The calculations seem to involve a strange kind of raffle, in which the first ticket is sold (for $5$ dollars). We check whether this is the winning ticket. If it is not, we sell another ticket, and check whether it is the winner. And so on.
You seem to be asking for the expected return. This is $5$ times $E(X)$, where $X$ is the total number of tickets until we reach the winning ticket.
The random variable $X$ has a distribution which is a special case of what is sometimes called the Negative Hypergeometric Distribution. (There are other names, such as Inverse Hypergeometric Distribution.)
The general negative hypergeometric allows the possibility of $r$ "winning tickets" among the $N$ tickets, and the possibility that we will allow sales until $k$ winning tickets have turned up. You are looking at the special case $r=k=1$ (I am using the notation of the link).
In your case, if the total number of tickets is $N$, of which only one is a winning one, we have $E(X)=\frac{N+1}{2}.$ Taking $N=100$, and $5$ dollars a ticket, the expected return is $5\frac{101}{2}$ dollars.
Remark: If the model I have described is not the model you have in mind, perhaps the question can be clarified.