Expected number of days before all magical seeds become apple trees This is a question that I came across recently.
At the end of day $0$, $6$ magical seeds are planted. On each day following it, each seed has a chance to magically transform into an apple tree with a probability of $\frac{1}{2}$. The outcomes of seeds are independent of each another.
What is the expected number of days for all six seed to have become apple trees?
My solution:
$E(n)$ - number of expected days from the point that there is only n seed(s) left.
So, $E(1)$ - number of expected days for the last seed to grow.
$E(1) = 1 + \frac{1}{2} E(1) \,or \,E(1) = 2$. This we anyway know from a coin flip analogy.
$E(2) = 1 + \frac{2}{4} E(1) + \frac{1}{4} E(2) \,or\, E(2) = \frac{8}{3}$.
This comes from the fact that if at the end of a day, two seeds are left, I have 3 possible events - i) both seeds become trees the next day ($+1$ day). ii) one seed becomes tree and one seed is left (probability $\frac{2}{4}$). So we further add expected number of days for $E(1)$. iii) None of the seeds become a tree (probability $\frac{1}{4}$). So we add further expected number of days for $E(2)$.
Similarly, $E(3) = 1 + \frac{3}{8} E(1) + \frac{3}{8} E(2) +  \frac{1}{8} E(3)$
$E(4) = 1 + \frac{4}{16} E(1) + \frac{6}{16} E(2) +  \frac{4}{16} E(3) + \frac{1}{16} E(4)$
$E(4) = 1 + \frac{4}{16} E(1) + \frac{6}{16} E(2) +  \frac{4}{16} E(3) + \frac{1}{16} E(4)$
$E(5) = 1 + \frac{5}{32} E(1) + \frac{10}{32} E(2) +  \frac{10}{32} E(3) + \frac{5}{32} E(4) + \frac{1}{32} E(5)$
$E(6) = 1 + \frac{6}{64} E(1) + \frac{15}{64} E(2) +  \frac{20}{64} E(3) + \frac{15}{64} E(4) + \frac{6}{64} E(5) + \frac{1}{64} E(6)$
This gives me an answer of $E(6) = \frac{55160}{13671}$. However the answer given is $(\log_2 6)$. I do not understand how the answer got into $\log$. When I calculate both, they are not same values.
Also, are there more generic and faster methods that I could use to get to the answer?

Let $X_i$ be the r.v. corresponding to the number of the day where the $i$-th seed become an apple tree. Obviously, for all $i=1, ..., 6$ and for all $n \geq 1$,
$$\mathbb{P}(X_i = n) = \frac{1}{2^n}$$
What you are asked to compute is
$$\mathbb{E}(\max(X_i))$$
But for all integer $n \geq 1$,
$$\mathbb{P}(\max(X_i)=n) = \mathbb{P}(\max(X_i) \leq n) - \mathbb{P}(\max(X_i) \leq n-1) = \prod_{i=1}^6 \mathbb{P}(X_i \leq n) - \prod_{i=1}^6 \mathbb{P}(X_i \leq n-1)$$
$$=\left( 1 - \frac{1}{2^n} \right)^6-\left( 1 - \frac{1}{2^{n-1}} \right)^6$$
And $$\mathbb{E}(\max(X_i)) = \sum_{n=1}^{+\infty} n \left[\left( 1 - \frac{1}{2^n} \right)^6-\left( 1 - \frac{1}{2^{n-1}} \right)^6\right]$$
