Time to first success for a simple event

Question

I beg your pardon for the silly question.

An urn contains $c$ elements, $\gamma$ of which are of type $G$.

We define the event $E_n^G$ as to get in $n$ independent trials, exactly $n$ times one element of kind $G$, so that $P(E_n^G)=\left(\frac{\gamma}{c}\right)^n$.

Clearly, the event $E_n^G$ cannot occur before the last trial $n$.

I learned that the time to first success can be seen as the number of trials needed to get, in average, the first success of a given event.

But then what is the expected "time to to first success" for the event $E^G_n$? Is it $n$? Yes/No? Why?

Answer 1

If you have an event $A$, and a number of trials $N$ (which can be $\infty$), then the definition of the "time to first success" of A is $$T_A = \min \{n \le N\;:\; A\;\text{is true at the $n-$th trial}\}$$

You can see that if the event $A$ does not occurs for the $N$ trials then $\{n \le N\;:\; A\;\text{is true at the $n-$th trial}\}$ is empty and then $T_A = \infty$

Let us assume that $\gamma < c$. If you define $E_n^G$ only for $n$ trials then the "time to first success" of $E_n^G$, will be $n$ with probability $\left(\frac\gamma c\right)^n$ and $\infty$ with probability $1-\left(\frac\gamma c\right)^n > 0$, in this case the expected value of the "time to first succes" is $\infty$. This can be easily generalized when the number of trials is finite. However if the number of trials is infinite you can see that getting $n$ times an element of $G$ is equivalent to getting $n$ times one time an element of $G$. Then $$\mathbb E \left[T_{E_n^G}\right] = n \mathbb E\left[T_{E_1^G}\right] = n\frac c \gamma$$