Did you ever ask what it takes to win 6-0 6-0? You clearly need to dominate your opponent. But by what margin? Out of ten points, how many points do you need to win on average for that result?

In an earlier post we established a relationship between the likelihood of winning a point and the likelihood of winning a game. We can now build on that. Instead of asking for the winning percentage of each point we can ask a simpler question: what percentage of games does one need to win? From there we can translate back to the percentage of points one needs to win.

In order to win 6-0 6-0, the player needs to win 12 games in a row. Say we want 80% confidence that we indeed win 6-0 6-0. So the equation is $y^{12}=80\%$ or $y=0.8^{1/12}$ which yields $y=0.982$. Therefore, the player needs a $98.2\%$ game winning percentage.

What does this mean with regard to the point winning percentage? Let us go back to the formula we have established:

$P_{wingame}(x) = x^{4} + 4\cdot x^{4}\cdot(1-x) + 10\cdot x^{4}\cdot(1-x)^{2} + (20\cdot x^{3}\cdot(1-x)^{3})\cdot(\frac{x^2}{2\cdot x^2-2\cdot x + 1})$

We are looking for a $98.2%$ game winning percentage, so the equation is:

$98.2\%= x^{4} + 4\cdot x^{4}\cdot(1-x) + 10\cdot x^{4}\cdot(1-x)^{2} + (20\cdot x^{3}\cdot(1-x)^{3})\cdot(\frac{x^2}{2\cdot x^2-2\cdot x + 1})$

A solver like Wolfram Alpha easily outputs the solution: $x=81\%$ (rounded).

This is quite surprising: In order to win 6-0 6-0 (with 80% confidence), it is sufficient to win roughly 4 out of 5 points!