how many fish are needed to have probability 2/3 of at least two fish in a bucket? 
Lets say we have $f$ fish and $b = 10$ buckets. You toss the fish independetly and uniformly into the buckets.
Define $A_{f} =$ "There exists a bucket that has at least two fish"
Let $p_{f} = Pr(A_{f})$
What is the lowest value of $f$ such that $p_{f} \ge 2/3$ ?

Here's my intuition:
$f = 1:$
When we are just about to throw the first fish, there are no fish in any buckets, so the probability that it will land in a bucket with another fish is $0$.
$f = 2:$
When we are just about to throw the second fish, the probability that it will land in a bucket with another fish is $1/10$ because there is one bucket with one fish. Similairly, there are $9/10$ other buckets with no fish in them, so  $9/10$ is the probability that it will land in an empty box.
$f = 3:$
Proability of at least two fish is $2/10$, probability of landing in empty bucket is $8/10$.
$...$
I think there is a pattern here: $p_{f} = {\frac{f-1}{10}}$
So for the probability of $2/3$, thats:
$2/3 = \frac{f-1}{10}$
$20/3 = f-1$
$\lfloor23/3\rfloor =  f$
$\lfloor23/3\rfloor =  7$ fish
It cannot be $\ge 10$ because if it were, then the probability would be $1$ (pigeonhole principle). So there must be a lower bound.
Which makes sense to me. Now my question is how can generalize this and turn it into a proper solution, assuming I am correct? If I am not, can anyone explain?

This is sometimes called the birthday problem.
In general if there are $f$ fish and $b$ buckets then the probability that no bucket has two fish is:
$(1-\frac{1}{b})(1-\frac{2}{b})\dots (1-\frac{f-1}{b})$
So the probability that there is at least one bucket is:
$1-(1-\frac{1}{b})(1-\frac{2}{b})\dots (1-\frac{f-1}{b})$.
So we want the smallest $f$ such that:
$(1-\frac{1}{10})(1-\frac{2}{10})\dots (1-\frac{f-1}{10})\leq \frac{1}{3}$
Notice:
$9>3.\overline 3$
$9*8=72>33.\overline 3$
$9*8*7=504>333.\overline 3$
$9*8*7*6=3024\leq3333.\overline 3$
So the answer is $5$ fish
