Mean of a Poisson Distribution From Two Relative Probabilities

Suppose we have good reason to suspect that a particular random variable
$X$
follows a Poisson distribution, so that
$Z \sim Po( \lambda )$
, where
$\lambda$
is the relevant parameter, typically a frequency rate. Then we can find
$\lambda$
given the relative probabilities of observations of
$X$
.
For a Poisson distribution,
$P(X=x)=\frac{\lambda^x e^{-\lambda}}{x!}$
.
Suppose that
$P(X=7)=2P(X=8)$
.
Then
$\frac{\lambda^7 e^{-\lambda}}{7!}=2 \frac{\lambda^8 e^{-\lambda}}{8!}$
.
Divide by
$e^{- \lambda}$
to give
$\frac{\lambda^7}{7!}=2 \frac{\lambda^8 }{8!}$
.
Multiply,y both sides by
$8!$
and divide by
$\lambda^7$
. Then
$8=2 \lambda \rightarrow \lambda =4$
.