Simple coin flip probability There's are many people saying either one of two answers for this:
If you flip a coin twice, what's the chance that both will be heads given 
that at least one will be heads?

The more "educated" answer is that the three possible outcomes are H/H, H/T, T/H so the chance is 33.33% The more obvious answer is that since one is heads, you should only count it as a 1 coin toss so it's a 50% chance.
I think 50% is correct even though the above explanation is wrong.
So, one of the coin flips is a guaranteed heads flip..
Case 1: The first flip is is the guaranteed heads: Outcomes: H/T, H/H
Case 2: The second flip is the guaranteed heads: Outcomes: T/H, H/H
H/H should be counted as 2 outcomes, like T/H and H/T are. They're both the same in quantity (i.e T/H, H/T is 1 heads 1 tails, H/H, H/H is 2 heads)
The thing that differs is the ORDER, in which case (guaranteed)H/H and H/(guaranteed)H should be counted as two different outcomes, therefore making it 50%. Am I wrong here?
 A: Let $E_{HH}$ be the event that both coin tosses are heads, $E_{HT}$ (resp. $E_{TH}$) resp. that the first is Heads and the second Tails (resp. the converse). Letting $E$ be the event that at least one coin flip is Heads, you get exactly $$E=E_{HH}\cup E_{HT}\cup E_{TH}$$ and this is a disjoint union.
You want to compute $\Pr[E_{HH}\mid E]$, which can be written
$$
\Pr[E_{HH}\mid E] = \frac{\Pr[E_{HH}\cap E]}{\Pr[E]} = \frac{\Pr[E_{HH}]}{\Pr[E_{HH}]+\Pr[E_{HT}]+\Pr[E_{TH}]}
$$
as $E_{HH}\cap E = E_{HH}$. As the coins are fair and independent, this is 
$\Pr[E_{HH}\mid E] = \frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}} = \frac{1}{3}.$
A: Your analysis is correct, but for a different question: Suppose you flip two coins.  You pick one of the coins at random, and examine it.  It proves to be heads.  The probability that the other coin is heads also is $1/2$, for the reasons you provide.
If you represent the two coin flips as binary variables $A$ and $B$, with heads being $1$ and tails being $0$, then the difference between the original question, and the question you really answered, is that the original question is asking for the probability
$$
P(A \land B \mid A \lor B) = \frac{1}{3}
$$
where $\land$ and $\lor$ are logical AND and OR, respectively, while the question you actually answered asks for the expression
$$
P(B \mid A) \cdot P(A) + P(A \mid B) \cdot P(B) = \frac{1}{2}
$$
A: I think, in your case, order counts: $H, T \ne T,H$ since they are two different events, and they have their own possibilities. Task only says that one $H$ is guaranteed, but doesn't say anything about, when is it guaranteed, therefore the answer is indeed $33,3$%.
A: The point is that there are more ways to get exactly one head if you have two coins and don't assume anything, compared to the number of ways to get exactly one head if you assume that the first coin (or second coin) is heads or tails. This is in essence why the probability is indeed $1/3$ and not $1/2$.
If you want a rigorous argument, just take the 4 equally likely and disjoint possibilities $TT, HT, TH, HH$. The law of conditional probability says that the probability of getting $HH$, assuming you have at least 1 head, is $P(HH)/(P(TH) + P(HT) + P(HH))$. So it's 1/3.
