What is the lowest possible winning score in snooker assuming that...
- The loser does not concede the frame
- No fouls are made
- No more than one ball is pocketed on any given shot
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Firstly, under these conditions, the lowest total score is 42 points (all reds sunk with no colour, plus all colours in sequence to end the frame).
Secondly, if a player scored exactly half of these points, the opponent scores the other half, and would tie the frame by definition. If they score less than half these points, the opponent must score more, and would win by definition. Therefore, the minimum winning score is one point more than half of 42, i.e. 22 points.
Finally, is it possible to score this number of points? Yes, clearly, for example
Player 1 sinks a red ball, and then misses the colour, while Player 2 misses only, until all reds are sunk (scores 15-0). Thereafter, Player 1 sinks the yellow and green in sequence and misses the brown (scores 20-0). Player 2 sinks the remaining balls in sequence and ends the frame (final scores 20-22)
Player 1 sinks a red ball and misses the colour, while Player 2 misses only, until all red balls are sunk (scores 15-0). Player 2 sinks the yellow, green, brown, pink and blue in sequence, and misses the black (scores 15-20). Player 1 sinks the black and ends the frame (final scores 22-20).
Other combinations that lead to a total of 22 for the winner, or equivalently a total of 20 for the loser, may be found by trial-and-error or via a constructive process (either depth-first or breadth-first tree searching would be recommended).
Technically the lowest possible score is 31, although it is highly unlikely that this would ever happen in real play. It requires all 15 reds to be potted, on the break (!) and the white to go in off. A foul shot and the next player would be awarded 4 points. That player would then pot all the colours for a further 27 points, 31 in total.