To determine the probability that the experiment will end in the fifth trial (with the second six occurring on the fifth roll), let us analyze the problem step by step.

Key Requirements

1. The die must be rolled exactly 5 times.
2. The fifth roll must be a six (since we need two consecutive sixes).
3. The fourth roll must also be a six (for the pair of consecutive sixes).
4. The first three rolls must not result in a six, as the streak of consecutive sixes must begin only on the fourth roll.

Probability of Each Roll

• The probability of rolling a six on any given roll is $\frac{1}{6}$.
• The probability of not rolling a six is $\frac{5}{6}$.

Probability Breakdown

1. First three rolls must not be six: $P(\text{First three rolls not six}) = \left( \frac{5}{6} \right)^3.$

2. Fourth roll must be a six: $P(\text{Fourth roll is a six}) = \frac{1}{6}.$

3. Fifth roll must also be a six: $P(\text{Fifth roll is a six}) = \frac{1}{6}.$

Combined Probability

The experiment ends on the fifth trial if these events happen in sequence. The combined probability is the product of all these individual probabilities: $P(\text{End on fifth trial}) = \left( \frac{5}{6} \right)^3 \cdot \frac{1}{6} \cdot \frac{1}{6}.$

Simplify this expression: $P(\text{End on fifth trial}) = \frac{5^3}{6^5} = \frac{125}{7776}.$

Final Answer:

The probability that the experiment ends on the fifth trial is: $\boxed{\frac{125}{7776}}$

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Would you like further clarification or details on any step? 😊

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Follow-up Questions

1. What is the general probability for the experiment ending in the $n$-th trial?
2. How does the probability change if the die is not fair (e.g., weighted six)?
3. Can you derive the expected number of rolls required to complete the experiment?
4. What happens if we need three consecutive sixes instead of two?
5. How would this problem generalize to a die with $k$ sides?

Tip:

For problems involving "successive" outcomes, it's often helpful to carefully examine the conditions leading to the final success.