17-card Chinese Poker hands

[markgritter]

Is there a better way to set either of the following hands

Comments

[hgfalling]

We usually play the following tiebreaker, which is a superior strategic game:

Expose the odd card of the deck. If it's a tie, then the player with the highest card of the suit of the odd card goes +1.
If a guy wins 3 hands, he is +3, two hands + 1 tie = +2, 3 hands plus a tie = +5, sweep = +7.

5c4c3c2cAc / 5h6d8hTdQs / 4h5d7s8c / JhJdKh

I would play:
AKQJT 4c3c2c6d8h 7sJd4h8c 555 i think.

I have the worst possible holding for jacks in front, one each of A/K/Q. This means my opponents will be less likely to make big full houses with these cards, and instead will play big pairs in front. Jacks are pretty bad anyway.

you played #2 right imo. i dont think it's right to improve your 9 low to beat all the 8s that nobody has because you have three of them.

[mgrape]

I think you meant, "If it's a tie, then the player with the highest card of the suit of the odd card in the 2-7 hand goes +1." So if the 52nd card is the Kd, and two players end up tied, the player with the highest diamond in the deuce hand goes +1. If it's 3-1 or a sweep, the diamond doesn't matter.

[hgfalling]

Yes, that's what I meant. The highest card of the suit in the 2-7 hand! This means that you sometimes weaken your deuce hand to get the tiebreaker.