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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 $(\lambda-1)\in\mathbb{N}$?

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You are making one dangerous assumption here: namely, that λ is an integer. The Poisson distribution makes perfect sense for non-integral λ, and in that case this result is not valid (well, it is valid in the sense that both probabilities are identically 0.) – Nick Peterson Jun 25 '13 at 4:51
@nrpeterson Thank you. I'll explicitly add the condition that lambda is a nonnegative integer. – Fred Jun 25 '13 at 4:52
When $\lambda$ is not an integer, indeed $P(X_\lambda=\lambda)=P(X_\lambda=\lambda-1)$ because both are zero. – Did Jun 25 '13 at 6:32

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