Razz -- worst hand wins on Absolute

[jbrennen]

Here's how I implemented a Razz hand ranker. We'll use the two hands in this discussion (Ac 9s 7c 3s Ad 3h 9h vs. 4s 2h 7s 5d 4d 5s 4c) as an example.

Go through the cards and replace each one with the number 1 to 13, based on its rank. If this rank has been seen before in the hand, add an extra 13 for each time it's been seen.

So Ac 9s 7c 3s Ad 3h 9h becomes:

Hero: 1 9 7 3 14 16 22

And 4s 2h 7s 5d 4d 5s 4c becomes:

Villain: 4 2 7 5 17 18 30


Now sort the hands from high to low. Note that by the way we constructed these, there will never be duplicates within a hand:

Hero: 22 16 14 9 7 3 1

Villain: 30 18 17 7 5 4 2

Strip off the highest two numbers in each hand:

Hero: 14 9 7 3 1

Villain: 17 7 5 4 2

Note that Hero's hand represents A,9,7,3,A. Villain's represents 4,7,5,4,2.

To figure out who wins (including in multi-way pots), just start at the left. Each time that one of the hands is not the lowest at a position, throw it out as a loser.

In this case, at the first position, Villain's hand has a 17 (representing a paired 4), while Hero's hand has a 14 (representing a paired Ace). Since Villain's hand is not the lowest at this position, throw it out. Obviously that leaves Hero's hand as the winner.


It's not that hard to get right, IMHO.

[RustyBrooks]

That's a nice way to think of it. When I redo the ranker in C I'll probably steal that. For now I like mine OK: I did it in a scripting language so I make heavy use of associative arrays, which makes my actual algorithm fairly straightforward - in C my algorithm would be fairly difficult.

[jbrennen]

[ QUOTE ]
That's a nice way to think of it. When I redo the ranker in C I'll probably steal that.

[/ QUOTE ]

Unfortunately, I just realized that I got it wrong, to my embarrassment.

(9-9-9-5-5-5-A) vs (K-K-K-K-3-2-A). My algorithm gets it wrong. Oops.

Obviously, the K-K-3-2-A should win over the 9-9-5-5-A, but my "algorithm" gets it wrong. Time to go fix my algorithm.

[chillrob]

I think for that you would just need to add an extra qualifier saying that two pair always loses to one pair.

[Fleshbot]

[ QUOTE ]

More follow-up from Absolute please.
We need a humble soft-spoken "oops" from them to give us closure.

Although their righteous incorrect e-mail responses have been far more entertaining.

[/ QUOTE ]

I'm still waiting for a followup email from them. Although I love the thread this has generated! Somehow I don't think the programmers at Absolute are regular 2+2-ers...

[thebottomline]

Ignore.

[TomBrooks]

I see the solution here as being very simple.

If no player at the showdown can produce a 5 card hand, meaning 5 cards of unique rank--then the best 4 card hand should win.

Simple. And Absolute got it right.

[Wetdog]

Simple, but wrong.

[Andy B]

This is razz, not badugi. Poker is played with five-card hands.

[TomBrooks]

OK. Then...

How is a pair scored? Higher than a King?

And is AA the highest pair or the lowest pair?

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If you count a pair as higher than the highest single card (i.e. a King)...

And if AA is the highest possible pair, 44752 is the winner.

But if AA is lowest possible pair, AA973 is the winner.

And note the way I wrote the hands would then be the correct way to write them using the conventional highest to lowest card (or pair.)

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Or is this whole discussion ongoing because nobody knows the applicable rule; or has this never come up before in the history of poker and a rule never made for it? It's hard to imagine that.