What is the expected number of red apples left when all the green apples are picked? We have 4 green apples and 60 red apples. Each time we pick one out without replacement. Then what is the expected number of red apples left when all 4 green apples are picked?

Two arguments:
I. Brute Force
For $i\in \{0,1,\cdots, 60\}$ let $p_i$ denote the probability that exactly $i$ red ones remain. Then the answer we want is $$E=\sum_{i=0}^{60}i\times p_i$$
But $p_i$ is easily computed:  we need the last apple drawn to be green, which has probability $\frac 4{64}$.  Conditioned on that, we need all three of the remaining green ones to be drawn ahead of that one. But there were $63-i$ chosen before the last green (out of a possible $63$), three of which were also green.  the number of ways to do that is $\binom {60}{60-i}=\binom {60}{i}$ out of a total number of combinations equal to $\binom {63}{63-i}=\binom {63}{i}$ Thus, the conditional probability is the ratio $$\frac {\binom {60}{i}}{\binom {63}{i}}=\frac {(63-i)\times (62-i)\times (61-i)}{63\times 62\times 61}$$  and $$p_i=\frac 4{64}\times \frac {(63-i)\times (62-i)\times (61-i)}{63\times 62\times 61}$$
Sanity checks: let's compute $p_{60}$.  The only way all the red apples can be left behind is if the first $4$ were green.  The probability of that is $\frac 4{64}\times \frac 3{63}\times \frac 2{62}\times \frac 1{61}$ which matches the result of the formula.  it is also a simple matter (at least with a machine) to confirm that $\sum p_i=1$.
Now it's easy to compute $E$ directly (using a machine, at least).  We get $$\boxed {E=12}$$
II.  Symmetry
Working backwards, the answer is the expected gap between the end of the string of apples and the last green apple drawn.  But the expected lengths of the gaps are all the same...there are $5$ gaps in all, hence $E=\frac {60}{5}=12$
To gain intuition on the symmetry, I find it useful to imagine that there was one blue apple in the mix.  You then arrange the, now $65$ apples in a circle.  It's clear I think that, in this context, the red strings should all have expected length $12$.  You obtain a linear  order for your original $64$ apples by deleting the blue one and straightening out the circle.
