= Solution
Let $(A_t,B_t)$ be a <Markov jump process> on integer populations, with rates
$$
(a,b)\longrightarrow(a-1,b)\text{ at rate }\kappa b,\qquad(a,b)\longrightarrow(a,b-1)\text{ at rate }\kappa a
$$
when $a,b>0$, where $\kappa>0$. Make both axes absorbing, so the fight ends when one population reaches zero. Start at $(N,b_N)$, with $b_N=\lfloor\alpha N\rfloor$; when $\alpha N$ is integral this is the printed initial condition exactly. The harmless rounding convention makes the model defined for every integer $N$.
Use the embedded <jump chain> $(A_j,B_j)$ rather than physical time. Its probabilities of the two losses are respectively $b/(a+b)$ and $a/(a+b)$, independently of $\kappa$. It reaches an axis after at most $N+b_N\leq2N$ steps; hold it constant thereafter. Put
$$
D_j=A_j^2-B_j^2.
$$
Before absorption its two increments are $-2a+1$ and $2b-1$, so
$$
\mu_j=\mathbb E[D_j-D_{j-1}\mid\mathcal F_{j-1}]=\frac{b-a}{a+b},\qquad |\mu_j|\leq1.
$$
After absorption set $\mu_j=0$. Consequently
$$
M_j=D_j-D_0-\sum_{r=1}^j\mu_r
$$
is a discrete-time <martingale>. All populations are at most $N$, so each martingale difference is bounded in absolute value by $2N+2\leq4N$. The total drift up to step $2N$ is at most $2N$. This is the <squared-population near-invariant for stochastic attrition>.
The two-sided <Azuma-Hoeffding inequality> states that a martingale with $m$ differences bounded by $C$ satisfies $\mathbb P(|M_m|\geq z)\leq2\exp(-z^2/(2mC^2))$. Its exponential estimate follows by applying the conditional <Hoeffding lemma> to each centered difference, iterating, and optimizing the exponential Markov bound. With $m=2N$ and $C=4N$ it gives
$$
\mathbb P(|M_{2N}|\geq\eta N^2)\leq2\exp(-\eta^2N/64).
$$
Let $c=\sqrt{1-\alpha^2}>0$, and let $X^N=A_{2N}/N$ be the final proportion. If the first population survives, then the second is zero and $D_{2N}/N^2=(X^N)^2$. Hence $|X^N-c|>\delta$ implies
$$
|D_{2N}-c^2N^2|=N^2|X^N-c|(X^N+c)>c\delta N^2.
$$
If the first population is extinguished, $X^N=0$ and $D_{2N}\leq0$. This can satisfy the indicated deviation event only when $\delta<c$, and then $|D_{2N}-c^2N^2|\geq c^2N^2>c\delta N^2$. Thus the same implication holds in both cases.
The initial rounding error obeys $|D_0-c^2N^2|=|\alpha^2N^2-b_N^2|\leq2N$. Combining it with the drift bound shows that the deviation event forces
$$
|M_{2N}|>c\delta N^2-4N\geq\tfrac12c\delta N^2
$$
for all sufficiently large $N$. Set $\eta=c\delta/2$ in the preceding concentration inequality. We conclude
$$
\boxed{\mathbb P(|X^N-\sqrt{1-\alpha^2}|>\delta)\leq2e^{-(1-\alpha^2)\delta^2N/256}}
$$
for sufficiently large $N$, and therefore
$$
\boxed{\limsup_{N\to\infty}\frac1N\log\mathbb P(|X^N-\sqrt{1-\alpha^2}|>\delta)\leq-\frac{(1-\alpha^2)\delta^2}{256}<0.}
$$
The <symmetric stochastic attrition process> thus has <exponential concentration of terminal stochastic attrition survivors>. The deterministic equations conserve $a^2-b^2$ and predict the same surviving fraction, but the martingale argument supplies the stronger exponential probability estimate required here.
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