Razz -- worst hand wins on Absolute

[Cashane]

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**The worst 5 card hand, from the 7 dealt, someone can have is two pair.

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Didn't read the rest, but wrong assertion. JJJ4444, the lowest hand is 444JJ (a full house).

[Praxising]

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**The worst 5 card hand, from the 7 dealt, someone can have is two pair.

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Didn't read the rest, but wrong assertion. JJJ4444, the lowest hand is 444JJ (a full house).

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Full houses don't count in Razz - that is, they are considered not to exist. So, 444JJ is actually two pair with a 4, written JJ444 which would win compared to JJ445. (Of course, if one did have a full house, the other player couldn't have the same 2 pairs - obviously. The illustration is theoretical.)

[MarkGritter]

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Full houses don't count in Razz - that is, they are considered not to exist. So, 444JJ is actually two pair with a 4, written JJ444 which would win compared to JJ445. (Of course, if one did have a full house, the other player couldn't have the same 2 pairs - obviously. The illustration is theoretical.)

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Wow... this is the first I've ever heard of that. Can you provide a source?

PokerStar's rules explicitly list 55532 as the worst hand from their examples (worse that KK432). So, at least in their implementation, trips exists and should be counted as worse than two pair. Presumably full houses exist as well, though we could ask.

In A-5 lowball draw, full houses and trips certainly do exist.

[SGspecial]

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Full houses don't count in Razz - that is, they are considered not to exist. So, 444JJ is actually two pair with a 4, written JJ444 which would win compared to JJ445. (Of course, if one did have a full house, the other player couldn't have the same 2 pairs - obviously. The illustration is theoretical.)

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Wow... this is the first I've ever heard of that. Can you provide a source?

PokerStar's rules explicitly list 55532 as the worst hand from their examples (worse that KK432). So, at least in their implementation, trips exists and should be counted as worse than two pair. Presumably full houses exist as well, though we could ask.

In A-5 lowball draw, full houses and trips certainly do exist.

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AFAIK full houses do exist, tho if Prax or anyone with a reliable source says otherwise I can be convinced. Trips, however, do not and the pokerstars example hand you quoted is impossible to play at showdown.

[electrical]

What possible Seven-card hand could make 55532? I don't think that's possible, assuming trips count as trips and not a pair of Fives with a Five kicker.

[RustyBrooks]

Actually I don't think it's possible to have trips in your hand without a full house. Not "illegal" just impossible. Say you have only 3 different ranks of cards in your hand (required if you want to have trips because you need XXXYZ, if you had XXXYY it's a full house)

so you need at least XXXYZ and you'll need 2 other cards as well.
* If you add 2 Ys or 2 Zs then you get XXXYYYZ who's best hand is XXYYZ
* If you add 1 Y and 1 Z then you get XXXYYZZ who's best hand is XXYYZ (assuming X < Y < Z)
* if you add anything else, like AA or AB you get non-trips like XXYZA or XYZAB

You can still make a full house with XXXXYYY. I have never heard this is not considered a hand by the rules of razz but I've never seen it happen either.

[jbrennen]

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You can still make a full house with XXXXYYY. I have never heard this is not considered a hand by the rules of razz but I've never seen it happen either.

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I always assumed that XXXXYYY played as a full house and lost to any non-full-house hand.

The odds against being dealt XXXXYYY in a 7 card game are about 214,397 to 1. On top of that, the odds that such a hand would actually stick around to see showdown have to be very small -- I think I'm being quite optimistic if I estimated that 1 in 500 of such hands would stick around to showdown either by being all in very early or by not having to call a bet on 5th street or later. (By 5th street, any such hand would be showing a pair and would be drawing to a pair at best.)

So figure somewhere around 100 million to 1 against such a hand actually reaching showdown... [img]/images/graemlins/shocked.gif[/img]

I've had my own two pair hand reach showdown, but that's only happened to me once, and such hands are a whole lot more likely than XXXXYYY. (In that hand, I had 2-9-K-K-9-K-2, I was the bring-in, got two callers, and they checked the hand down all the way.)

[Praxising]

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PokerStar's rules explicitly list 55532 as the worst hand from their examples (worse than KK432).

In A-5 lowball draw, full houses and trips certainly do exist.

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This isn't A-5 lowball draw, though; it's Razz. And Poker Stars is wrong, as I just emailed them. Of course, like Absolute, they might insist they are right.

As SG said, the PS example can't be made as a showdown hand in Razz from any 7 cards.

But - when you make a Razz hand, you take uniquely-ranked cards and group them, to get the lowest hand, so, if you have only four unique cards (8765) and put the next lowest card available in front: so from 8765888 you end up with 88 765. So, even if there could be such a hand as 55532, it would still be the three lowest cards (532) preceeded by the 55.

But it really doesn't matter as it's all theoretical and not practical. Or, hey, I'm wrong. Been known to happen.

[Xhad]

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This isn't A-5 lowball draw, though; it's Razz.

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Isn't razz->stud as A-5 lowball draw->draw high?

If not, can you provide a source?

[jbrennen]

The reason that trips don't "play" in Razz has nothing to do with any of sort of rule which says that trips don't exist or don't count. It has to do with the fact that Razz is a 7-card game, and any 7 card hand which can make a valid trips hand (XXX-Y-Z) can also make a two pair hand which plays lower. In other words, there is no 7 card Razz hand where you would choose to play trips as your 5-card low.