Roll 1d6: How many tries will it take to get at least one ace?

In order to answer this question, we have to quantify a willingness to be wrong. Tolerance for risk, or aversion to misadventure is a factor in the equation. Science types often express this as a level of certainty, or confidence in the outcome.

It also turns out that the math is best done in a mirror universe type of way, where we look at the chance of failure instead of the chance of success.

$$numberOfTries=log(chanceOfMisadventure)/log(chanceOfFailureEachTry)$$

All you have to do is decide how much of a chance of misadventure you're willing to gamble in exchange for accuracy of the prediction.

I happen to like 95% certainty, an allowance to be wrong once in twenty, but your mileage may vary.

## Example 1

I want to be 95% sure (5% wrong) of getting at least one ace (5/6 chance of failure):

$$tries = log(5\%)/log(5/6)=16.4 \ tries$$

So it will cost me whatever the price is for 16.4 tries to get at least one ace, at my chosen level of confidence designated as chance of misadventure.

## Example 2

There is a 1% chance of success; how many times for 95% certainty of at least one success?

$$tries = log(5\%)/log(99\%)\approx 300 \ tries$$

A more in-depth coverage of loot drop mechanics can be found here.