Poisson Distribution: Find 2 x values that give the same probability I have a problem where I am supposed to assume $\lambda = k$, $k \in \mathbb{Z}$ and show that there is always 2 $x$ values that give the same probability in a poisson distribution and express them by $k$. Am I on the wrong track with this? And could someone explain what it means?:
\begin{equation*}
 \begin{aligned}
  & \lambda = k \Rightarrow f(x) = \frac{k^x}{x!}e^{-k}\\\\
  & f(a) = f(b) \\\\
  & \frac{k^a}{a!}e^{-k} = \frac{k^b}{b!}e^{-k} \Leftrightarrow
    \frac{k^a}{a!} = \frac{k^b}{b!} \Leftrightarrow \frac{k^a}{k^b} =
    \frac{a!}{b!} \Leftrightarrow k^{a - b} = \frac{a!}{b!} 
    \Leftrightarrow k = \frac{ln(a!) - ln(b!)}{a - b}
 \end{aligned}
\end{equation*}
 A: Observe that, for $x = 0, 1, 2, 3, 4, \cdots$, we have
Poisson probabilities as
$$e^{-\lambda}, \quad\lambda e^{-\lambda}, \quad\dfrac{\lambda^2}{2}e^{-\lambda}, \quad\dfrac{\lambda^3}{6}e^{-\lambda}, \quad\dfrac{\lambda^4}{24}e^{-\lambda},\cdots $$
whose values for $\lambda = k = 1, 2, 3, 4,\cdots$ are
$$\lambda=1;\quad \underbrace{ e^{-1}, \quad 1 e^{-1}}, \quad\dfrac{1}{2}e^{-1}, \quad\dfrac{1}{6}e^{-1}, \quad\dfrac{1}{24}e^{-1},\cdots $$
$$\lambda=2;\quad e^{-2}, \quad \underbrace{ 2 e^{-2}, \quad\dfrac{2^2}{2}e^{-2}}, \quad\dfrac{2^3}{6}e^{-2}, \quad\dfrac{2^4}{24}e^{-2},\cdots $$
$$\lambda=3;\quad e^{-3}, \quad  3 e^{-3}, \quad \underbrace{ \dfrac{3^2}{2}e^{-3}, \quad\dfrac{3^3}{6}e^{-3}}, \quad\dfrac{3^4}{24}e^{-3},\cdots $$
That is, when $\lambda$ takes integer values, two successive values of x will have equal probabilities. By equating $f(x)$ to $f(x+1)$, we observe that, $x = \lambda-1$. This means that, if $\lambda=k$, the two values of $x$ for which the Poisson distribution will have equal  probabilities are $x=k-1$ and $x=k$. 
A: A Poisson random variable $X$ with integer mean $\lambda$ has largest
probabilities at $P(X = \lambda) = P(X = \lambda - 1):$
$$P(X = \lambda) = e^{-\lambda}\lambda^{\lambda}/\lambda! =
e^{\lambda}\lambda^{\lambda-1}/(\lambda-1)!\times\frac\lambda\lambda = P(X = \lambda - 1).$$
Look at $\frac{P(X = k)}{P(X = k-1)}$ to establish that the equal probabilities occur at a 'double mode'. 
