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there are more complicated systems than that, but I never found blackjack particularly interesting. poker is much richer in strategy and you are taking money off other players, so casino's will never get angry at you for winning! ;).

Re: 2012 HSC MX2 Marathon He explained his method to me through PM and I don't think it is valid, but we will see if someone can post a valid solution using MX2 only techniques (I don't think this is possible).

10+1 play money chips. It is not legal for most people on here to play for real money. So the league is more for bragging rights I guess, your results are tracked.

No worries. No one is obliged to play on any given night, just if anyone wants, there will be a tournament on every night at 10pm. Depending on turnouts I will make more.

There is no upper limit, it will start on multiple tables if need be.

The first tournament is at 10pm tonight!

No, it currently has 5 of us on it, playing with play money.

Did so, you should be able to find and join our game through the home games section.

Nope, go to home games and you should be able to check out the group, I added you.

Follow the instructions at top of thread to join the club, then you sohuld be able to navigate to the club tables pretty easily.

yep anyone who wants to play today just drop in at the p00n table, its a cash game so u can enter and leave as u wish.

Awesome, going to grab coffee then I will create a mini tournament when I return. Unless you guys are more in the mood for a cash game?

amazeballs being the word used to describe what I think of your name.

Done hahaha your name is amazeballs.

K, need to caffeinate myself.

Of course, we aren't playing for money or anything.

Hardcore. Winner gets kunis.

I am sitting at the Heads up P00n table if anyone wants to play.

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Re: 2012 HSC MX2 Marathon The obvious method that comes to mind is looking at the sector contour with one side being part of the positive real axis, and an angle of pi/4 radians. The integral along the circular arc tends to zero which implies that half of the Gaussian integral is equal to the...