Jerry looks under the chair and, seeing both coins says, "At least one of them is tails." What is the probability that Tia wins? 
Tia and Jerry are tossing two coins. Tia wins when both coins turn up tails. The coins are tossed but roll under a chair. Jerry looks under the chair and, seeing both coins says, "At least one of them is tails." What is the probability that Tia wins?

Because it says at least I know I should be doing $1 -$ however, I'm not sure how to apply this, the odds of the other coin being tails is $1/2$, so wouldn't you say the probability of Tia winning be $50\%$, because it's based off that one coin?
 A: Be careful! Just because a problem says "at least" one does not mean you do "$1-$"! There are problems where "at least one" has to be approached directly.
There are a couple of ways to think about this.


*

*If I call $N$ the number of heads in two tosses, then the probability of interest is
\begin{align*}
P(N = 2|N\geq 1)&= \frac{P(N=2, N\geq 1)}{P(N \geq 1)}\\
&=\frac{P(N=2)}{1-P(N=0)}\\
&=\frac{(1/2)^2}{1-(1/2)^2}\\
&=\frac{1}{3}
\end{align*}

*Alternatively, notice that the possible outcomes are
$$\text{(H,T),(H,H),(T,H),(T,T)}$$
all equally likely. They tell us that we have at least one head, hence our outcome space has reduced to 
$$\text{(H,T),(H,H),(T,H)}$$
So, the probability of interest is
$$\frac{\text{# of winning outcomes}}{\text{# possible outcomes}} = \frac{1}{3}.$$
A: HINT:
There are $3$ equally likely options: TT, TH, HT.
A: These things come down to phrasing.  A priori there are $4$ equally likely states of the world:  $HH,HT, TH, TT$ .  Jerry's comment is probably best interpreted as "$HH$ did not occur."  In that case there are three equally likely outcomes remaining and, as Tia wins in exactly one of these, the answer is $\frac 13$.
It's very different if Jerry says "the coin on the left came up $T$."  Then your argument holds and the answer is $\frac 12$.  Otherwise said, the possible states would then be $TH$ or $TT$.
