Let me work through your reasoning and demonstrate exactly why it doesn't work.
Just now, I asked my friend to flip 3 coins, one at a time. I then asked her if at least one of the coins came up tails. She said yes.
What's the probability that all three coins came up tails?
Define $A$ as the first coin which came up tails. Define $B$ as the first coin which is not $A$, and then define $C$ has the remaining coin.
If I make your reasoning more explicit, it says something like:
The probability that $A$ is tails, $P(A = T)$, is $1$. The remaining two coins are equally likely to be heads or tails, so $P(B = T) = 1/2$ and $P(C = T) = 1/2$. Finally, the three coins are independent, so the probability that all three coins are tails is the product of these: $P(A = T, B = T, C = T) = 1 \cdot 1/2 \cdot 1/2 = 1/4$.
However, this reasoning isn't correct. We have defined $B$ in an "unfair" way, so that $B$ is actually more likely to be heads than tails.
(What's so unfair about the definition of $B$? If the first coin comes up heads, then $B$ comes up heads, because, according to our definition, $B$ is the first coin. But if the first coin comes up tails, $B$ may come up heads anyway.)
We can see the actual probabilities of $B = T$ and $C = T$ by writing out the seven possible outcomes (all of which are equally likely):
- $HHT$: $B = H$, $C = H$
- $HTH$: $B = H$, $C = H$
- $HTT$: $B = H$, $C = T$
- $THH$: $B = H$, $C = H$
- $THT$: $B = H$, $C = T$
- $TTH$: $B = T$, $C = H$
- $TTT$: $B = T$, $C = T$
By looking at this table, we can see that $P(B = T) = 2/7$ and $P(C = T) = 3/7$. The two events aren't independent, either: $P(B = T)P(C = T) = 6/49$, whereas $P(B = T, C = T) = 1/7 = 7/49$.