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<strong>Figure 1.</strong> Probability $\mathbf{P}(p, T)$ on the left and expected length $\mathbf{E}(p, T)$ on the right for N-point tiebreakers.
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<strong>Figure 1.</strong> Probability $\mathbf{P}(p, T)$ for N-point tiebreakers on the left and $\mathbf{P}(p, M)$ for the official tennis match on the right.
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</p>
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</div>
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<p>In the plot in Figure 1 we can see how $\mathbf{P}(T; p)$ follows a "S" shape. On the right part of the plot, where $p > 0.5$, </p> we observe that $\mathbf{P}(T; p) > p$, meaning that the advantage is amplified.
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<p>
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In Figure 1, the curve for $\mathbf{P}(T; p)$ has an "S" shape. On the right side where $p > 0.5$, we observe that
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$\mathbf{P}(T; p) > p$, so the stronger player's advantage is amplified.
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</p>
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</section>
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<sectionclass="space-y-3">
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<h2class="font-serif text-2xl">Fairness</h2>
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<p>
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While all tiebreakers aplify the advantage of the stronger player, it is clear that $T_7$ will be more "fair" than $T_2$. To quantify this "fairness", we look at the steepness of the line tangent to
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$\mathbf{P}(p, T)$ at $p = 0.5$, this is the dashed line in the left plot of Figure 1. When this line is very steep, it means that a small edge for Alex results
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in a large advantage. We call this steepness $\mathbf{F}(T)$.
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While all tiebreakers amplify the advantage of the stronger player, $T_7$ is clearly more "fair" than $T_2$.
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To quantify this "fairness" we look at the steepness of the tangent to $\mathbf{P}(p, T)$ at $p = 0.5$ (the dashed
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line in the left plot of Figure 1). A steeper line means that a small edge for Alex turns into a much larger advantage.
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We call that steepness $\mathbf{F}(T)$.
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</p>
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<p>
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Formally, $\mathbf{F}(T)$ is the derivative of $\mathbf{P}(p, T)$ evaluated at $p=0.5$. Intuitively, let
While we can all agree that there is a tradeoff between match length and fairness, we would like to quantify it better. In Figure 2, we plot $\mathbf{E}(0.5, T)$ against $\mathbf{F}(T)$. All the dots are tiebreaker of different length, we can see that as the tiebreakers become longer (more up), they also become more fair (more right).
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While we can agree that match length trades off with fairness, we would like to measure this relationship more precisely.
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In Figure 2 we plot $\mathbf{E}(0.5, T)$ against $\mathbf{F}(T)$. Each dot corresponds to a tiebreaker of a different
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length: as the tiebreakers grow longer (higher on the axis) they also become fairer (further to the right).
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</p>
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<p>
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The question that now rises naturally is: how does the real tennis game structure compare, with bet of 3 sets, each set being 6 games, tiebreakers and so on? You can see it as the star in Figure 2. Surprisingly, it lies under the tiebreaker curve. This means, that a tiebreaker of 91 points, will be just as fair as a normal tennis match, but on average it would take a bit more to play, and a tiebreaker of 87 points will be just as long as a tennis match on average, but it will be less fair.
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The natural question is how the official tennis format compares: best-of-three sets with first-to-six games and
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tiebreakers. This format is shown as the star in Figure 2. Surprisingly, it falls below the tiebreaker curve,
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meaning a 91-point tiebreaker would be as fair as a normal tennis match but would take slightly longer, while an
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87-point tiebreaker would last about the same as a tennis match but be less fair.
Before being faced with this result, I thought tiebreakers would have been more efficient than the official tennis format, maybe less exiting to watch, but for sure simpler. This motivated me to try to undestand what is that makes tennis matches better, and is there a tennis match which is even more efficient?
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Before seeing this result, I assumed tiebreakers would be more efficient than the official format—maybe less exciting
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to watch. That motivated me to understand what makes tennis matches better and whether there
After some thinkering, I came up with a match structure that is considerably more efficient. The match would work as follows. It starts with a score of $0$. If the first player wins the point, the score goes up by $1$, if the second one wins instead, it goes down by $1$. When the score reaches $N$, the first player won, when it reaches $-N$, the second one does.
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After some tinkering I devised a match structure that is significantly more efficient. It starts at a score of
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$0$. Whenever the first player wins a point the score increases by $1$, and if the second player wins it decreases
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by $1$. When the score reaches $N$, the first player wins; if it reaches $-N$, the second player wins.
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</p>
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<p>
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We will call this match structure $O_N$, where $N$ and $-N$ are the scores to be reached respectivly from the two players.
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Using our <aclass="underline hover:no-underline" href="../probability-calculator/">Probability Calculator</a>, we observe a very satisfing fact:
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We call this match structure $O_N$, where $N$ and $-N$ are the target scores for the two players. Using our
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<aclass="underline hover:no-underline" href="../probability-calculator/">Probability Calculator</a>, we observe a very
Even though the name $O_N$ is of course inspired by "optimal", I never managed to prove this. I have a strong intuiton that there is a bound that cannot be beaten, and that such bound is achieved by $O_N$. Such bound is
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$$
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\mathbf{E}(T; 0.5) \geq \mathbf{F}(T)^2.
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$$
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Even though the name $O_N$ is inspired by "optimal", I have not been able to prove it. I have a strong
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intuition that there is a lower bound that cannot be beaten and that $O_N$ achieves it:
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$$
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\mathbf{E}(T; 0.5) \geq \mathbf{F}(T)^2.
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$$
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</p>
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<p>
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Of course, I would love to either prove this conjecture, or to find a counterexample. If you have any cool ideas, you can use my <aclass="underline hover:no-underline" href="../probability-calculator/">Probability Calculator</a> to test your ideas. If you find a counterexample, or a proof, please let me know! Contact me at <aclass="underline hover:no-underline" href="mailto:marco.milanta@gmail.com">marco.milanta@gmail.com</a>.
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