Phil Mickelson incurs bizarre penalty for hitting moving ball at US Open
Last Updated: 18/06/18 1:42am
Phil Mickelson ran up a sextuple-bogey 10 in a bizarre incident which soured his 48th birthday celebrations on day three of the US Open.
Mickelson was putting down the hill for bogey at the par-four 13th having earlier seen an early chip from the back of the green run off the front edge, but what happened next will be replayed for many months to come.
The six-time runner-up saw his putt run wide of the hole and it was heading off the green once again, but he suddenly broke out into a trot and caught up with his ball, knocking back towards the hole while it was still on the move.
Mickelson's actions even caught playing-partner Andrew Johnston by surprise, but deliberately striking his ball before it had come to rest incurred a two-stroke penalty as a breach of rule 14-5.
The USGA leaderboard initially showed Mickelson had made an eight, which became a nine swiftly followed by a 10, and the tournament organisers then issued a statement clarifying the ruling.
"During play of the 13th hole Phil Mickelson made a stroke on the putting green at the time his ball was moving. As a result, he incurred a two-stroke penalty for a breach of Rule 14-5. His score for the hole was 10."
There were initial rumours Mickelson had withdrawn from the event, while the USGA refused to comment on whether the five-time major champion would face any misconduct charges for his behaviour.
But senior rules official John Bodenhamer did confirm to Sky Sports that Mickelson would not be disqualified, as the breach of Rule 14-5 results in a maximum two-stroke penalty.
When the left-hander eventually holed out, he was 10 over par for the day and +16 for the tournament, while Beef was soon on the same score when he dropped six shots in four holes from the 12th.
