What is the highest theoretical break in snooker?

free ball...pot red n black=8 then 147

Opponent fouls and gives you a free ball, then you pot any colour which is worth 1 and take the black to score 7. Break 8 - 15 reds on the table still, clear as normal.

Got it - 162

Technically possible but highly unlikely to occur.
The first player plays a succession of misses so the 2nd gets 159 points in aways. The 2nd players misses too, gives the first player a free ball and 4 away. The first player gets a break of 155 from the 147 + free ball + black. Scores are now level so black is respotted, which he pots too resulting in 155 +7 = 162
Break = 155 +7 = 162.

The simplest explanation is normally correct, Occam :^0

???? As soon as the player trying to escape from the snooker needed snookers to win then the miss rule would no longer apply.

Also, are you mixing up 'total score' with 'break'?

Aah, you're wrong!
Because when he goes above 147 behind, he'd need a snooker, so no miss would be called!

(I must get out more)

technically the highest break possible is infinite

Also, John Virgo would have spontaneously combusted by then, probably leading to an evacuation of the arena which would end the break.

and he's INCHH PERRRFECT ON THE BLACK !!!!!!!! :0

I had heard that before (infinite).

How?

I don't think a respotted black would count towards a break, as your break finishes when you pot the black to draw. Then you have to draw to see who plays next. It's not a continuation of the break.

There is a way which involves knocking balls off the table but I can't remember how.

Occam's Razor 03 May 14:35
I had heard that before (infinite).

How?

something to do with the player and the ref permanently mistaking a colour for a red and realising their mistake after the next colour has been potted

Thats the one!

Thank you.