Drop chance calculator

No. Each attempt is independent, so the chance of the item never dropping is 0.99 multiplied by itself a hundred times, which is 36.6 percent. That leaves 63.4 percent, not 100.

The number barely moves when you scale it: 0.1 percent over a thousand tries and 5 percent over twenty both land near 63 percent as well. It is the same limit showing up every time, and it is why grinding exactly one over the drop rate feels unlucky so often.

Being unlucky past that point is not rare either. After two hundred tries at 1 percent, more than one player in eight still has nothing.

For a single drop it is one minus the chance of failing every time: 1 - (1 - p) raised to the number of attempts. For more than one drop it is the binomial tail, which adds up the chance of getting zero, one, two and so on up to one less than what you want, and subtracts that from one.

The arithmetic is done in logarithms because the binomial coefficient of a hundred thousand attempts is far past the largest number a browser can hold. Done directly it would overflow to infinity and the result would come out as an error instead of a probability.

In a plain random drop, no. The dice have no memory, so the chance on your two hundredth attempt is exactly the same as on your first, no matter how long the streak has been. Expecting it to be otherwise is the gambler's fallacy.

Some games do add pity timers or bad luck protection, where the rate rises after a number of failures. This calculator assumes a fixed rate, so if your game has such a system the real chance is better than the number here.

It is the average across many players in your situation, not a prediction for you. Two hundred attempts at 1 percent gives two expected drops, and no single player receives two point zero of anything.

It is still the most useful single number for planning, because it scales: double the attempts and the expectation doubles, which the probability of at least one does not do.