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Tuesday, December 18, 2007
Worst possible 0/8 hand
This was a forum trivia question. Everyone pretty much homed in on Quad Deuces as the answer. I threw in a follow up question of "What is the worst O/8 hand without duplication of rank?" Anyone? Bueller?
Yeah, that seems worse that quad deuces. I don't have the time or inclination to run the numbers, but it seems weird to say that a starting hand with a made pair in it (such as 2222) could be the worst starting hand. I'm going to guess 2-6-10-A rainbow.
Any hand with A2 in it will give you the nut low if there is one possible and they don't get counterfeited.
So far 269K rainbow seems to be a leading candidate. No wheel possible. No flush possible. A low is possible albeit a bad one. I wonder if having only one low card 8 or lower to eliminate any hope at a low is worth clumping up more cards that could potentially straighten out for high.
It starts with a 2 for sure, and of course, it's a rainbow. I think that removing *any* possibility of a low is more important than spreading out the connectivity. You don’t want your hand randomly winning half the pot because you are the only one with a qualifying low.
So then, what’s more important, spreading out the connectivity of the three high cards, or keeping them as low as possible? I’ll run some tests against a random hand using ProPokerTools.
KJ92: 38.25% QJ92: 37.60% TcJh9s2d 37.30%
OK so my answer is JT92 rainbow, which is 37.3% against a random hand. Beat that!
Another way to run it would be the worst hand against three players playing the top 10%. This might make “crappy but possible” low hands bad enough to compete, since the best hands will have it crushed. Here is the EV for some hands against three players playing top 10% hands:
Very interesting. The hand that was worst against a random hand is the best of this awful bunch against three opponents with good hands. So I guess it depends on how selective your opponents are when asking yourself what the worst O8 hand is.
Marshall’s hand (K984) against 1 random hand, then against three good hands:
K984 vs. random: 43.06% K984 vs. 3 top 10%: 19.15%
I don't think there's any good reason not to make that 4 a 2 as long as neither one has connectivity with any of the other cards and no low is possible:
Well I'm glad we got that settled. I'll be sure to add ALL of these hands to the unplayable list. Up until now I've been thinking that KT94 rainbow was DA NUTZ.
The reason for choosing a 2 over a 4 would be so that full houses involving the lone low card are more likely to be trumped by a higher full house. For example a board of 99234 gives more chances for 92 to lose.
Was that just a long-winded way of saying, "2s are lower than 4s, and therefore worse?" If so, genius.
"you've been dealt quad 2s in Omaha?"
I have. And I have never been dealt quads in any other four- or five-card starting-hand game, ever. It was hard not to turn it face-up immediately and howl with laughter.
18 comments:
maybe like K 8 4 9 rainbow or something?
Yeah, that seems worse that quad deuces. I don't have the time or inclination to run the numbers, but it seems weird to say that a starting hand with a made pair in it (such as 2222) could be the worst starting hand. I'm going to guess 2-6-10-A rainbow.
Any hand with A2 in it will give you the nut low if there is one possible and they don't get counterfeited.
So far 269K rainbow seems to be a leading candidate. No wheel possible. No flush possible. A low is possible albeit a bad one. I wonder if having only one low card 8 or lower to eliminate any hope at a low is worth clumping up more cards that could potentially straighten out for high.
It starts with a 2 for sure, and of course, it's a rainbow. I think that removing *any* possibility of a low is more important than spreading out the connectivity. You don’t want your hand randomly winning half the pot because you are the only one with a qualifying low.
So then, what’s more important, spreading out the connectivity of the three high cards, or keeping them as low as possible? I’ll run some tests against a random hand using ProPokerTools.
KJ92: 38.25%
QJ92: 37.60%
TcJh9s2d 37.30%
OK so my answer is JT92 rainbow, which is 37.3% against a random hand. Beat that!
K962: 45.78%
That's a fucking MONSTER! Definitely "no possible low" is the way to go against a random hand.
Well, that's good to know that being 3-connected in O8 isn't necessarily a good thing.
And totally didn't think about the low possibility on my 2-6-10-A suggestion. Whoops.
Stupid lack of O8 knowledge.
can we run my k 8 4 9?
I figured that the low being possible would be a big factor. It was either that or K 4 9 10 for me
Another way to run it would be the worst hand against three players playing the top 10%. This might make “crappy but possible” low hands bad enough to compete, since the best hands will have it crushed. Here is the EV for some hands against three players playing top 10% hands:
KT62: 17.38%
K962: 17.67%
K972: 17.85%
Q972: 17.86%
KT72: 18.05%
QT62: 18.5%
Q962: 18.76%
QT82: 20.03%
JT92: 20.90%
Very interesting. The hand that was worst against a random hand is the best of this awful bunch against three opponents with good hands. So I guess it depends on how selective your opponents are when asking yourself what the worst O8 hand is.
Marshall’s hand (K984) against 1 random hand, then against three good hands:
K984 vs. random: 43.06%
K984 vs. 3 top 10%: 19.15%
Marsh's other hand suggestion:
KT94 vs. random: 39.11%
KT94 vs. 3x 10%: 20.42%
I don't think there's any good reason not to make that 4 a 2 as long as neither one has connectivity with any of the other cards and no low is possible:
KT92 vs. random: 38.53%
KT92 vs. 3x 10%: 19.49%
Well I'm glad we got that settled. I'll be sure to add ALL of these hands to the unplayable list. Up until now I've been thinking that KT94 rainbow was DA NUTZ.
As for Bob’s earlier thought that a hand with a made pair can’t be that bad off…check it out.
2222 vs. random: 8.02%
2222 vs. 3x 10%: 1.02%
Quad deuces, which I have been dealt at 08 before, is probably going to be the worst hand preflop by almost any measure.
Let’s see, the best scenario for it would pretty much be against one of those other bad hands:
2222 vs. KT63: 10.47%
The reason for choosing a 2 over a 4 would be so that full houses involving the lone low card are more likely to be trumped by a higher full house. For example a board of 99234 gives more chances for 92 to lose.
Like I said: KT63 = Da NUTZ. Look at that, it beats quad 2s!
But seriously: you've been dealt quad 2s in Omaha? Sheesh. How many times have you been dealt quads in Omaha?
"The reason for choosing a 2 over a 4..."
Was that just a long-winded way of saying, "2s are lower than 4s, and therefore worse?" If so, genius.
"you've been dealt quad 2s in Omaha?"
I have. And I have never been dealt quads in any other four- or five-card starting-hand game, ever. It was hard not to turn it face-up immediately and howl with laughter.
...except for the fact that deuces quad up at a higher percentage than quad fours.
However, the quadding up does not happen at a rate high enough to offset the times when a boat with a deuce in it falls to a higher boat.
How's that job hunt going, cowboy?
Not well enough. I really could use a job so I have some time set aside during the day to run O/8 simulations and blog about them.
It’s better if you get a job so you can have some free time to comment about the comments on a blog. VERY good use of your time.
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