Consider a Poisson process with arrival rate $\lambda$ arrivals per unit time. Given a window of time $W$ and a total of $k$ events, what is the upper bound of the probability that no three events happen in that window? Said another way, if events are numbered $1,2,\ldots,k$, what is the upper bound of $$P(\operatorname{gap}(j,j+2) \geq W \text{ for all } 1 \leq j \leq k-2)?$$

The problem in estimating this for me has been that while each successive gap is distributed exponentially, $\operatorname{gap}(1,3)$ and $\operatorname{gap}(2,4)$ are dependent. So, my best upper bound so far is obtained by just completely ignoring half the gaps: $$ P\bigl(\operatorname{gap}(j,j+2) \geq W\ \text{ for }\ 1 \leq j \leq k-2\bigr) < P\bigl(\operatorname{gap}(j,j+2) \geq W\ \text{ for }\ 1 \leq j \leq k-2,\ j \text{ odd}\bigr). $$ The right hand side is approximately $P(\operatorname{gap}(1,3) \geq W)^{(k-1)/2} = (\exp(-\lambda W)(1+\lambda W))^{(k-1)/2}$

because $exp(-\lambda W)(1+\lambda W)$ is the probability that 0 or 1 arrivals happened in duration W, and there are $(k-1)/2$ such odd gaps.

Is there a better upper bound?