Efficient bowling

If you knock down 100 pins over the course of a game, you are guaranteed to have a score that’s at least 100. But what’s the minimum total number of pins you need to knock down such that you can attain a score of at least 100?

Indeed, you can in fact obtain a score of just 100 by knocking down exactly 100 pins, for instance, by getting a total of 9 pins in each of the first 9 frames, for a score of 81, then doing ever so slightly better and bowling a 9, 1, and 9 in the 10th frame. Note that in the last frame you don't really end up getting any "bonus" points than the number of pins that you knock down (you knocked down a total of 19 pins and your score went up by 19 points).

On the other hand, consecutive strikes gives you the most amount of bonus points, so let's look at how much points you end up with by hitting $n \leq 10$ strikes in a row and then getting all gutterballs ($0$'s) for the rest of the game. With the trailing gutterballs the last strike is score as just 10 points. The second-to-last strike (if applicable) gets only 10 points of bonus from the last strike and then a gutterball, so it's score is 20 points. Any strike before the second to last one receives a full 30 points since it is followed by two trailing strikes. So the score of $n$ consecutive strikes followed by gutterballs is \begin{align*}X_n &= 10 + 20 \min \{ 1, \max \{ n- 1, 0 \} \} + 30 \max \{ n-2, 0 \} \\ &= \begin{cases} 10, &\text{if $n=1$}\\ 30(n-1), &\text{if $n = 2, 3, \dots, 10.$}\end{cases}\end{align*} At this point we can stop for the purpose of the Classic Fiddler, but just for completeness, if we have 11 strikes in a row followed by a gutterball, we have $X_{11} = 290,$ and a perfect 12 strikes yields $X_{12} = 300.$ So we see that we can certainly get more than 100 by getting 5 consecutive strikes, but probably need 4 consecutive strikes plus some other trailing pins to most efficiently get to 100.

Let's assume that our first 4 frames were strikes, and let's assume that we are being efficient and want to take frames 6 through 10 off by devising ever more childish ways of gutterballing or foot faulting or other bowling alley hijinx. In this case the outcome of our total score depends on the values of our two balls in frame 5, say $b_1$ and $b_2.$ In particular, the final score for frame 3 will be $20+b_1,$ while the final score for frame 4 will be $10+b_1+b_2,$ and the score for frame 5 will be $b_1+b_2,$ so the overal final score of your entire round is $$S(b_1,b_2) = 60 + (20+b_1) + (10+b_1+b_2) + (b_1+b_2) = 90 + 3b_1 + 2b_2,$$ since the first two frames will have already been scored at their maximal values of 30, each. Here we want to then solve the integer linear program \begin{align*}\min \,\, & b_1+b_2 \\ \text{s.t.} \,\, & 3b_1 + 2b_2 = 10 \\ & b_1, b_2 \in \{0,1,2,3,4,5,6,7,8,9\}.\end{align*} There are really only two possible feasible integer solutions to choose from $b_1 = b_2 = 2$ and $b_1 = 0$, $b_2 = 5,$ so it is easy to determine that in fact $\hat{b}_1=\hat{b}_2=2$ is the optimal solution so that the minimum number of pins you need to knock down to attain a score of 100 is 44 pins.