Red Spade Open ($1,000,000 Guaranteed)
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Posted on 01 November 2013 by "T".
Red Spade Open returns on November 3 at the world's biggest online poker room. This special tournament ($55 buy-in) has a $1,000,000 guaranteed prize pool - of which the winner will take home at least $200,000! If you eliminate one Team PokerStars Pro you will received a $100 bounty, and if you knock out two or more you'll get a PokerStars Caribbean Adventure (PCA) package worth $15,468!
You can buy-in directly for Red Spade Open for $55, or win your seat for much less via one of many satellites running from October 25 until the tournament begins. Good luck!
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5 comments for "Red Spade Open ($1,000,000 Guaranteed)"
| noonlion||01/11/2013 11:02:31 GMT|
| This sounds quite good as there will be low level satellites to get in.|
The fact you can win a PCA package just by eliminating two pros is pretty sick, you could finish in the first hour and potentially do that, or win and do it, pretty sweet either way.
What sort of player fields are there in these? I'm thinking about 40k people...
| pochui||01/11/2013 11:05:36 GMT|
| The fact you can win a PCA package just by eliminating two pros is not sick at all imo- the thing is, i really feel that you won't be able to do this technically...if you happen to knock out one pro, i am pretty sure you won't be seated at a table where any other pro is playing...|
| takingdrugs||01/11/2013 13:12:20 GMT|
| Yeah I'm In you've got to be lucky to share a table with one pro let alone two|
| 249peanut||04/11/2013 12:34:44 GMT|
| Looking forward to actually sitting at a tourney knoowing the big timmers are present.will be interesting to watch peaple use thier skills to bump them out. Even the pros are capable of a muck or two.|
| doubletop777||11/11/2013 09:58:28 GMT|
| this sounds like a great tourney. $55 entry fee to potentially win $200,000 is not a bad risk reward! would love to play against some of the pros and win the bounty. would be very hard to knock them out and to knock two of them out would be nearly impossible.|
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