Fifteen Reds, Fifteen Blacks & all the colours would yield a break of 147, but if a foul is committed before any balls have been potted & a free ball is awarded a break of 158 is possible..
I reckon 162 is possible; if player A fouls the black before any balls are potted, then player B is awarded 7 points. If, with a free ball, as you say, he then nominates the black as a red, pots it and then pots the re-spotted black (as a black), he has 15 points. If he clears the table for 147, his total score would be 162. Agree or disagree?
The highest possible score is infinate, someone could conceivably carry on fouling for hours. The highest possible break is 155 and not 158 as I previously stated. Points awarded as fouls do not count as part of the break.
The highest break that can be made under normal circumstances is 147. To achieve that, the player must pot all 15 reds, with the black after every red, followed by potting the six remaining colours. This "maximum break" of 147 rarely occurs in match play.
If an opponent fouls before any balls are potted, and leaves the player a free ball, the player can then nominate a colour and play it as a red ball. Then, black can be nominated as the next colour. This means it is actually possible to score the value of 16 "reds" and blacks, which equals 155 points. This has never been done. The highest break in tournament play is 149, the highest break in professional matchplay is 148. (see also highest snooker break).
You didn't say wh4ther you meant in a single break or in a frame, but the answers seem to have covered both options. 155 for a break, no limit for a frame score.
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