2016 is absolutely awesome 11

For every nonnegative integer $n$ , let $D_n$ be the number of dice on the table after $n$ steps.

At the start, $D_0 = 1$ .

Given $D_n$ , we are told that $D_{n +1}$ is $D_n + 1$ with probability $6^{-D_n}$ and is $D_n$ otherwise.

Hence the expected value of $6^{D_{n +1}}$ , given $D_n$ , is

E(6^{D_{n +1}}\mid D_n) = 6^{-D_n}. 6^{D_n +1}+(1 - 6^{-D_n})
.6^{D_n} = 6^{D_n} + 5 .

Averaging over $D_n$ shows that $E(6^{D_{n +1}}) = E(E(6^{D_{n +1}}\mid D_n)) = E(6^{D_n} + 5) = E(6^{D_n}) + 5$ .

Each expected value is $5$ more than the previous one, so we have:

$E(6^{D_{2016}}) = E(6^{D_0}) + 2016. 5 = E(6^1) + 10080 = 6 + 10080 = \boxed{10086}$ .

Let be $P_{n,d}$ the probability of having $d$ , $1 \leq d \leq n+1$ , dices at step $n$ . It follows that

$\displaystyle P_{n,d}=\begin{cases} P_{n-1,d}\cdot \frac{6^d-1}{6^d}+P_{n-1,d-1}\cdot \frac{1}{6^{d-1}}, & \mbox{for} \ 1 \leq d \leq n \\ \frac{1}{6^{n(n+1)/2}}, & \mbox{for} \ d=n+1 \end{cases}$

So, the expected value of $6^d$ after $n$ steps is

$\displaystyle E_n=\sum_{d=1}^{n+1} 6^d P_{n,d} = \sum_{d=1}^{n} 6^d P_{n,d} + \frac{6^{n+1}}{6^{n(n+1)/2}}$

If we expand the last sum and substitute $P_{n,d}$ , we get

$\displaystyle \sum_{d=1}^{n} 6^d P_{n,d}=\sum_{d=1}^{n}\bigg[ P_{n-1,d}\cdot (6^d-1)+6P_{n-1,d-1}\bigg]=\sum_{d=1}^{n} 6^d P_{n-1,d}-\sum_{d=1}^{n} P_{n-1,d} + 6\sum_{d=1}^{n} P_{n-1,d-1}=E_{n-1}-1+6\bigg(1-\frac{1}{6^{(n-1)n/2}}\bigg)$

Hence

$\displaystyle E_n=E_{n-1}+5-\frac{6}{6^{n(n-1)/2}}+\frac{6^{n+1}}{6^{n(n+1)/2}}$

Given that $E_0=6$ , solving the recurrence we get

$\displaystyle E_n=5n+6$

Eventually

$\displaystyle E_{2016}=5\cdot 2016+6=\boxed{10086}$