Nate should only do zero-or-everything bets? Can Nate do better than betting everything on the last coin if he has only seen Tails in the previous rounds? He gets 1+(1/n) from this.

If $B$ is a symmetric invertible matrix you can use diagonalization to do roughly the same calculation and get $\lim \frac{1}{n}\sum_{k=0}^{n-1}(I-n^{-1}B)^k=B^{-1}(1-e^{-B})$.

$\frac{1}{n}\sum_{k=0}^{n-1}(1-(b/n))^k = \frac{1}{b}[1-(1-(b/n))^n]$ by summing the geometric progression. Now if you write this expression as $\frac{1}{b}[1-e^{n\log(1-(b/n))}]$ it is easy to see it has the limit $(1-e^{-b})/b$.

If you calculate this limit for scalars you get an exponential and your conjectured limit does not seem to hold.

Let: $n$ be the number of first class, $m$ the number of second class and $\mu_{nm}$ the stationary measure. Then I think $$\mu_{nm} =\lambda^n \sum_{i=0}^m A_{im} n^i$$ where $\lambda$ is the smaller root of $0=x^2-(a_1+a_2+1)x+a_1$. You can see this if you first solve the balance equations in the $m=0$ case then $m=1$ cases.

I'm sorry I should have written $-1\leq z\leq 0$. The integral I wrote describes all the ways tomorrows state could be $z<0$ when $z=(x+u)/2$ where $u$ comes from the uniform distribution and $x$ from an arbitrary pdf $f$. In this case $u$ would have to range from $-1$ to $2z+1$ and $x$ would be the symmetric (about $z$) opposite of this. The probability density of each of these $(x,u)$ pairs is $(1/2)\times f(x)$. Integrating gives the probability of $z$ tomorrow. At a stationary distribution this probability equals $f(z)$ - hence the integral equation.

If you think of this as a discrete-time continuous-state Markov process. What you are looking for is its stationary distribution. It is not too difficult to write down the recursion the stationary distribution satisfies. In the case $A,B=1/2$ for example, this is $f(z)=\int_{-1}^{2z+1}f(x)dx$ ($0\leq z\leq0$) where $f(.)$ is the pdf of the stationary distribution. Or $F'(z)=1/2F(2z+1)$ in terms of the cdf. This is difficult to solve explicitly as it is a pantograph equation, but maybe this is a start.

Doesn't $P=x_1(x_1+x_2)$ give $A=1$, $B\in[0,1]$ and hence violate this inequality?

For discrete distributions this seems not to work try (0.7 0.15) and (0.15, 0) as the two rows of a bivariate distribution where X and Y take two values.

The paper "Basic properties of strong mixing conditions. A survey and some open questions" by Richard C. Bradley seems to have a similar condition on p.108.

Maybe edit the question then to make this clear?

Suppose $A$ has only two non-zero elements, $a_{12}$ and $a_{21}$, satisfying $\alpha=a_{12}d_1$ and $\alpha=a_{21}d_2$ (where $D$ has $(d_1,d_2,\dots,d_n)$ on the diagonal), then isn't the bottom of this equal to zero?

Ken Binmore's book Fun and Games has a really nice chapter (12) on poker games where he deals with Borel's, von Neumann's and Nash's versions of poker. This might be a good place to start looking.

I wrote it as a system of equations $\frac{\nu_i}{1-\alpha_i}=\sum_j\frac{\alpha_i\nu_j}{1-\alpha_j}$ (here the sum runs over all $j$. Then changed variable $x_i=\nu_i/(1-\alpha_i)$ and observed that $x_i$ has to be proportional to $\alpha_i$.