To solve this problem, we need to calculate the probability of getting a double six in multiple independent throws of two dice.

Step 1: Single Throw Probability

The probability of rolling a double six in a single throw of two dice is: $P(\text{double six}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

Step 2: Probability of Not Getting a Double Six

The probability of not getting a double six in a single throw is: $P(\text{not double six}) = 1 - \frac{1}{36} = \frac{35}{36}$

Step 3: Probability of Not Getting a Double Six After $n$ Throws

The probability of not getting a double six after $n$ independent throws is: $\left( \frac{35}{36} \right)^n$

Step 4: Probability of Getting At Least One Double Six

The probability of getting at least one double six in $n$ throws is the complement: $P(\text{at least one double six}) = 1 - \left( \frac{35}{36} \right)^n$

We want this probability to be at least 0.50 and 0.90.

Step 5: Solving for $n$

1. For $P \geq 0.50$:

$1 - \left( \frac{35}{36} \right)^n \geq 0.50$ $\left( \frac{35}{36} \right)^n \leq 0.50$ Taking the natural logarithm of both sides: $n \ln\left( \frac{35}{36} \right) \leq \ln(0.50)$ $n \geq \frac{\ln(0.50)}{\ln\left( \frac{35}{36} \right)}$

Calculating:

$n \geq \frac{\ln(0.50)}{\ln\left( \frac{35}{36} \right)} \approx \frac{-0.6931}{-0.0286} \approx 24.24$

Rounding up, we get $n = 25$.

2. For $P \geq 0.90$:

$1 - \left( \frac{35}{36} \right)^n \geq 0.90$ $\left( \frac{35}{36} \right)^n \leq 0.10$ Taking the natural logarithm of both sides: $n \ln\left( \frac{35}{36} \right) \leq \ln(0.10)$ $n \geq \frac{\ln(0.10)}{\ln\left( \frac{35}{36} \right)}$

Calculating:

$n \geq \frac{\ln(0.10)}{\ln\left( \frac{35}{36} \right)} \approx \frac{-2.3026}{-0.0286} \approx 80.50$

Rounding up, we get $n = 81$.

Final Answers

• For $P \geq 0.50$: $n = 25$ throws
• For $P \geq 0.90$: $n = 81$ throws

Your answer for $0.50$ is correct, but for $0.90$, the correct number of throws is 81.

Would you like any further details or have any questions?

Here are 5 related questions you might find interesting:

1. How would the probability change if we increased the number of dice?
2. What if we wanted a probability of at least 0.95 for a double six?
3. How does the expected value of the number of throws relate to these probabilities?
4. How would the calculations change if we used loaded dice?
5. Can we use a different approximation method, like Poisson distribution, for this problem?

Tip: When solving probability problems involving multiple trials, always consider using complementary probabilities and logarithmic transformations for easier calculations.