Forrest has a bag of $n$ coins. You know that $k$ are biased with probability $p$ (the biased coins have probability $p$ of being heads). Let $F$ be the event that Forrest picks a fair coin; let $B$ be the event that he picks a biased coin. Forrest draws 3 coins from the bag but doesn't know which are biased/fair. (There is no replacement/repetition of the coins.)
What is the probability that the 3rd coin he draws is biased?
There are 4 possible cases: FFB, BFB, FBB, BBB, and hence the probability would be
$$ \frac{k(n-k)(n-k-1)}{n(n-1)(n-2)} + 2 \cdot \frac{k(k-1)(n-k)}{n(n-1)(n-2)} + \frac{k(k-1)(k-2)}{n(n-1)(n-2)}$$
Evaluating this yields $\frac{k}{n}$, but there is apparently a simpler way using symmetry:
Since we don't know which coins were picked before the 3rd coin, we can just think of the 3rd coin being biased as being equivalent to the 1st coin being biased, which gives us $k/n$.
Unfortunately, this argument doesn't make much sense to me at all. The probabilities are all dependent, so how does this argument actually work, when we ignore the previous cases entirely by switching the 3rd choice to be the 1st choice instead? Could someone help me understand why this argument is true / why this way of thinking works? (Conceptual or mathematical explanations would both work.) Also, in which (other) probability situations can I apply these way of thinking?
Suppose that instead of drawing 3 of the coins out, Forrest instead takes all the coins out and arranges them in a line. Any individual coin will have a $\frac{k}{n}$ probability of being biased, regardless of where it is in the line, so if he then picks up the first three, ignores two of them and just looks at the third, then it should still have that same $\frac{k}{n}$ probability.