To determine how many times you would need to make a small wish to get a legendary item with a 1% chance (or a probability of 0.01), we can use the concept of expected value and probabilities.
The probability of not getting a legendary on a single wish is:
\[ P(\text{not legendary}) = 1 - P(\text{legendary}) = 1 - 0.01 = 0.99 \]
If we make \( n \) wishes, the probability of not getting a legendary in \( n \) attempts is:
\[ P(\text{not legendary in } n \text{ wishes}) = 0.99^n \]
We want to find \( n \) such that the probability of getting at least one legendary is high enough. The probability of getting at least one legendary in \( n \) wishes is:
\[ P(\text{at least one legendary in } n \text{ wishes}) = 1 - P(\text{not legendary in } n \text{ wishes}) = 1 - 0.99^n \]
To find the expected number of wishes needed to have a certain probability of getting at least one legendary, we can set \( P(\text{at least one legendary in } n \text{ wishes}) \) to be equal to 0.5 (50%) as a common threshold for "likely".
So we can set up the equation:
\[ 1 - 0.99^n \geq 0.5 \]
Rearranging gives:
\[ 0.99^n \leq 0.5 \]
Taking the natural logarithm on both sides:
\[ \ln(0.99^n) \leq \ln(0.5) \]
This simplifies to:
\[ n \cdot \ln(0.99) \leq \ln(0.5) \]
Since \( \ln(0.99) \) is a negative number, we divide by it (which reverses the inequality):
\[ n \geq \frac{\ln(0.5)}{\ln(0.99)} \]
Calculating this gives:
\[ n \geq \frac{-0.693147}{-0.010050} \approx 68.92 \]
Thus, you would need to make approximately 69 wishes to have at least a 50% chance of obtaining at least one legendary item. If you want to ensure at least one legendary wish with a higher probability, you might decide to make more wishes accordingly.
1 answer