What's the score?

[Shockerz]

Blue, pink and black left and opponent's just fowled and left a free ball.

I choose the pink, play it onto the blue and pot both the blue and pink, what's score do I get, 5 or 10?

[DeanH]

Blue, pink and black left and opponent's just fowled and left a free ball.

I choose the pink, play it onto the blue and pot both the blue and pink, what's score do I get, 5 or 10?

You get 5. Blue stays in pocket, pink spotted.

[Shockerz]

Yeah I figured five but it's one of them you never see until you're ref'in!!

[DeanH]

What's the score?


The only time when the value of multiple balls being potted in a single strike are accumulated is with Reds when Reds are on.

[SnkrRef]

Dean was spot on as usual! Five points scored, blue stays down, and pink respotted.

[DeanH]

Am I correct in the following scenario:
Red on, snookered, Blue selected as free ball, Blue struck onto a Red, both Red and Blue potted, score of 2, Red remains off table, Blue spotted.
As I mentioned before the only time ball values are accumulated are when Reds are on, the Free ball has the value of a Red for this stoke. Yes?

[jrc750]

Am I correct in the following scenario:
Red on, snookered, Blue selected as free ball, Blue struck onto a Red, both Red and Blue potted, score of 2, Red remains off table, Blue spotted.
As I mentioned before the only time ball values are accumulated are when Reds are on, the Free ball has the value of a Red for this stoke. Yes?

I would agree with that

[SnkrRef]

Am I correct in the following scenario:
Red on, snookered, Blue selected as free ball, Blue struck onto a Red, both Red and Blue potted, score of 2, Red remains off table, Blue spotted.
As I mentioned before the only time ball values are accumulated are when Reds are on, the Free ball has the value of a Red for this stoke. Yes?

Yes, that's right. This has been in the rules for as long as I can remember, and the dubious logic proffered is that, at the start of a frame, there are 15 reds, so it is perfectly feasible to pot more than one red in one stroke, whereas with the colours there is only one of each, so it is never possible to pot more than one. Not sure I really agree with that logic, but who are we to disagree?

[t.lavery55]

Yes, that's right. This has been in the rules for as long as I can remember, and the dubious logic proffered is that, at the start of a frame, there are 15 reds, so it is perfectly feasible to pot more than one red in one stroke, whereas with the colours there is only one of each, so it is never possible to pot more than one. Not sure I really agree with that logic, but who are we to disagree?

The opposing argument being that you should score for every ball you pot, (as is always the case apart from this exception), and that therefore you ought to score for both the free ball and ball on, if both are potted in one shot. It would arguably be more logical to stick to this principle instead.