What Is The Lowest Score Possible For A Clearance In Snooker ?

[johnny.5]

what is the lowest score technically possible for a clearance in snooker ? from a single visit.
I make it 37
Am i correct ?

[jim360]

Who's being uppity? On the other hand, that's the second time today that you've called me flat-out wrong when I've been quickly able to produce evidence showing nothing of the sort. It's impressive that you can still spin that to be my problem, rather than yours. Don't call people wrong so bluntly if you haven't actually checked your own argument -- a lesson I should have thought was obvious.

[New Judge]

What do you mean by "clearance"?

My definition of a clearance is when all the remaining balls are potted. Thus the lowest possible clearance is seven - when only the black remains. Of course it could be argued that the lowest possible is zero. If a player approaches the table with six or less points between him and his opponent and with only the black remaining, if the black is potted by means of a foul stroke (scoring zero points for the player at the table and seven penalty points for his opponent) then the frame is over.

However, if the question really is what is the lowest clearance when all balls are still on the table the answer is, as quoted, 37. All 15 reds potted in one stroke followed by a yellow (17). Then 20 for all the colours bar the black (potted with a foul stroke).

A popular question is “what is the lowest total score possible in a frame?” The answer to this is 31. All reds potted by means of a foul stroke (4 penalty points). Then 27 for the colours – total 31. Thus the lowest winning score (in similar circumstances) is 16.

[Jackdaw33]

Sorry, but I think you are all talking a load of balls.

[New Judge]

So come on, jd, what's your solution which we are all missing?

[grumpy01]

As the question is asked the answer must be 72.This would be 15 red's followed by 15 yellow's then the 6 colours in sequence.Thus 15+30+2+3+4+5+6+7 = 72

[New Judge]

But you can pot more than one red at a time, grumpy.

Whilst I accept that potting 15 reds with one shot is about as improbably as it gets (but is nonetheless a valid consideration when discussing the question theoretically) it is not at all unusual to pot two reds with one shot. You've also neglected the fact that the final black can be potted by means of a foul stroke and hence score zero. (Admittedly the opponent receives seven points and that would count towards the aggregate score, but not the winner's).

[JJ109]

If you are counting 37 with an answer where the black is potted and he goes in off with a foul, then why isn't the answer 1?

you pot all 15 reds, the colours and the white all with the same shot!
Obvious foul, but you have potted all the balls in one stroke!

[New Judge]

Wow!

That really is "thinking outside the box! :-)

But shouldn't the answer in that case be zero (apart from the seven away for potting the black when it is not the ball "on"?

[dannyk13]

Black would be re-spotted so it would not amount to a clearance.

[New Judge]

"Black would be re-spotted so it would not amount to a clearance."

Why would it, danny? The black is only re-spotted when the scores are equal after it is scored or fouled. The scores wouldn't be equal as the opponent would have seven points.

[dannyk13]

It would be re-spotted when the cue ball went in-off.

[dannyk13]

Sorry NJ. Just checked the rules and you are correct, the game would end with the foul.

[New Judge]

Why just the black?

All of the colours would have been potted by means of a foul stroke. Thinking this through, all six of them should be re-spotted. Taking a more plausible example along the same lines, a player pots the final red and with the same stroke pots a colour (any one, doesn't matter). Foul stroke, no points for the red and points away depending on the colour illegally potted. That colour is re-spotted. If he should happen to pot two colours as well as the red, both of them would be re-spotted. So if he pots all six colours along with the red.....? (And the foul stroke does not have to involve an in-off. Simply potting a colour along with the red of course constitutes a foul).

So JJ's solution is not really a runner. I would suggest that he gives seven away and all the colours would be respotted.

[Ninnynanny]

"Theoretically" if one the break-off all 15 red balls were all potted, then all the colours were potted, one after another by the white ball the score would be 42

[jim360]

^^ 44, because you need at least one colour before clearing all the colours, which would be the yellow.

I still maintain that a clearance can't be called a proper clearance if it ends with a foul shot.

[Flonska]

Ninnynanny . . .

42? . . . not sure about that . . . .

If all 15 reds are potted in a single stroke - the score would be 15. The next ball with the lowest value colour would be yellow - 2. If potted, the score would be 17. The yellow would be returned to the table and potted again - score = 19 followed by Green 3, Brown 4, Blue 5, Pink 6 and finally Black 7. That would make 44 - wouldn't it?

[-Talbot-]

jim360

I still maintain that a clearance can't be called a proper clearance if it ends with a foul shot.



I agree, that's why the answer is ...



7

[aelmpvw]

next time i pot all fifteen reds from the break i'll go for any colour bar the yellow