rule question

[alabadi]

i was playing a friend last week and this situation arose.

there were 2 reds left on the table, i put my friend in an awkward situation, he tried getting out of it however fouled by not hitting the red.

when i came to the table there was one red that was close to a pocket but i was snookered on that ball. i could see the other red clearly but it was on a cushion.

i thought i have read or heard this somewhere before, maybe on here. is it a free ball because i can not hit one of the reds, although i can hit the other red both extreme sides.

i have looked at the rules and cannot figure out is this true. all i can gather is that a free ball is given after a foul and the ball on cannot be hit at both extremes. when it says the ball on does it mean any of my choice.

i thought it was a free ball my friend said no way because i can hit the other red.

any help appreciated

Alabbadi

[krill]

Your friend is correct, it is not a free ball because you could see the one red on both sides. As long as you can see a red on both sides it is not a free ball.

[Terry Davidson]

Your only viable choice is to put him in again or if you're playing by the rules then 'foul and a miss' and replace the balls.

You must be snookered on ALL reds for a free ball

Terry

[SnkrRef]

This is how the rule starts off:

17. Snookered
The cue-ball is said to be snookered when a direct stroke in a straight line to every ball on is wholly or partially obstructed by a ball or balls not on. If one or more balls on can be struck at both extreme edges free of obstruction by any ball not on, the cue-ball is not snookered.

So clearly you didn't have a free ball.

Terry refers to replacing the balls under the F&M rule, but obviously this needs to be tempered with 'was it a good enough attempt'? Failed attempts at getting out of a snooker don't automatically lead to a F&M call, at least not in the amateur game. The referee should have regard for the difficulty of the snooker and the ability of the player.

[DeanH]

This is how the rule starts off:

17. Snookered
The cue-ball is said to be snookered when a direct stroke in a straight line to every ball on is wholly or partially obstructed by a ball or balls not on. If one or more balls on can be struck at both extreme edges free of obstruction by any ball not on, the cue-ball is not snookered.

So clearly you didn't have a free ball.

Terry refers to replacing the balls under the F&M rule, but obviously this needs to be tempered with 'was it a good enough attempt'? Failed attempts at getting out of a snooker don't automatically lead to a F&M call, at least not in the amateur game. The referee should have regard for the difficulty of the snooker and the ability of the player.

in the OP I don't think there was a referee, so this could be "if there is a referee, they would have regard..."

alabadi - as has been said as you could see a red clearly, i.e. able to hit both extreme edges, there was no Free Ball option. The rule that SnkrRef quotes shows what other have been saying, that "every ball on" is the phrase of importance in your scenario, just because the ball you want to go for cannot be seen does not come into it but every Red (Reds being on) has to be considered.

[SnkrRef]

in the OP I don't think there was a referee, so this could be "if there is a referee, they would have regard..."

In games without an independent referee, the opponent (non-striker) is deemed to be the referee!

[DeanH]

In games without an independent referee, the opponent (non-striker) is deemed to be the referee!

very true
but how often does this actually come true with "friends" just playing

[tommygunner1309]

A pal, last week, asked me why you cannot snooker behind a free ball. I have never had the reason explained to me, but I told him, that, if after a foul, he was snookered on the reds and he nominated blue as a free ball, that blue became a red for the purpose of that stroke, and therefore you cannot snooker the reds with another red.
I wonder if any of you have another explanation that makes sense for this rule?

[mythman69]

A pal, last week, asked me why you cannot snooker behind a free ball. I have never had the reason explained to me, but I told him, that, if after a foul, he was snookered on the reds and he nominated blue as a free ball, that blue became a red for the purpose of that stroke, and therefore you cannot snooker the reds with another red.
I wonder if any of you have another explanation that makes sense for this rule?

You can't snooker behind a free ball because the rules say you can't. Except when there is only the pink and black left on the table.

You can use the free ball to snooker your opponent as long as another ball closer to the cue ball also snookers the ball on. Last night I used a free ball to snooker one of two reds left on the table as the other red (closer to the cue ball) was already snookered behind the black.



For more info see The Statman's great Free Ball posts:
Free Ball – how to judge if you have one
Free Ball – what you can (and can't) do with it

[SnkrRef]

It seems a little odd that you can't snooker behind a free ball (generally), but that's been the rule for at least the last 25 years.

Imagine this scenario. Last red is hanging over a middle pocket. All colours bar the blue (which is on its spot) are close to the cushions. Having failed to hit the red, player A has now left the cue ball close to the blue but on the opposite side to the red, thereby snookering him, and a 'Free Ball' is called. The obvious shot would be to roll up tight to the blue, as there's no pot on any colour, and it's almost impossible to get the cue ball safe... but it's not allowed. So basically Player B is forced into asking Player A to play again.

Maybe therein lies the logic: if you don't like the position left you can always ask the player to play again, so you shouldn't gain an additional advantage by being able to snooker behind the free ball.

[DandyA]

You can use the free ball to snooker your opponent as long as another ball closer to the cue ball also snookers the ball on. Last night I used a free ball to snooker one of two reds left on the table as the other red (closer to the cue ball) was already snookered behind the black.

if there are two reds left on the table, I don't think distances matter do they? as long as the free ball is only snookering one red and some other ball is snookering the other red, it doesn't matter which is closer ...

[SnkrRef]

if there are two reds left on the table, I don't think distances matter do they? as long as the free ball is only snookering one red and some other ball is snookering the other red, it doesn't matter which is closer ...

No they don't. If different reds are snookered by different balls not on, then there is no effective snookering ball.

I think though, that the post you quoted had two scenarios in the same paragraph. First, where the nominated ball would still snooker you on (say) the last red, but there is another ball no on which is closer to the cue ball, so that will be the effective snookering ball and not the nominated ball.

He went on to refer to the use of the 'no effective snookering ball' rule when different reds are snookered by different colours.

[mythman69]

No they don't. If different reds are snookered by different balls not on, then there is no effective snookering ball.

I think though, that the post you quoted had two scenarios in the same paragraph. First, where the nominated ball would still snooker you on (say) the last red, but there is another ball no on which is closer to the cue ball, so that will be the effective snookering ball and not the nominated ball.

He went on to refer to the use of the 'no effective snookering ball' rule when different reds are snookered by different colours.

Thanks for clearing that up!

[DandyA]

No they don't. If different reds are snookered by different balls not on, then there is no effective snookering ball.

I think though, that the post you quoted had two scenarios in the same paragraph. First, where the nominated ball would still snooker you on (say) the last red, but there is another ball no on which is closer to the cue ball, so that will be the effective snookering ball and not the nominated ball.

He went on to refer to the use of the 'no effective snookering ball' rule when different reds are snookered by different colours.

Thanks for clearing that up!

yep, many thanks from me too SnkrRef ...

a slightly different question ... say there's one red on the table, you nominate a free ball and, after your shot, both the free ball and another colour both snooker the red and both of them are exactly the same distance from the cue ball ...

is that a foul or not ... and why?

[SnkrRef]

yep, many thanks from me too SnkrRef ...

a slightly different question ... say there's one red on the table, you nominate a free ball and, after your shot, both the free ball and another colour both snooker the red and both of them are exactly the same distance from the cue ball ...

is that a foul or not ... and why?

Yes, it would be a foul. This is covered by 2.17(b)(ii)

17.
(b) If the cue-ball is so obstructed from hitting a ball on by more than one ball not on:
(i) the ball nearest to the cue-ball is considered to be the effective snookering ball; and
(ii) should more than one obstructing ball be equidistant from the cue-ball, all such balls will be considered to be effective snookering balls.