Mean of geometric distribution is odds? Context: I mean the $P(X=k)=(1-p)^k p$ not the $P(Y=k)=(1-p)^{k-1} p$. Apparently the mean of the 1st kind of geometric is $\frac{1-p}{p}$ instead of $\frac{1}{p}$ for the 2nd kind of geometric. I believe the relation between the 2 kinds is $Y=X-1$ s.t. $P(Y=k)=P(X=k+1)$.
Definition: Afaik, if the probability of an event $A$ is $P(A)=p, 0 \le p<1$, the the odds of $A$ is $\text{odds}(A) = \frac{p}{1-p}$
Question: So what's the meaning/intuition that the odds of $A$ is the mean of $Y$ ?

If $p$ denotes the probability on success then $n$ trials will contain approximately $np$ successes and approximately $nq$ failures (where $q:=1-p$).
So against $1$ success there will stand $\frac{nq}{np}=\frac{q}{p}$ failures, corresponding with the expectation of the number of failures needed to arrive at a success.
This makes clear why the odds of failures shows up in this context.
