Two Time Bracelet Winner JC Tran Heads List Of Final Nine Players At This Years WSOP
Ever since this years World Series of Poker (WSOP) started way back on Wednesday 29th May we have brought you news on all the winners and losers at this years event. In total 61 bracelets have been awarded and last night in Las Vegas the final nine players were determined for this years final table.
When play resumes in November two time bracelet winner JC Tran will be the one to beat as he is sitting top of the chip pile with an impressive 38million in chips, and in second place is one time bracelet winner Amir Lehavot from Israel and he will return to the felt with an impressive looking stack of 29.7million chips.
The next three players all have in excess of 25million chips, but there is just 650,000 separating the three of them. In third place is Marc McLaughlin from Canada and he has 26.525million in chips and he will be looking to beat his best finish (3rd in Event 32 2011) at the WSOP when he returns in November. Just behind McLaughlin in fourth place is Jay Farber from America with 25.975million and in fifth is fellow American Ryan Riess with 25.875million, both of whom will be in search of their first ever bracelet win.
In sixth place is Frenchman Sylvain Loosli and he has a decent sized stack of chips which stands at 19.6 million and like the three players immediately above him in the standings he will be looking to win his first ever WSOP bracelet. The final three players all have plenty of work to do if they are to walk away with the first prize come November, but a quick double up or two will see them back in the thick of the action.
Michiel Brummelhuis from the Netherlands lies seventh and has 11.275million in chips, with American Mark Newhouse in eighth place and he has just 7.35miilion chips to his name. The final November Niner is another American in David Benefield who has just 6.375million chips, but he is guaranteed to walk away with at least $733,224 when play resumes in November.