Expected number of turns or rolls to end "death roll" game In the game World of Warcraft, I came across this game called "death rolling". (See similar post here Death rolls - 50/50?).
The game has $2$ players. The first player types /roll $N_0$ which generates a random integer $N_1$  in the interval $[1,N_0]$. The second player similarly generates a random integer $N_2$ in the interval $[1,N_1]$. This process continues iteratively alternating between Player 1 and Player 2 until one of the players rolls $1$, and this player is deemed the loser of the game.
What is the expected number of rolls $m$ in this game in terms of $N_0$?
EDIT: I think this question is actually a duplicate with Average Length of Random Number Generation with Decreasing Range of Numbers . The answer provided there makes pretty good sense, but I don't quite follow the final step. Can anyone help me understand their justification for the final step?
 A: The probability that the game continues, on any roll, is $\frac{N_1-1}{N_1}$. The probability that it ends is $\frac{1}{N_1}$.
You can use the above to get the probability that the game ends on turn $x$.  It has to not end, not end, not end, $x-1$ times, and then end on the last turn.  Then you can plug into the summation formula.  To calculate the expected value, you need to find $\sum x p(x)$.
A: Starting with the answer by heropup, call $a_n = E[M_n]$, giving the recurrence:
$\begin{align*}
  a_n
    &= \frac{n}{n - 1} + \frac{1}{n - 1} \sum_{1 \le k \le n - 1} a_k
\end{align*}$
Write as:
$\begin{align*}
  a_{n + 1}
    &= \frac{n + 1}{n} + \frac{1}{n} \sum_{1 \le k \le n} a_k
\end{align*}$
Multiply by $n$, subtract the expression for  $a_n$ from the one for $a_{n + 1}$ to get rid of the troublesome sum:
$\begin{align*}
  n a_{n + 1} - (n - 1) a_n
    &= (n + 1) - n + a_n \\
  n a_{n + 1} - n a_n
    &= 1
\end{align*}$
The last one is a linear recurrence of the first order. Add the initial condition $a_1 = 1$ this gives:
$\begin{align*}
  a_n - a_{n - 1}
   &= \frac{1}{n} \\
  a_n
   &= \sum_{1 \le k \le n} \frac{1}{k}
\end{align*}$
This is just the harmonic numbers:
$\begin{align*}
  a_n
    &= H_n \\
    &= \gamma + \ln n + O(1/n)
\end{align*}$
where $\gamma = 0.5772156649$ is the Euler constant.
