)=- 5\times36\)

$$ 3y-195=-180 $$

$$ 3y=195 - 180=15 $$

$$ y = 5 $$

Wait, but if \(y = 5\), that means the player wins 5 points? But let's check the number of outcomes again. Wait, the number of ways to roll a 2 is 1 ( (1,1) ), number of ways to roll 3: 2, 4:3, 10:3, 11:2, 12:1. So 2 + 3+3 + 2+1 = 11. Number of ways to roll 5:4, 6:5,7:6,8:5,9:4. 4 + 5+6 + 5+4 = 24. 11+24 + 1=36, which is correct.
Wait, but the expected value is a loss of \(\frac{5}{3}\). Let's recalculate the expected value with \(y = 5\):

$$ \frac{11\times5+24\times(-5)+1\times5}{36}=\frac{55 - 120 + 5}{36}=\frac{-60}{36}=-\frac{5}{3} $$

Yes, that works. So the player will win 5 points by rolling a 2. Wait, but let's check the logic again. The number of outcomes for 2 is 1. The winning outcomes (3,4,10,11,12) have 11 outcomes, losing (5,6,7,8,9) have 24 outcomes, and 2 has 1 outcome. When we calculate the expected value, with 2 giving + 5 (winning 5 points), the calculation gives the correct expected value of losing \(\frac{5}{3}\).

Answer:

win, 5