Ten Pin Bowling Score

[JonR]

I played ten pin bowling last night, and scored 180.
What I remember of my scorecard was thus:

'X' 'X' 'a,b' '9/' '9/' 'X' 'X' 'X' 'c,d' 'e,f'

I know that I had 25 on the first frame, so a=5, and that I ended up with 180, and that frame 9 ended 175, so e+f = 5 (neither were zero)
and I know that the machine missed a bit, and that I had 88 at some point.
can anyone help me work out my other scores?

[Cosmologicon]

What does it mean, "the machine missed a bit"?

[OverBored]

I don't think it is possible. I need to check my working, but I get 2b + 2c + 2d = 31

Nevermind that

Edit: Okay, proper answer

Spoiler:

We know a is 5, so we fill that in. e+f is 5, but we cannot gain any more information about that.
The score: (25) + (15+b) + (5+b) + (19) + (20) + (30) + (20+c) + (10+c+d) + (c+d) + (5) = 180

2b + 3c + 2d = 31

I don't know what "missed a bit" means, so I'll ignore it. Only other info is that 88 was achieved at some point.

A little bit of checking shows that this means that 84 + 2b = 88
b = 2

Unfortunately this gives 3c + 2d = 27. This is only possible if you got a spare for c,d. This completely changes all working so either there is a mistake in your memory or in my working. If it is possible you got this spare, see the next spoiler.

Spoiler:

b still has to equal 2:
The score: (25) + (15+2) + (5+2) + (19) + (20) + (30) + (20+c) + (10+c+d) + (c+d+e) + (5) = 180

3c + 2d + e = 27
c + d = 10
e<5
Therefore c+e = 7

possible values of a-f:
a = 5, b = 2, c=3, d=7, e=4, f=1
a = 5, b = 2, c=4, d=6, e=3, f=2
a = 5, b = 2, c=5, d=5, e=2, f=3
a = 5, b = 2, c=6, d=4, e=1, f=4

[JonR]

What does it mean, "the machine missed a bit"?

usually it adds up the score as you go along, when you get a spare, it waits one ball and then works out your spare value, when you get a strike it waits 2 balls and then works out your strike value, and 'the machine missed a bit' by not working out the value of the strike in frame 6until after frame 9 instead of after frame 8 when strike plus two balls had been played.
... makes perfect sense.

anyway, I also had a memory recall, and I got 88 on frame 5, frame 3 was 5 2, frame 9 was a 9 and a zip, and frame 10 was 3 2.
running total of 25, 42, 49, 69, 88, 118, 147, 166, 175, 180.

it was not possible to work out with just the information given.

thanks anyway.

@Overboard - nice working, but when you state that 3c+2d = 27 (or something like that), you miss that if d =0 then c =9, which works, without a spare.

[Xar]

The "machine missing a bit" is actually a superstition thing where until you stop getting strikes it will not complete the scores. Like adding as you go along is unlucky or something.

[Andymeo]

The "machine missing a bit" is actually a superstition thing where until you stop getting strikes it will not complete the scores. Like adding as you go along is unlucky or something.

Actually it's because 1 strike is worth 10 points, 2 strikes in a row is worth 20 points and 3 or more strikes in a row is worth 30 points for each strike. This is why the machine has to wait to see your next score because if you get another strike your score increases.

[Adacore]

The "machine missing a bit" is actually a superstition thing where until you stop getting strikes it will not complete the scores. Like adding as you go along is unlucky or something.

Actually it's because 1 strike is worth 10 points, 2 strikes in a row is worth 20 points and 3 or more strikes in a row is worth 30 points for each strike. This is why the machine has to wait to see your next score because if you get another strike your score increases.

No, if you get more than 3 strikes in a row the next score doesn't matter for the computing of the first score. In the example given, there's a sequence of 3 strikes, followed by a score of 'c'. The first strike could be computed from merely those 3 strikes (10+10+10=30), but if I'm understand right the machine waits until you stop making strikes, giving a score when the 'c' is bowled.

[JonR]

I never knew that - thanks, and it makes sense too, in an odd sort of way.

[SeaCalMaster]

I'ma necro this thread, because someone is WRONG on the Internet. Can't have that, right?

Spoiler:

We know a is 5, so we fill that in. e+f is 5, but we cannot gain any more information about that.
The score: (25) + (15+b) + (5+b) + (19) + (20) + (30) + (20+c) + (10+c+d) + (c+d) + (5) = 180

2b + 3c + 2d = 31

I don't know what "missed a bit" means, so I'll ignore it. Only other info is that 88 was achieved at some point.

A little bit of checking shows that this means that 84 + 2b = 88
b = 2

Unfortunately this gives 3c + 2d = 27. This is only possible if you got a spare for c,d. This completely changes all working so either there is a mistake in your memory or in my working. If it is possible you got this spare, see the next spoiler.

Spoiler:

3c + 2d = 27 has a solution where c = 9 and d = 0. As it happens, this does not give a spare.

[Snark]

@Overboard - nice working, but when you state that 3c+2d = 27 (or something like that), you miss that if d =0 then c =9, which works, without a spare.

That's been noticed.