To solve the problem, we need to find the least value of \( n \) such that the probability of getting at least one doublet when two dice are thrown \( n \) times is greater than \( 80\% \).
### Step-by-step Solution:
1. **Understanding the Probability of a Doublet**:
- When two dice are thrown, the total number of outcomes is \( 6 \times 6 = 36 \).
- The favorable outcomes for getting a doublet (i.e., both dice showing the same number) are \( (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \), which gives us 6 favorable outcomes.
- Therefore, the probability \( P \) of getting a doublet in one throw is:
\[
P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}
\]
2. **Calculating the Probability of Not Getting a Doublet**:
- The probability of not getting a doublet in one throw is:
\[
P(\text{not doublet}) = 1 - P(\text{doublet}) = 1 - \frac{1}{6} = \frac{5}{6}
\]
3. **Finding the Probability of Not Getting a Doublet in \( n \) Throws**:
- The probability of not getting a doublet in \( n \) throws is:
\[
P(\text{not doublet in } n \text{ throws}) = \left(\frac{5}{6}\right)^n
\]
4. **Finding the Probability of Getting At Least One Doublet**:
- The probability of getting at least one doublet in \( n \) throws is:
\[
P(\text{at least one doublet}) = 1 - P(\text{not doublet in } n \text{ throws}) = 1 - \left(\frac{5}{6}\right)^n
\]
5. **Setting Up the Inequality**:
- We want this probability to be greater than \( 80\% \) or \( \frac{4}{5} \):
\[
1 - \left(\frac{5}{6}\right)^n > \frac{4}{5}
\]
6. **Rearranging the Inequality**:
- Rearranging gives:
\[
\left(\frac{5}{6}\right)^n < \frac{1}{5}
\]
7. **Taking Logarithms**:
- Taking logarithms on both sides:
\[
n \log\left(\frac{5}{6}\right) < \log\left(\frac{1}{5}\right)
\]
- Since \( \log\left(\frac{5}{6}\right) \) is negative, we can reverse the inequality:
\[
n > \frac{\log\left(\frac{1}{5}\right)}{\log\left(\frac{5}{6}\right)}
\]
8. **Calculating the Values**:
- Using logarithm values:
\[
\log(5) \approx 0.6990, \quad \log(6) \approx 0.7781
\]
- Therefore:
\[
\log\left(\frac{5}{6}\right) = \log(5) - \log(6) \approx 0.6990 - 0.7781 = -0.0791
\]
- And:
\[
\log\left(\frac{1}{5}\right) = -\log(5) \approx -0.6990
\]
- Now substituting these values:
\[
n > \frac{-0.6990}{-0.0791} \approx 8.83
\]
9. **Finding the Least Integer Value of \( n \)**:
- The least integer value of \( n \) that satisfies this inequality is:
\[
n = 9
\]
### Conclusion:
Thus, the least value of \( n \) (denoted as \( \lambda \)) for which the probability of getting at least one doublet is greater than \( 80\% \) is:
\[
\lambda = 9
\]
