Independence of coin flips 
A fair coin is tossed three times in succession. If at least one of
  the tosses has resulted in Heads, what is the probability that at
  least one of the tosses resulted in Tails?

My argument and answer: The coin was flipped thrice, and one of them was heads. So we have two unknown trials. The coin flips are all independent of each other, and so there is no useful information to be derived from the fact that one of them was heads. The probability of getting at least one Tails in these two trials is $ \frac 12 + \frac 12 - \frac 14 = \frac 34 $. 
The given answer: $ \frac 67 $. The answer proceeds as follows: Initially the sample space consists of 8 events. We now know that one of those events can't happen (TTT can't happen because one of them were heads).6 of the remaining 7 events have at least one tail, and so the probability is $ \frac 67 $.
Why is my answer wrong? What am I missing?
 A: The answer $\frac{6}{7}$ is correct. Whatever the issue in your reasoning, I feel it must lie in the statement "there is no useful information" in saying one of the 3 flips was heads. To me it seems there is a difference between reading off the results of coins already flipped and predicting (as you do in your argument) the results of future flips based on previous ones. Clearly, as you mention, each flip is independent, but perhaps information about some of the flips is useful if all of the flips have already been made.
To illustrate this point, I think it helps to take this problem to the extreme. Suppose we perform a million coin flips. I don't tell you the results of each coin flip, but I do tell you that 999,999 of the flips yielded heads. I now ask you to guess the result of the remaining flip. What should you answer? (I recommend thinking about this yourself, then look at my answer below).
|
|
|
|
|
|
|
|
Answer: You may suspect that you have a 50-50 shot, but that is not the case. The crucial point is I did not tell you which of the 999,999 coins were heads. Had I made this specification it would be equally likely. However, there are, in fact, one million possible sequences of flips in which only one is tails. Here they are:
\begin{equation}
T\, H\, H\, H\, ...\, H \\
H\, T\, H\, H\, ...\, H \\
H\, H\, T\, H\, ...\, H \\
.  \\
.  \\
.  \\
H\, H\, H\, H\, ...\, T \\
\end{equation}
In contrast, only one sequence has all heads. Assuming sequences of flips are equally likely, we must conclude that guessing tails is the appropriate choice. Notice that we cannot solve this problem correctly by reducing the size of our string of flips to one, similar to your method. We had to consider the flips as part of a larger sequence. I hope this helps!
A: Well, let's look at the problem. It's asking the odds of flipping any tail given that you flipped at least one head, or $P(T>0|X>0)$  Using the definition of conditional probability, we can say that $P(T>0|X>0)=P(T>0,X>0)/P(X>0)$. Then, using some identities, we have that $P(T>0,X>0)=1-P(T=0)-P(X=0)$ and $P(X>0)=1-P(X=0)$.
To put it in words, the probability that our coin toss triple contains one head and one tail is 1 minus the odds of it containing no heads or no tails, and the probability that it held at least one head is 1 minus the odds that it contained no heads.
Then, doing the remaining math:
$$\frac{1-P(T=0)-P(X=0)}{1-P(X=0)}=\frac{1-\frac{1}{8}-\frac{1}{8}}{1-\frac{1}{8}}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$$
Your problem is that you specifically looked at the case when one head was flipped- you did not account for other cases.
A: 
The coin flips are all independent of each other, 

Here's the error in your reasoning.
The coin flips are not conditionally independent under constraint that at least one is a head.  

Let $H_k$ be the event that trial $k$ shows a head, $T_k$ be the event that it is a tail, and the evidence that at least one among them shows a head is: $E = H_1\cup H_2\cup H_3$
Obviously under that condition, each particular trial will still have a non-zero probability that it might be a tail.  
$$\mathsf P(T_1\mid E)~\mathsf P(T_2\mid E)~\mathsf P(T_3\mid E) >0$$
However, there the conditional probability that all trials are tails is zero under the condition that at least one of them is a head.
$$\mathsf P(T_1\cap T_2\cap T_3\mid E)=0$$
