As I take your question, you have a large number of locations (say $N$), of which $7$ of these contain a particular prize. You want to determine the probability that if a user randomly selects $x$ of these locations, then they will receive all $7$ prizes.
There is a special notation in mathematics called the binomial coefficient $\binom{n}{k}$, which gives the total number of ways to choose $k$ objects out of a total $n$. In this case, the total number of locations a user can choose is
$$\binom{N}{x}$$
And if we require that $7$ of these choices are the 'prizes', then since the user must visit these locations, the remaining number of choices is
$$\binom{N-7}{x-7}$$
and hence the probability is simply
$$\frac{\binom{N-7}{x-7}}{\binom{N}{x}}$$
or $0$ if $x<7$.
In Excel, we can find the value $\binom{n}{k}$ with COMBIN(n,k), and hence you would want COMBIN(N,x)/COMBIN(N-7,x-7). Note that for this to work you require $x$ to be between $7$ and $N$, otherwise Excel may return an error.
UPDATE (in order to reflect the correction given in the comments)
Consider calculating the probability that a user receives all the prizes on the $n^\text{th}$ visit to a location, which we call $P(n)$. Since there are 7 possible prizes, the number of 'prize sequences' a user could encounter in $n$ locations is $7\times7\times\cdots\times 7=7^n$.
Now, the hard part is determining how many of these sequences involve the user finding the final prize at the $n^\text{th}$ location. To find this, we choose one of the 7 prizes to be the 'last discovered', and then count the number of weighs a user can find all remaining $6$ prizes in the first $n-1$ locations. Mathematically, this is equivalent to the number of surjections from $[n-1]$ to $[6]$ (don't worry about this if you don't understand it). In any case, the probability will be
$$P(n)=\frac{7\times 6!S(n-1,6)}{7^n}=\frac{6!S(n-1,6)}{7^{n-1}}$$
where $S(a,b)$ refers to the Stirling Numbers of the Second Kind. Since a user must visit at least $7$ locations, the average number of visits for a user will be
$$\mu=\sum_{n=7}^{\infty}nP(n)=\sum_{n=7}^{\infty}n\frac{6!S(n-1,6)}{7^{n-1}}$$
Doing some numerical calculation, I find this average to be $\approx 18.15$. I'm not sure Excel has a native option for finding these Stirling Numbers, but an explicit formula is given on the Wikipedia page.
