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Please clarify this for me, is the highest possible break in snooker 147, or is it possible with the free ball rule that the highest possible break is in actual fact 155? An explanation would be much appreciated.
Question
#36067. Asked by puzzlerguy.
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Roget
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I think the answer is 155. If I'm wrong somebody will point it out.
The first player to the table plays a foul shot and leaves the cue ball snookered from the reds. The second player can nominate any colour as a red, which he sinks for one point. The colour is put back on the table. As a result of sinking the 'red' he can now sink the black (7 points) and the black is put back on the table.
There are now all the reds and colours on the table for 147 points plus the extra 8 he scored at the start. 147+8=155
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fosse4
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Penalty points are added before the player takes his first shot so are not included in the Break , if the first shot is a free ball then it can only count as a red,which if followed by a black and both balls are re-spotted will add the extra 8 to the maximum 147 making the break score of 155.
But no-one really cares it's if the player can pot the 15 reds and blacks then the colours to clear the table.
The opposite of your question is what is the minimum score that will win a game (very hypothetical but very probable answer)
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sequoianoir
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22 wins (brown, blue, pink and black)
Loser = 15 reds + yellow & green
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