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USGA clarifies ruling on why Phil Mickelson wasn’t disqualified

Official statement makes distinction between making a ‘stroke at the ball’ as opposed to trying to ‘deflect or stop the ball.’

Phil Mickelson eyes the shot on the 4th

Phil Mickelson eyes the shot on the 4th hole at the final round of the 2018 U.S. Open Championship at Shinnecock Hills Golf Club in Southampton on Sunday. Photo Credit: James Escher

The USGA on Sunday issued a further statement on their ruling in regard to Phil Mickelson’s deliberate putting of a moving ball on the 13th hole Saturday. There was some greater detail, but it again reinforced their determination that Mickelson did not do anything that would cause him to be disqualified. Part of the statement is as follows:

“There appears to be some continued uncertainty about the basis of the ruling with Phil Mickelson during the third round of the 118th U.S. Open, and we would like to further clarify previous statements. During play of the 13th hole Mickelson made a stroke on the putting green at his ball which was moving. As a result, he incurred a two-stroke penalty for a breach of Rule 14-5; the stroke made at the moving ball also counted. His score for the hole was 10. Rule 14-5 does not include a serious breach clause or disqualification as part of the penalty statement.

“Rule 1-2 (which could lead to disqualification) did not apply in this situation because Mickelson made a stroke at the ball (defined as the forward movement of the club with the intention of striking at and moving the ball) as opposed to another act to deflect or stop the ball in motion, which are two acts covered by Rule 1-2.”

In other words, Mickelson was able to play Sunday. He shot 69 and finished 16 over par for the tournament.

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