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  Poker - Question about Low Chicago (Low Spade in the Hole)
 
  #1  
17-03-2008, 7:56 PM
TheMighty0z
New Member
 
Location: Danvers, MA
Plays at: Full Tilt
Likes: NLHE
Posts: 3
Question about Low Chicago (Low Spade in the Hole)

Hi, I'm a new member to the forum, and wanted to ask the opinion of everyone here. There is a group that I play poker with that gets together once or twice a year for a dealers choice poker night. They've been playing for years before I was invited (about 4 years ago). One of the games commonly called is Low Spade in the Hole (Low Chicago). Usually when this game is played the 2 of spades would be the lowest card in the hole, which would split the pot with the best hand. They have always played that the ace of spades could be played as a low card in the hole to split the pot, but that the ace could not be played both high and low in the same hand. If you played the ace of spades as a low card in the hole it could only be used as a low card in the hand. I've seen the ace of spades played as both low spade in the hole, and the low part of a straight (5-4-3-2-A) in the same hand in the past as they have chosen to play the variant of the game.

Now, I didn't invent this variation and am not asking for approval for their house game.

The problem that came up is as follows:

During our last poker night some one else called Low Spade in the Hole as the game. My hand was 9-9-A-A-x, with the ace of spades in the hole. According to the way that they have chosen to play the game I saw my hand as two pairs (nines over aces) with low spade in the hole. Since I was calling the ace as a low card in the hole they were essentially ones.

Two players were adamant that the pair of aces could not count because, and I quote: Aces are NEVER a low pair in poker. I disagreed, arguing that the ace had paired and that was not nullified by playing the ace of spades low. I had played the ace low only in the hand, just as they had specified in their house rules. The arguing players responded that their position was not a house rule, but a universal rule that aces are never low in poker. Now, they are clearly wrong to state that aces are NEVER low cards in poker (A-A is the low pair in Razz and Ace to Five Lowball, as well as playable low in Seven Card Stud Hi-Low and Omaha Hi-Low). But if there are any Low Chicago players on the forum I'd like to know what their thoughts are on this matter.
 

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  #2  
17-03-2008, 10:53 PM
zachvac
2 More Years
 
Location: Cleveland, OH
Plays at: Pokerstars
Likes: NL Hold 'em
Posts: 4,578
Haven't played the game but I'm pretty sure in most games the high and low are separate, as in you can play an A as low for the low and high for the high. But as laid out you run into problems like this. I would aggree with you, with the caveat that if someone had for example 9922x they would beat you. The other question is when do you have to declare your hand. Does the person with the A of spades get to see everyone else's cards before declaring?

Like say you had AA884 and the 4 was a spade. If someone else had the 2 or 3 of spades you would obviously play the A as the low spade but if no one had the 2 or 3 of spades and you needed the AA to play as high pair to win the high you would use the 4 as your low spade. See my question?
  #3  
17-03-2008, 10:57 PM
smd173
CardsChat Regular
 
Plays at: PokerStars
Likes: Razz
Posts: 850
I've never played Low Chicago, but I do play Hi/Lo games alot. You mention A-A as low pair in Razz, but that's really stretching the interpretation since AA in Razz is not good at all. And in Razz and A-5 Lowball you are still looking at singular Aces being considered low, not a pair. Also in Stud8 or O8, Ace can act as hi or low at the same time, which is not the case in the rules you have stated for Low Chicago.

Think of it this way...if you did not have the Ace of spades, would you still be declaring your hand as Nines over Aces? Of course not.

Considering you had the Ace of spades, you should be happy you won half the pot and hope for the scoop in the future.
  #4  
18-03-2008, 12:23 AM
K_Kahne_Fan
CardsChat Regular
 
Likes: Hold-Em
Posts: 1,039
I'm guessing your two pair was the high hand as well? If so, the "in the hole" card should have NO bearing on the 5-card hand. As far as I'm aware it's a seperate hand altogether. If you held.

[blank]

and no one had the , your would be low in the hole AND THEN you look at the best 5 cards in your hand, which if your 9922 won, then you would sweep the pot. As long as you (and everyone) always called A as low, then it's just like having 9911x, but your 1c is low in the hole as well. Long story short, it's 2 seperate games in 1.
  #5  
18-03-2008, 12:30 AM
squarer
Junior Member
 
Plays at: bodog
Likes: holdem
Posts: 29
i once got into a similar argument over 5 of a kind, playing wild cards. I had 5 and the other player said that was not a real poker hand. I countered that since when is 4 of a kind (A A A with a 2(wild card) a real poker hand and to show me the rules on that. Of course my point was, once you deviate from the strict rules, all bets are off, and such rules should be clearly stated before playing such nonsense.
  #6  
18-03-2008, 1:14 AM
K_Kahne_Fan
CardsChat Regular
 
Likes: Hold-Em
Posts: 1,039
I had a similar situation with 5 of a kind with a wild. If you look at any __ wild slot machine 5 of a kind is only beat by a true royal...

Quickly Found Page on Google (replace "joker" w/ wild card of choice)

Another Variation
  #7  
18-03-2008, 3:46 AM
TheMighty0z
New Member
 
Location: Danvers, MA
Plays at: Full Tilt
Likes: NLHE
Posts: 3
Zach,
In their variation you have to declare the ace high or low when you show your hand (round the table style). Yes, I would agree that 9-9-A-A-x would lose in this case to 9-9-2-2-x as the aces would be the lowest possible pair.
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