What is the largest number of strikes you can get in bowling, yet still lose to someone who gets no strikes, and only that same number of spares?
That is, the loser gets X strikes. Winner gets no strikes, but gets X spares.
[edit: below is incorrect]
We have the Sultan of Spares vs the Strikey the Loser.
The maximum score if there are no strikes but X spares:
suppose each open frame is a roll of 9 followed by a gutter ball, and each spare is a 9 rolled by a 1
each open frame could score 9,
each spare scores 10 plus the next roll (a 9) or 19
So the maximum strikeless game with X spares is 19X + 9(10-X) = 10X + 90
max(Sultan of Spares) = 10X + 90
In a game with strikes, each strike scores 10 plus the 2 next rolls
To minimize the score with X strikes, assume no spares. In fact assume all rolls that are not strikes are gutter balls.
For X<6, ie up to 5 strikes, the following frame (2 rolls) can be gutter balls: score is 10X, obviously less than 10X + 90.
For example if X = 5, Strikey the Loser can score as little as 50 but the Sultan of Spares can score as high as 140.
6 strikes, spaced out as much as possible, say in frames 1,3,5,7,9,10 with every other roll a gutter ball.
Strikey scores 70; Sultan of Spares: 150
7 strikes in frames 1,2,4,5,7,8,10. Strikey 100 vs Sultan 160
8 strikes will require at least one '3 in a row'.
1,2,3,5,6,8,9,10 Strikey 150 vs Sultan 170
9 strikes: 1,2,3,4,6,7,8,9,10 Strikey at 210 finally beats the Sultan of Spares 180.
So the answer is you can get up to 8 strikes and still lose to someone who gets no strikes but does get 8 spares.
Edited on January 24, 2024, 4:33 pm
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Posted by Larry
on 2024-01-24 12:07:35 |
If I understand scoring in bowling ...
