Question

in this game there are item boxes and for the best item its a 0.2% chance of getting it how many item boxes do i need to get it

Answer 1

To determine how many item boxes you need to open to have a certain probability of getting the best item with a 0.2% chance (or 0.002 in decimal form), you can use the concept of expected probability.

If you want to know how many boxes you need to open to have at least a certain probability of getting the item, you can use the formula for the complement probability:

1. **Let \( p \) be the probability of success (getting the item)**: \( p = 0.002 \).
2. **Let \( n \) be the number of attempts (item boxes opened)**.
3. The probability of **not** getting the item in one attempt is \( 1 - p = 0.998 \).
4. The probability of **not** getting the item in \( n \) attempts is \( (1 - p)^n = (0.998)^n \).
5. Therefore, the probability of getting at least one item in \( n \) attempts is \( 1 - (0.998)^n \).

To find \( n \) for a certain probability threshold (let's say you want at least a 50% chance of getting the item), you can set up the inequality:

\[
1 - (0.998)^n \geq 0.5
\]

Solving this:

\[
(0.998)^n \leq 0.5
\]

Taking the natural logarithm of both sides:

\[
n \cdot \ln(0.998) \leq \ln(0.5)
\]

Now, calculate \( n \):

\[
n \geq \frac{\ln(0.5)}{\ln(0.998)}
\]

Calculating the values:

- \( \ln(0.5) \approx -0.693147 \)
- \( \ln(0.998) \approx -0.002002 \)

Now plug in these values:

\[
n \geq \frac{-0.693147}{-0.002002} \approx 346.257
\]

Thus, you would need to open approximately **347 item boxes** to have at least a 50% chance of getting the item with a 0.2% drop rate.

You can adjust the percentage of success you're aiming for by repeating this calculation with a different threshold.