Dominant bowling

You and your opponent each play a single game in which you both knock down the same number of pins. However, your scores are quite different.

Your opponent remarks, “Given only the information that we knocked down the same number of pins in our two games, there’s no way the difference between our scores could have been any greater!” What is this difference between your two scores?

As we saw in the Classic Fiddler, there is a way to knock down exactly 100 pins and get a final score of exactly 100 points. As part of that we saw that any pins knocked down in the 10th frame incur no bonus points, and we can attain any number from 0 up to 30 total pins knocked down in the 10th frame.

We will extend the result of getting a score equal to the total number of pins knocked down to all values 0 to 119. For any number less than 90, it should be relatively trivial to find some combination with no strikes or spares, so that there would be no way of getting any bonus points. For any score $S \in \{91, 92, \dots, 111\},$ you will need your extra ball in the 10th frame, so you can for instance get 81 points in the first 9 frames and then score $S-81$ in the 10th frame.

For any score $S \in \{112, 113, \dots, 119\},$ we have to get a bit trickier and layer on consecutive frames of gutterball then spare ($-/$). The fact that the spare is followed by a gutterball means that the spared frame still receives no bonus points. So for instance you can get a score of $S=119$ by $8$ consecutive gutterball then spare frames, followed by gutterball then 9 in the 9th frame, then three strikes in the 10th frame.

The minimum score for hitting all 120 pins down is 130, since if you gutterball then spare the 9th frame, then the first of the three strikes in the 10th frame will be added to the score in the 9th frame. So at a certain point, you must engage accrue some bonus points, but thankfully that is not until you get to 120 pins. So we have established the minimum score per number of pins knocked down as $$m(P) = \begin{cases} P, &\text{if $P \leq 119$;} \\ P+10, &\text{if $P = 120.$}\end{cases}$$

Now it is time to establish the highest score possible to attain given a certain number of pins knocked down. Here again we will look to the Classic Fiddler where we observed that the total score of $n$ consecutive strikes followed by all gutterballs is $$X_n = \begin{cases} 10, &\text{if $n=1$;}\\ 30(n-1), &\text{if $n=2,\dots, 10$;}\\ 180+10n, &\text{if $n=11, 12.$}\end{cases}$$ So just by observation we see that if we only focus on multiples of 10, we get the maximum score per number of pins knocked down as $$M(P) = X_{P/10}$$ whenever $10 \mid P.$ For completeness, we can fill in the gaps and note similar to the solution for the Classic Fiddler that if $20 \leq P \leq 100$ and the next two balls are $b_1$ and $b_2$ that we are trying to find \begin{align*}M(P) = \max \,\, & X_{\lfloor P / 10 \rfloor} + 3b_1 + 2b_2 \\ \text{s.t.}\,\, & b_1 + b_2 = P - 10 \lfloor P / 10 \rfloor \\ & b_1, b_2 \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\end{align*}, which has optimal solution $\hat{b}_1 = P - 10 \lfloor P/10 \rfloor$ and $\hat{b}_2 = 0,$ that is, $$M(P) = X_{\lfloor P / 10 \rfloor} + 3\left(P - 10 \lfloor P / 10 \rfloor\right) = 3P-30, \,\, 20 \leq P \leq 100.$$

Now let's go back and clean up the corner cases when $P \lt 20$ and $P \gt 100.$ For $P \leq 10,$ we have $M(P) = P.$ If we have $10 \lt P \lt 20,$ then the optimization problem becomes \begin{align*}M(P) = \max \,\, & 10 + 2b_1 + 2b_2 \\ \text{s.t.}\,\, & b_1 + b_2 = P - 10 \\ & b_1, b_2 \in \{0,1,2,3,4,5,6,7,8,9\},\end{align*} which for all feasible solutions gives $$M(P) = 10 + 2(P-10) = 2P-10, \,\, 10 \lt P \lt 20.$$ Whenever $100 \lt P \lt 110,$ then the optimization problem becomes \begin{align*}M(P) = \max \,\, & 270 + 2b_1 + b_2 \\ \text{s.t.}\,\, & b_1 + b_2 = P - 100 \\ & b_1, b_2 \in \{0,1,2,3,4,5,6,7,8,9, 10\},\end{align*} which for all feasible solutions give $$M(P) = 270 + 2(P-100) = 70 + 2P, \,\, 100 \lt P \lt 110.$$ And finally, it is clear that $$M(P) = 290+(P-110) = 180+P, for 110 \lt P \leq 120.$$ So putting this altogether we get the maximum score per number of pins knocked down is $$M(P) = \begin{cases} P, &\text{if $0\leq P \lt 10;$}\\ 2P-10, &\text{if $10 \leq P \lt 20$;}\\ 3P-30, &\text{if $20 \leq P \lt 100$;}\\ 2P + 70, &\text{if $100 \leq P \lt 110$;} \\ P + 180, &\text{if $110 \leq P \leq 120$.}\end{cases}$$

So having exhaustive determined the minimum and maximum scores for a given number of pins knocked down the only thing left to do is to determine that largest possible margin of victory given that the two bowlers knocked down the same number of pins, that is $$V(P) = M(P) - m(P) = \begin{cases} 0, &\text{if $0 \leq P \lt 10;$} \\ P-10, &\text{if $10 \leq P \lt 20$;} \\ 2P-30, &\text{if $20 \leq P \lt 100;$} \\ P+70, &\text{if $100 \leq P \lt 110;$} \\ 180, &\text{if $110 \leq P \leq 119$;} \\ 170, &\text{if $P = 120.$}\end{cases}$$ Scrutinizing the margin of victory function, we see that the largest possible margin of victory is 180 points, which can be attained $P \in \{110, 111, \dots, 119\}.$