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So Alice and Bob decide to go bowling. They play one round. By the end of the game, Alice and Bob have each managed to knock down the exact same number of pins in total. However, Alice won, because some pins are worth more than others.

The question is, what is the largest possible difference in score between Alice and Bob? And can you prove that the number you have is in fact the largest possible?

In case you don't know, the scores in bowling work as follows:

  • You play 10 frames; each frame you get 2 rolls.
  • If you don't knock down all the pins between the 2 rolls, you get 1 point per pin.
  • If you knock down all the pins on your second roll; you get a spare. For that frame, you get 10 points PLUS an additional point for each pin you knock down in your next roll (Meaning that those pins are worth 2 points each).
  • If you knock down all the pins on your first roll; you get a strike. For that frame, you get 10 points PLUS an additional point for each pin you knock down in your next TWO rolls.
  • If you get a spare on frame 10; you get a single roll for frame 11 to determine how much to add. Those pins only add to frame 10; they don't count themselves as well.
  • If you get a strike on frame 10; you get 2 extra rolls. Again, pins you knock down on those rolls only add to your regular frames (10 or 9 & 10); they don't count for themselves as well.
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  • $\begingroup$ As this is my first post on this network; please feel free to advise me on better tags to use; or better formatting or wording for my question! $\endgroup$ – GendoIkari Feb 9 '18 at 16:46
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    $\begingroup$ Welcome to Puzzling! Your tags seem right, and I think this is a nice, interesting first question :) $\endgroup$ – Quintec Feb 9 '18 at 16:50
  • $\begingroup$ As described, if I knock down all 10 pins in the first roll of a frame and again in the second roll, I only get 10 for the frame but I have a spare and a strike. I don't think this is quite right... $\endgroup$ – Peter Taylor Feb 10 '18 at 13:31
  • $\begingroup$ @PeterTaylor - A strike or a spare ends the frame immediately, and the next ball is the first ball of the next frame. The tenth frame is 'special', in that if you throw a strike or a spare, it will be a three-ball frame instead of a two-ball frame, to allow for the 'extra' score earned by the strike/spare. $\endgroup$ – Jeff Zeitlin Feb 12 '18 at 12:39
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I believe the maximum difference is

290 - 110, 180.

I got this by

Reasoning as follows: Strikes make each pin worth 3 instead of 1, so we want as many of those as possible to maximize the difference. We should have 11 strikes(the first 9 frames and the first 2 on the tenth frame), to make each strike worth the full 30 points. This makes the score 290. This knocks down 110 pins. For the minimum, 10 pins also need to be knocked down in each frame, and the smallest score we can get from this is knocking down 0 pins in the first roll, and then all 10 in the second roll(to minimize the bonus given through the spare). In the last frame, we can roll a 0, a spare, and then a strike to minimize the score of the spare in the 9th frame. This gives a total score of 110(10 in each frame except for the last one, 20). Subtracting these gives the answer.

P.S.

If Bob was really able to pull this off, he's probably as good as Alice.

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  • $\begingroup$ Might want to mention that the 11 strikes need to be consecutive, and that the miss must be the 12th roll. $\endgroup$ – Bass Feb 9 '18 at 18:50
  • $\begingroup$ So this is the number my friend and I came up with... but I'm not sure I see an actual proof that it can't be higher. $\endgroup$ – GendoIkari Feb 9 '18 at 21:38
  • $\begingroup$ 11 consecutive strikes is the maximum number or points you can score while every pin is worth at least 2 points. Any further points would have to be scored by knocking down 1-point pins. A pin is always worth at least one point, so making more points couldn't possibly help. Coincidentally, 110 is the maximum number of 1-points pin you can score, so the 111th pin would be worth 2. Therefore the solution is unique. $\endgroup$ – Bass Feb 9 '18 at 21:55
  • $\begingroup$ That's what i tried to convey in my solution, sorry if it wasn't clear. You want to maximize the point difference between pins, and have as many pins as possible. Maximum point difference = 2, maximum pins for that is 110. $\endgroup$ – Quintec Feb 9 '18 at 22:15
  • $\begingroup$ The point difference isn't 2 for all the pins, of course, because the answer isn't 220. The value of a "regularly scoring" pin is 2 if there is a spare preceding it, or if exactly one of the two rolls preceding it is a strike. The value of a pin is 3 only if there are two strikes preceding it. Therefore, the first 10 pins are worth only 1 point, and the pins 11-20 are worth 2. (Also, at the end, pins 101-110 are worth only 2, because the extra pins only count as bonuses). It's impossible to avoid the one-point pins at the beginning, since you need to make the first 2 strikes somehow. $\endgroup$ – Bass Feb 10 '18 at 6:35

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