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Pike
08-25-2005, 02:44 PM
Does anyone know what the minimum age for someone playing in the US is for the WCOOP? On the website it says:

The minimum age to play in any PokerStars event is 18 or the legal age of majority in your country, whichever is older.
TomHimself
08-25-2005, 02:46 PM
18
08-25-2005, 02:49 PM
I believe you only need a credit card. Which you cant legally get til your 18. Didnt some cocky 17 yr old make to the final table one year, through his mom's credit card??
godofgamblers
08-26-2005, 01:14 AM
It says or legal age of majority in your country though...I was thinking about playing in a few and I'd be real disappointed if I luckily made a final table and found out Im DQ'd because I'm only 19 and its 21 for the US...
shaniac
08-26-2005, 02:23 AM
[ QUOTE ]
It says or legal age of majority in your country though...I was thinking about playing in a few and I'd be real disappointed if I luckily made a final table and found out Im DQ'd because I'm only 19 and its 21 for the US...

[/ QUOTE ]

You really don't have to worry about Pokerstars dq'ing you. There are hundreds of underage American players who will be participating in the WCOOP with no fear of penalty.
Pike
08-26-2005, 05:28 AM
I emailed them and they replyed that being 18 was fine for US citizens.
Jeremy517
08-26-2005, 11:25 AM
[ QUOTE ]
I believe you only need a credit card. Which you cant legally get til your 18. Didnt some cocky 17 yr old make to the final table one year, through his mom's credit card??

[/ QUOTE ]

I think you can get one before you are 18, if you have a someone cosigns. Granted it was 14 years ago, but I had my first CC when I was 16.
godofgamblers
08-26-2005, 11:40 AM
Thanks for the replies, time to start satelliting.
shaniac
08-26-2005, 05:06 PM
[ QUOTE ]
but I had my first CC when I was 16.

[/ QUOTE ]

And what does getting a credit card at 16 have to do with gambling online at 16?

The only way to circumvent the 18+ rule on Stars is to pretend to be someone else, who is of-age. That's something that could get you dq'd, I imagine, though it would tough-to-impossible to prove.