Tiger Woods has reined supreme over the golfing world for five years, holding the number one spot since 2005, but his position is now under threat from the supremely consistent Lee Westwood and Phil Mickelson.
The WGC-Bridgestone Invitational tees off today, Woods defending the title he has won seven times in Ohio. Should he fail to do so, he could lose his number one position to either Mickelson or Westwood.
Here’s how. If Westwood wins and Woods is outside the top two, he will become world number one. If he finishes second, Woods fails to finish better than tenth and Mickelson DOESN’T win the tournament, he will become world number one. If Mickelson wins the tournament, he will become world number one. If Mickelson finishes in the top four and Woods fails to finish in the top 37 (yes, THIRTY SEVEN), he will become world number one.
It’s a bit complicated, but basically the odds on favourite to be the best golfer in the world come Sunday is Lee Westwood – despite never winning a major.
There’s a few stumbling blocks. First of all, while Westwood is in fantastic form, Woods LOVES this course. If he is to regain some of the form he has lost since his fall from grace, it could well be here at Firestone. Then consider this – Westwood will tee off next to Woods for the first two rounds. That’s pressure, playing alongside the man you are trying to depose!
It’ll be compulsive viewing, and you’ll want to keep an eye on Mickelson too, of course. For me, I really hope Westwood does it, his game deserves it and it might just give him that boost to win that first elusive major.















