Where $X_\lambda$ is a Poisson random variable with mean $\lambda$,$$P(X_\lambda=k\in\mathbb{N}) = \frac{\lambda^{k}e^{-\lambda}}{k!} $$ When it happens that $(\lambda-1)\in\mathbb{N}$, $P(X={\lambda})\ {\equiv}\ P(X={\lambda}-1)$ since $\frac{\lambda^{\lambda}}{\lambda!}\ {\equiv}\ \frac{\lambda^{\lambda-1}}{(\lambda-1)!}\ $.
Consider a simple case where a website has a mean of 3 hits per day. The probability of getting 2 hits per day would be equal to that of getting 3 hits per day. My intuition is that the mean will always be greater than the mean of the modes due to the skewness of the distribution.
Still, what is an intuitive explanation behind the specific fact that the probability that X equals ${\lambda}$ is always the same as the probability that X equals ${\lambda -1 }$ when $(\lambda-1)\in\mathbb{N}$?