What constitutes a 'Free ball'

[bluenose1940]

Hope this isn't too stupid a question but I don't know the answer and would appreciate some help/guidance please.

Your opponent fouls (4 away) and there are 3 reds remaining on the table, one of them is completely hidden behind the black and the other two reds are touching each other and, you cannot hit the extreme left-hand edge of the ball on the left of the pair because the black is in the way but, you can hit the extreme right-hand edge of the red that is to the right of the pair, you cannot hit the other edges of these two reds as they are touching each other, is this deemed a 'free Ball'?

I hope that you experts can understand what I am trying to say. If I knew how to post a drawing it might help to explain it more clearly.

My opponent likened it to breaking off, as he rightly says, you cannot hit the extreme edge of any of the reds at that point but a free ball is not awarded! My thoughts on this were that a foul had not been committed before the break-off!

Your thoughts or the confirmation of the rule will be much appreciated. Thank you.

[snookerdad]

A red cannot be snookered by another red. Therefore, assume that one red is not there. Are you snookered? Then assume the other red is not there. Are you snookered? If the answer is 'no' to one of these questions, it's not a free ball.

Tim Dunkley (World Snooker coach)

[t.lavery55]

You can only be snookered on a ball on (red) by a ball not on (a colour). From the description you have given above, it sounds like the black is only obstructing one of the reds, and it is the reds themselves that are preventing you from being able to hit both extreme edges of the other two reds. If this is the case, then you are not snookered, so it is not a free ball. The same applies with the break-off shot, as again, it is the reds themselves that are preventing you from hitting both edges of the reds, so you are not snookered.

But if, in the scenario you describe above the black is obstructing the path to at least one edge of all three reds, then it is snookering you, so you have a free ball.

[SnkrRef]

Indeed a red cannot obstruct another red to give a free ball. Imagine the scenario that on the break off the player miscues and fails to disturb the pack of reds. If the cue ball comes to rest towards one side of the table, then although no red can be struck on both fine edges, it is not a free ball because the extreme left (or right) red could be struck if the other reds weren't there in front of it.

[bluenose1940]

Thank you for making that clear for me, much appreciated.