If a player is 50% as good as I am at a game, how many games will it be before she finally wins one game? This is a real life problem. I play Foosball with my colleague who hasn't beaten me so far. I have won 18 in a row. She is about 50% as good as I am (the average margin of victory is 10-5 for me). Mathematically speaking, how many games should it take before she finally wins a game for the first time?

A reasonable model for such a game is that each goal goes to you with probability $p$ and to her with probability $1-p$. We can calculate $p$ from the average number of goals scored against $10$, and then calculate the fraction of games won by each player from $p$.
The probability that she scores $k$ goals before you score your $10$th is $\binom{9+k}9p^{10}(1-p)^k$, so her average number of goals is
$$\sum_{k=0}^\infty\binom{9+k}9p^{10}(1-p)^kk=10\frac{1-p}p\;.$$
Since you say that this is $5$, we get $10(1-p)=5p$ and thus $p=\frac23$. The probability that you get $10$ goals before she does is
$$\sum_{k=0}^9\binom{9+k}9\left(\frac23\right)^{10}\left(\frac13\right)^k=\frac{1086986240}{1162261467}\approx0.9352\;,$$
so her chance of winning a game should be about $7\%$, and she should win about one game out of $1/(1-0.9352)\approx15$ games – so you winning $18$ in a row isn't out of the ordinary.
