So the question is, if two consequtive coin tosses occured and we know, in the aftermath, that at least of them resulted in a tails, what is the probability that they were both tails?
The common answer to this question is 1/3, with the arguement that there are four possible outcomes for two coin tosses: (HH, HT, TH, TT) and that by by knowing at least one tails, we eliminate HH so the sample space becomes (HT, TH, TT) in which only TT would satisfy the condition, hence 1/3.
However, here is what I think the correct way to represent this problem would be:
The sample space for two coin tosses = (HH, HT, TH, TT) The sample space for two coin tosses, knowing that one yielded a tails= (HT, TH, TT, TT), where bold indicates that we know about that outcome. Obviously, in this way, the likelyhood of TT is 1/2.
Another way to think about this, is that on a probability tree, either we know about the first even giving a T, or we know abou the second event giving a T. Either way, there's a 1/2 chance that the other event lead to a T.
I tagged this as fake proof, because it goes against every other answer I've seen to this question, on this board or elsewhere, so I expect to be proven wrong, but I haven't seen this argument made anywhere else for the answer being 1/2.
Here is another argument giving the answer $1/3$: Letting $TT$ stand for the event "Two tails" and $T$ stand for "At least one tails", we have $$ P(TT \mid T) = \frac{P(TT\cap T)}{P(T)} = \frac{P(TT)}{P(T)} = \frac{1/4}{3/4} = \frac{1}{3} $$
Yet another argument is this: Let a lot of people do two coin flips, and tell all who got two heads to leave the room. You're now in the "$\hphantom{}\mid T$" situation. Only one third of the people there flipped two tails.
What you're thinking about might be this: Pick a coin that was tossed tails. Then the probability that its partner was also tails is $1/2$. The difference between the two is that the $TT$ pairs are twice as likely to be picked in this way, compared to the "pick a random pair and examine it" way.
Because, in the end, that's what the conditional probability $P(A\mid B)$ means. It means "Repeat the entire experiment (e.g. flip two coins, or pick a family with three kids, or look at the top card of a deck) until you have a sample fulfilling $B$. What is the probability that it fulfills $A$ as well?"