Rolling with Advantage

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I was trying to figure out how to quantify the benefits of rolling with Advantage in D&D and I’m coming up short.

For anyone unfamiliar with the idea, it’s simply rolling a 20-sided die twice and taking the more favorable roll.

Correct me if I’m wrong, but with a single roll my probability of success would be:

Rolls resulting in success ÷ total possible rolls.

My question, and I apologize for the rambling, is how would I figure out the aggregate chance of success with a second roll?

In: Mathematics

8 Answers

Anonymous 0 Comments

Check for failure. If you need to roll a 6 to beat the check, then you fail 5/20 times, or 25% of the time.
If you roll with advantage, you square that chance of failure. (5/20)*(5/20)=fail 6.25% of the time.

If it’s a straight coin flip, then a normal roll is 1/2 and an advantaged roll would be 1/2 * 1/2, or 1/4 chance of failure.

The benefit is very different depending on what you need to roll.

Edit: math is hard

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