D&D Probability of 2 Attempts

Question

I wanted to get the full probability of 2 attempts made at 60% chance of success.

I was looking at a different chain of math and found my probability to hit an enemy is 60% per each attack but I was wondering how it would look at all the outcomes and the probability of it.

6/10 * 6/10 = 36%

of both attempts failing and both attempts succeeding with a 64% chance of at least 1 attempt succeeding?

I didn't understand how it would look as a failure adding up to 136%

Answer 1

Scenario 1: Both Hit

6/10 * 6/10 = 36%

Scenario 2: Hit first, miss second

6/10 * 4/10 = 24%

Scenario 3: Miss first, hit second

4/10 * 6/10 = 24%

Scenario 4: Miss Both

4/10 * 4/10 = 16%

Notice how this total adds to 100%. If you define success as at least one hit, then you may add scenarios 1-3 or take 1-P(Scenario 4).

Answer 2

You got 136% by adding $60\%+40\%+36\%$, i.e. $P(S) + P(F) + P(SS)$. This is incorrect because these outcomes do not partition sample space (full set of outcomes).

Your sample space are the following outcomes: $$ \Omega = \{SS, SF, FS, FF\} $$ The probability of these 4 events must sum to 1. Here, $P(SS)=0.36$, assuming outcomes are independent and two failues, $P(FF) = 0.24$. You can compute the rest.