Originally Posted by CrayBish
hmmm but (let's take same example as in OP) "554KK / AA333" Obv, it's 5 cards but technically aren't we really counting the non-paired cards only? Therefore, 45K vs A3...since the double cards cancel each other out?
I'm not really a razz player....(so I may be totally wrong the way I view it...therefore don't bite my head off!)...
Nope, the paired cards definitely
don't cancel each other out - you have
to play a five-card hand. In the 554KK vs AA333 example it's two pair (kings and fives with a four) against a full house (threes full of aces). Two pair is a terrible
Razz hand, but it's still lower than a full house, so it wins.
Note that for the two players to be playing those two awful hands, they'd have to have two even worse cards in their hands that they weren't using. So it would actually have been 5554KKK vs AAA3333 or something.
Originally Posted by dtexhol567
1) 55566 beats 44477?
2) 55666 beats 44477?
3) Above is really in effect 65 beats 74, since duplicates don't count?
4) 55666 ties with 55566? (65 =65)
Cheers, checking sanity now.
1 - 44477 wins, because fours-full is a lower hand than fives-full.
2 - 44477 wins again, because fours-full is even lower than sixes-full
3 - Definitely not, see above
4 - 55566 wins
All you need to do is think of it in regular high-hand-wins poker terms, and then reverse the rankings. The hand that would have won in stud hi or NLHE loses in Razz.
This is all largely meaningless, BTW, because it's very rare that you'll actually come across a real-life Razz situation where these kinds of hands are going up against each other. The hand ranking situation that trips most people up is actually this one:
56789 beats A234T, because you work from the highest card down. In this case the first hand is nine-high and the second hand is ten-high. So even though the second hand has four other cards that are lower than all
of the cards in the first hand, it loses because the ten is higher than any card in the first hand.