Edited for (hopefully) clarity.
There are three cards on the table facedown, two with the [o] symbol and one with [x]. You must find [x] to "win" and you have two attempts. If you do not find the [x] after your first try, the card you picked is removed and there are now only two cards. One [o] and one [x].
First attempt:
[o] [o] [x]
1/3 chance, 33.3%
Second attempt (assuming first one was unsuccessful):
[o] [x]
1/2 chance, 50%
What is the final probability of picking the [x]?
I thought that Chance AB = Chance A * Chance B, but that would be 1/3 * 1/2 = 1/6, which seems way too low to be correct.
On the other hand 1/3 + 1/2 = 5/6
I also further confused myself by trying to calculate the odds of not picking the [x] using both above methods.
2/3 * 1/2 = 1/3, which is obviously incorrect, since it does not add up to a 1 if you sum the 1/6 from above
But then 2/3 + 1/2 = 7/6 ???? yeah...
Obviously I have no idea what I am doing, can someone please untangle me? Thank you.
You're close but I think you are conflating two different ideas. Namely, if $A$ and $B$ are independent then it is true that $P(AB) = P(A)P(B)$ however this is NOT the probability we want to calculate. We want to calculate the probability that we get our desired outcome. Let's call this event $X$. So we can get $X$ on our first try, let's call this event $C$, or we can get $X$ on our second try, let's call this event $D$. We now note that $X = C \text{ or } D$, also written as $X = C\cup D$, hence what we really want to find is $P(C\cup D)$. There are a few easy ways going about this so I'll go over one such way. We first note that since $C$ and $D$ can't occur at the same time then (this is called the property of mutual additivity) then we have that $$P(C \cup D) = P(C) + P(D)$$ You correctly pointed out that $P(C) = \frac{1}{3}$ so it begs the question, what is $P(D)$? Informally, for $D$ to occur we have to fail the first guess and get the correct outcome on the second. Hence $$P(D) = P(\text{first guess is incorrect})(\text{second is correct}) = \frac{2}{3}\frac{1}{2} = \frac{1}{3}$$ and hence $$P(C \cup D) = P(C) + P(D) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$$