Conditioned binomial dominates unconditioned with different parameter

Question

Let $X \sim \text{Bin}(n,p)$ and $Y \sim \text{Bin}(n-1,p)$ with $n >1, p \geq 1/2$ and $X,Y$ are independent. I'd like to show $$(X\mid X \geq 1) \succeq_{sd} 1 + Y.$$ Here $(X \mid \cdot)$ is the conditioned random variable and $\succeq_{sd}$ denotes stochastic dominance, formally: $$\mathbf P[X \geq k \mid X \geq 1] \geq \mathbf P[1+ Y \geq k], \qquad \forall k\geq 1.$$

The intuition is that with $X$ conditioned to be larger than 1 we are guaranteed to have one success. Moreover, with $p \geq 1/2$ the conditioning only makes the remaining $n-1$ Bernouli r.v.s in $X$ more likely to be 1. So, they ought to dominate an independent $\text{Bin}(n-1,p)$.

Note this isn't true in the case $n=2$ and $p=\epsilon$ for small enough $\epsilon$.

Answer 1

This does not seem to be true. Take $p=2/3$ and $n=k=1000$. Then, you need that $$ \frac{(2/3)^{1000}}{1-(1/3)^{1000}} \geq (2/3)^{999}, $$ which is clearly false.