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Osrs Wiki Drop Calculator

OSRS Drop Formula:

\[ P = 1 - (1 - p)^k \]

(0-1)
attempts

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1. What is the OSRS Drop Probability Formula?

The OSRS (Old School RuneScape) drop probability formula calculates the overall chance of receiving at least one drop after multiple attempts. It's based on the complementary probability of not receiving the drop across all attempts.

2. How Does the Calculator Work?

The calculator uses the OSRS drop probability formula:

\[ P = 1 - (1 - p)^k \]

Where:

Explanation: The formula calculates the probability of getting at least one successful drop by subtracting the probability of never getting the drop from 1.

3. Importance of Drop Probability Calculation

Details: Understanding drop probabilities helps players estimate how many attempts they might need to obtain rare items, plan their gameplay time, and set realistic expectations for item acquisition.

4. Using the Calculator

Tips: Enter the drop probability per attempt (as a decimal between 0 and 1) and the number of attempts. All values must be valid (p between 0-1, k ≥ 1).

5. Frequently Asked Questions (FAQ)

Q1: What does a 1% drop chance mean after 100 attempts?
A: With a 1% (0.01) drop chance per attempt, after 100 attempts you have approximately a 63.4% chance of receiving at least one drop.

Q2: How many attempts are needed for a 90% chance at a rare drop?
A: For a drop with probability p, you need approximately log(1-0.9)/log(1-p) attempts. For a 1/1000 drop, this would be about 2300 attempts.

Q3: Does killing multiple monsters at once affect the probability?
A: No, the formula assumes independent attempts. Each kill is a separate chance at the drop, regardless of how many monsters you kill simultaneously.

Q4: Are drop chances always independent in OSRS?
A: Most drop mechanics in OSRS use independent probabilities, but some special items or events may have different mechanics.

Q5: What's the probability of getting exactly one drop in k attempts?
A: This follows the binomial distribution: \( \binom{k}{1} \times p \times (1-p)^{k-1} \), which is different from the "at least one" probability calculated here.

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