Probabilities: Meeting people The probability of women meeting a man is $m$. Let's look at the perspective of a specific man. The probability of meeting him is $\tilde m$. 
Say women look twice for men. Then (assuming $\tilde m$ is so small that $\tilde m \tilde m$ is negligible), the probability of a woman meeting the specific man is $\tilde m + \tilde m$. The probability of meeting both this specific man, and any other man, is (given two searches)
$$ \tilde m m  + m \tilde m$$ 
What is the probability of a woman meeting the specific man and at least another man, given that she searches $s$ times? As I showed above, the probability is $2m \tilde m$ for $s=2$.
For $s=3$, this would be $6\tilde m m + 3\tilde m m m $. I lack finding a general formula here, and these equations get messier the higher $s$ gets (and I am actually writing down all the potential combinations). Could someone give me some pointers?

The question doesn't seem to add up. If $\tilde m\tilde m\approx 0$ I don't understand why it's not true that $mm\approx 0$ or $\tilde m m\approx 0$. Also the $6\tilde m m$ should be, instead $6\tilde m m (1-\tilde m-m)$
I'll try to solve your question with some considerations and changing some notation for easiness.
Let there be two men $A$ and $B$. For any search, $X_A$ is the event of finding man $A$ (equivalent for $B$), and $P(X_A)=a$ and $P(X_B)=b$. Let $X_A^s$ finding him after $s$ searches.
$$P(X_A^s\cap X_B^s) = \sum_{i_A=1}^{s-1}\sum_{i_B=1}^{s-1}\frac{s!}{i_A!i_B!(s-i_A-i_B)!}a^{i_A}b^{i_B}(1-a-b)^{s-i_A-i_B}
$$
What I did there was taking into account all the ways where there can be man $A$ and man $B$ found at least once, multiply it by the probability of that happening and then adding all together.
For $s=8$
$$P(X_A^8\cap X_B^8) = \sum_{i_A=1}^{7}\sum_{i_B=1}^{7}\frac{8!}{i_A!i_B!(8-i_A-i_B)!}a^{i_A}b^{i_B}(1-a-b)^{8-i_A-i_B}
$$
Simplify for the final answer (I'll think about how to do it, if possible)
EDIT (simplifying)
If we assume $a=b$
$$P(X_A^8\cap X_B^8) = \sum_{i=2}^6 \binom{8}{i}a^i(1-a)^{8-i}
$$
From here, using Wolfram Alpha:
$$P(X_A^8\cap X_B^8)=14a^8-56a^7+140a^6-224a^5+210a^4-112a^3+28a^2$$
And using your notation:
$$14\tilde m^8-56\tilde m^7+140\tilde m^6-224\tilde m^5+210\tilde m^4-112\tilde m^3+28\tilde m^2$$
