Asymptotics of system of linear equations I have a system of linear equations as follows.

$$M(p) = 1+\frac{n-p-1}{n}M(n-1) + \frac{2}{n} N(p-1) + \frac{p-1}{n}M(p-1)$$
   $$N(p) = 1+\frac{n-p-1}{n}M(n-1) + \frac{p}{n}N(p-1)$$
   $$M(1) = 1+\frac{n-2}{n}M(n-1) + \frac{2}{n}N(0)$$
   $$N(0) = 1+\frac{n-1}{n}M(n-1)$$

$M(p)$ is defined for $1 \leq p \leq n-1$.  $N(p)$ is defined for $0 \leq p \leq n-2$.  What is $M(n-1)$?
 A: As I noted in a comment, we can reformulate the problem as follows: We're looking for two particular elements of a set with n elements. In each time step we uniformly randomly draw one element with replacement. If we draw the same irrelevant element twice before finding both of our elements, we have to start over. What's the expected time before we find the two elements?
Incidentally, I suspect you would have gotten an answer long ago if you'd posed the question in that or a similar form (your comments seem to indicate that this is how you arrived at it) rather than cloaked in a mysterious system of linear equations.
The expected time to draw the same irrelevant element twice goes as $\sqrt n$, and the probability to find the two elements in this time goes to zero for $n\to\infty$, so asymptotically we almost always have to restart, so we don't need to conditionalize on all except the last search being known to restart.
Thus, asymptotically, the expected time to find the two elements is just the expected number of restarts times the expected time until a restart. The expected time until a restart goes as $\sqrt\frac{\pi n}2$ (see Wikipedia). The probability to find the two elements we're looking for before a restart after $t$ draws goes as $(t/n)^2$. Thus we need $\langle t^2\rangle$, the expected value of $t^2$. This is $\langle t\rangle^2+\operatorname{Var}(t)=\frac{\pi n}2+\operatorname{Var}(t)$. Unfortunately I don't know $\operatorname{Var}(t)$ or how to calculate it, and your results seem to indicate that it's not $o(n)$ so we can't neglect it. However, it's reasonable to assume that if it's not $o(n)$, then it goes as $n$, and in that case $\langle t^2\rangle$ would go as $\alpha n$ with some unknown constant $\alpha$.
So the probability of a successful run goes as $\frac\alpha n$, so the expected number of runs goes as $\frac n\alpha$, and the expected duration of each of them goes as $\sqrt{\frac{\pi n}2}$, so in total the expected time goes as $\sqrt{\frac{\pi n^3}{2\alpha^2}}$, which agrees with the functional dependence you found. From numerical results it seems that $\alpha=2$, which would mean that the expected time goes as $\sqrt{\frac{\pi n^3}8}\approx0.627n^{3/2}$, which agrees well with your plot but not so well with your coefficient $0.7$.
