This weekend Jordan Spieth will attempt to become the youngest golfer to win the career Grand Slam when he tees off at the US PGA Championship. But when he tees off at the first, does his reputation as a ‘brilliant player’, defined as some who has the as the capacity to produce ingenious results as well as the capacity to make mistakes, give him the advantage (e.g. Speith’s final round at The Open last month)? Or does the ‘steady player’ have a better chance of taking home the Wanamaker Trophy?

In 1945, G. H. Hardy presented a model which challenged the traditionally-held doctrine that the brilliant player had the advantage. Instead, he argued that the consistent golfer had the better chance of winning over 72 holes. The two players in Hardy’s model are equal in performance on average, the difference being that the brilliant player has a higher standard deviation whereas the consistent player has a standard deviation of 0. Simply put, the ‘brilliant player’ has more room for error making victory less likely.

Now, Jeehoon Kang in a new paper for The Mathematical Gazette offers an alternative to Hardy’s model. This model takes into account the dependency of shots, different probabilities for each outcome and explores flaws in Hardy’s model. Ultimately, though, Kang concludes that what golfers really need, as described in the Glasgow Herald (1935), is ‘Brilliant Steadiness’.

Kang’s paper will be freely available until September.

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