Question 1177711
Let's solve this problem step-by-step.

**1. Define Random Variables**

* **X:** The outcome of the first roll (X can be 1, 2, 3, 4, 5, or 6).
* **Y:** The number of wins (getting a "3") in the subsequent rolls.

**2. Probability of Winning on a Single Roll**

* The probability of rolling a "3" on any single roll is 1/6.

**3. Conditional Expectation**

We need to find the expected number of wins given the outcome of the first roll.

* If X = 1, we roll the die once more. The expected number of wins is 1 * (1/6) = 1/6.
* If X = 2, we roll the die twice more. The expected number of wins is 2 * (1/6) = 2/6.
* If X = 3, we roll the die three more times. The expected number of wins is 3 * (1/6) = 3/6.
* If X = 4, we roll the die four more times. The expected number of wins is 4 * (1/6) = 4/6.
* If X = 5, we roll the die five more times. The expected number of wins is 5 * (1/6) = 5/6.
* If X = 6, we roll the die six more times. The expected number of wins is 6 * (1/6) = 6/6 = 1.

**4. Expected Value of X**

* Since the die is fair, each outcome of the first roll has a probability of 1/6.

**5. Total Expected Wins**

We need to find the expected value of Y, which is the sum of the conditional expectations multiplied by the probability of each outcome of X:

* E(Y) = (1/6) * (1/6) + (2/6) * (1/6) + (3/6) * (1/6) + (4/6) * (1/6) + (5/6) * (1/6) + (6/6) * (1/6)
* E(Y) = (1/36) * (1 + 2 + 3 + 4 + 5 + 6)
* E(Y) = (1/36) * 21
* E(Y) = 21/36
* E(Y) = 7/12

**Therefore, the expected number of wins from this experiment is 7/12.**