Question

how many times do you have to roll to get
a 0.1 out of 99.9

Answer 1

To address your question, we first need to clarify your scenario: Are you asking how many times you need to roll a die (or perform some action) to achieve a result that has a probability of 0.1 (10%) when the outcome is 99.9% out of another total or context?

If you're asking how many times you need to perform a trial (like rolling a die) to have at least a 10% chance of success (with 'success' defined as achieving some outcome that has a probability of 0.1), we can use the binomial probability formula.

To achieve at least a 10% chance of getting at least one success when each individual trial (roll) has a 0.1 probability of success, you can use the complement probability:

1. The probability of failure in one roll is 0.9.
2. The probability of failing in \( n \) rolls is \( 0.9^n \).
3. Therefore, the probability of at least one success in \( n \) rolls is \( 1 - 0.9^n \).

We want this probability to be at least 0.1:

\[
1 - 0.9^n \geq 0.1
\]

Solving for \( n \):

\[
0.9^n \leq 0.9
\]

Taking the logarithm of both sides:

\[
n \log(0.9) \leq \log(0.1)
\]

Since \( \log(0.9) \) is negative, we can simplify:

\[
n \geq \frac{\log(0.1)}{\log(0.9)}
\]

Now calculate those values:

\[
\log(0.1) \approx -1 \quad \text{and} \quad \log(0.9) \approx -0.045757
\]

Thus,

\[
n \geq \frac{-1}{-0.045757} \approx 21.85
\]

So, you would need to roll approximately 22 times to have at least a 10% chance of getting at least one success (with a 10% success rate on each trial).

If you were referring to something else, please clarify your question!