First lets define some useful terms:

$p$ = total population =

$s$ = people seen =

$f$ = num friends =

### Approach

Since each way of seing $s$ people is equally likely, we can use the "Equally Likely Events" probability calculation:

$P(E) = \frac{|E|}{|S|}$

Where $S$ is the sample set (all the ways of seing $s$ people) and $E$ is the event set (all the ways of seing $s$ people where at least one is a friend).

One way to approach this problem is to directly count all ways you see 1 or more friends. That is hard. You would have to count the ways you could see exactly one friend, then exactly two friends and so on. It is much easier to calculate the ways that you see zero friends. If we can calculate the probability of seeing zero friends our answer is just one minus that number.

### Prob that you don't see friends

Let the sample space ($S$) be the set of ways that you could see $s$ people. The size of the sample space is: the total population choose the number of people seen.

The event space ($E$) is the set of ways that you could see no friends. The size of the event space is: the number of non friends (aka population - friends) choose the number of people seen.

Thus the probability of not seeing a friend is:

$\text{prob(not seen)} = \frac{ \left( {\begin{array}{*{20}c} p - f \\ s \\ \end{array}} \right) } { \left( {\begin{array}{*{20}c} p \\ s \\ \end{array}} \right) } $

### Prob that you see friends

Now the probability that you see at least one friend is 1 minus the probability that you see no friends.

$\text{prob(seen)} = 1 - \frac{ \left( {\begin{array}{*{20}c} p - f \\ s \\ \end{array}} \right) } { \left( {\begin{array}{*{20}c} p \\ s \\ \end{array}} \right) } $

That is equal to

### Isn't that suprising?