What is the probability of getting an even number from a Poisson random draw? Below is a graph showing the probability of drawing an odd number (y-axis) from a Poisson distribution with a given expected value (x-axis)
x = seq(0,1e4,1)         // range of values to explore 
lambdas = seq(0,4,0.01)  // expected value of the Poison distribution

FracOdd = numeric(n+1)       // response variable
for (i in 1:length(lambdas)) 
{
   FracOdd[i] = sum(dpois(x,lambdas[i])[seq(2,length(x),2)])  // calculate probability of drawing an odd number
}

plot(y=FracOdd,x=lambdas, type="l", lwd=3, xlab="Expected value", ylab="Probability of drawing an odd number")   // plot the data


It seems that the probability of drawing an odd number from a Poisson distribution with non-infinite mean is always lower than the probability of drawing an even number.


*

*What is the intuition for why the probability of drawing an odd number from a Poisson distribution always below 0.5?

*Is the function I drew numerically, easy to derive analytically?

 A: Yeah, it's reasonably well known.
$$\begin{align}\mathsf P(\text{Even}) ~&=~ \sum_{k=0}^\infty \dfrac{\lambda^{2k} e^{-\lambda}}{(2k)!}\\[1ex] ~&=~ e^{-\lambda}\cosh(\lambda)\\[1ex] ~&=~ e^{-\lambda}\tfrac 12(e^{\lambda}+e^{-\lambda})\\[1ex]~&=~ \tfrac12(1+e^{-2\lambda})
\\[2ex]\mathsf P(\text{Odd}) ~&=~ \sum_{k=0}^\infty \dfrac{\lambda^{2k+1} e^{-\lambda}}{(2k+1)!} \\[1ex]~&=~ e^{-\lambda}\sinh(\lambda)\\[1ex] ~&=~ e^{-\lambda}\tfrac 12(e^{\lambda}-e^{-\lambda})\\[1ex]~&=~ \tfrac12(1-e^{-2\lambda})\\[2ex]e^{-\lambda}(\cosh\lambda-\sinh\lambda)~&=~e^{-2\lambda}\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 0}^{\infty}{\expo{-\lambda}\lambda^{2n} \over \pars{2n}!}} =
\expo{-\lambda}\sum_{n = 0}^{\infty}{\lambda^{n} \over n!}\,{1 + \pars{-1}^{n} \over 2}
\\[5mm] = &\
{1 \over 2}\,\expo{-\lambda}\
\underbrace{\sum_{n = 0}^{\infty}{\lambda^{n} \over n!}}_{\ds{\expo{\lambda}}}\ +\
{1 \over 2}\,\expo{-\lambda}\
\underbrace{\sum_{n = 0}^{\infty}{\pars{-\lambda}^{n} \over n!}}_{\expo{-\lambda}}
\\[5mm] = &\
\bbx{1 + \expo{-2\lambda}\over 2} \\ &
\end{align}
