Ult TX Hold'em Strategy
However, I have 2 questions on the ideal strategy listed:
1) Why would I raise 2x on a flush draw (9 chances in the last 2 cards) and not on the open-ended straight draw (8 outs)? Can you go over the specific odds of both to justify the difference.
2) I know you state that you specifically only are looking at dealer "outs" and considering one card, but the dealer does have two cards coming and I think that the "21 outs" is too high if considering both cards. Assuming that I have no pair in the 7 cards, but I have the highest kicker in my 2 card hand. That means that the dealer has 15 outs. Are there not 570 out of a possible 990 2 card hands that the dealer is holding that can beat my hand?
These sites have been great. Oooh, bonus question---does the odds change for more players at the table than just one on one with the house? Initial thought is no, but if they are playing optimally (big assumption) their actions may indicate certain cards out of play.
Danke!!!
Flush pays 3 to 2, Straight 1 to 1, for the Blind bet. Plus In 7 card, it's not that much harder to get a flush than a straight, while the former beats the latter [for the heads-up bet with dealer]. Non-mathematical answer there for you, LOL.Quote: ZimdogI've been following the strategy posted on wizard of odds and using their simulator. Much appreciated....
However, I have 2 questions on the ideal strategy listed:
1) Why would I raise 2x on a flush draw (9 chances in the last 2 cards) and not on the open-ended straight draw (8 outs)? Can you go over the specific odds of both to justify the difference.
https://wizardofodds.com/games/poker/
The 21 outs is intended to be simple strategy and is not perfect. Looking at the outs with one dealer card is how you use it. You can find links to more complicated and complete strategies at the bottom of the Wizard pageQuote:2) I know you state that you specifically only are looking at dealer "outs" and considering one card, but the dealer does have two cards coming and I think that the "21 outs" is too high if considering both cards. Assuming that I have no pair in the 7 cards, but I have the highest kicker in my 2 card hand. That means that the dealer has 15 outs. Are there not 570 out of a possible 990 2 card hands that the dealer is holding that can beat my hand?
don't quote me, but I think this has been asked before and the answer is that more players don't matter. On the other hand, you may be able to scan the cards of other players, or chat about your hands is especially likely [camera doesn't pick that up LOL].Quote:
These sites have been great. Oooh, bonus question---does the odds change for more players at the table than just one on one with the house? Initial thought is no, but if they are playing optimally (big assumption) their actions may indicate certain cards out of play.
If you have the highest available kicker on a non-paired board, you're right, the dealer has only 15 outs, and you should bet 1x. If you have the second highest, the dealer has 19 outs, and you should bet 1x. A queen as 3rd highest kicker takes a little more examination. If the board is paired and contains no aces or kings, the dealer has 19 outs, so you play 1x. If the board is not paired and contains no aces or kings, the dealer has 23 outs, so you fold the queen. If the board is not paired, and contains an ace and a king, that's 21, right on the line of betting or folding. (This might be where it would help to know what's in other people's hands....lol).
So, that's an example of how it works. It doesn't take into account straight/flush draws, or several other situations that might come up to complicate your decision; it's just demonstrating the application of the concept. The 21 outs, however, does expect that you're already playing the aggressive optimal strategy (like 4x any ace, which I see a LOT of players not doing), then advises you what to do if you've not already bet.
Welcome to the board!
And I was trying some of the other links as mentioned but some are broken. Did not find a more in depth analysis of the odds.
Quote: Zimdogwe need to calculate the probability that neither of his cards is one of those 21 cards.
Your making this too hard. Just count the outs and don't worry that the dealer has two cards ... unless you want to re-invent the wheel. What I mean is, sure, you can come up with your own strategy if you want to.
If the links are broken, I'd just google the subject, I'm pretty sure you can find the other strategies out there.
Personally, the outs thing is all I am going to do. I find that I already have trouble "solving" in the time alotted for that. The dealers really want to keep it going. I sometimes do tell them they are "going too fast for me" as it is. I suppose you could be even more assertive about slowing it down, and deal with how that is going to make waves - how fast it is dealt is important to the house.
Possibly, you could "slow it down" so to speak by playing so much that the game just slows down for you by getting so deep into the whole thing. This clearly can happen in all games. As for myself, I don't play all that often, nor is that going to change.
And yes I have been getting familiar with how many outs a board presents....5 unique ranks, 15 outs; one pair, 11 outs; 3 of a kind, 7 outs; 2 pair, 7 outs. Then checking how many other cards and adding 4 times that number.
1.) Open-ended 4-straight: aside from the 15 cards that win with a pair, there aree the 4 cards at each end of the Straight to consider. Therefore there are 23-outs available to dealer. If holding a K2, K3, or K4 off-suit or pocket Deuces, this is a fold.
2.) 4-Flush on the board: same as 1.) except there are 9 additional outs.
3.) BOTH 1 & 2: Nightmare board. Best rule of thumb provided by the Strategy analyzer sums up as " Raise if you pair the middle board rank or better, else fold". So a board 3456Q thats a Q346 Flush would raise 55 or better, and fold any hand ranked lesser. 789TK is a pair of 9's or better, else fold.
Comments: Rule 1 and 2 can sum up as "raise 55 or better, include 44 if that pair is not the bottom pair".
An inside-Straight (gut-shot) DOES add 4 more outs, and any pair should raise... high-card is tricky. Without an Ace on the board, fold the high-card hands. With an Ace, raide K2, K3, or K4 off-suit AND raise pocket Deuces.
Regards
98
Quote: ZimdogAnd that's all for and good. I was just trying to understand why 21 outs was chosen for the optimal strategy.
And yes I have been getting familiar with how many outs a board presents....5 unique ranks, 15 outs; one pair, 11 outs; 3 of a kind, 7 outs; 2 pair, 7 outs. Then checking how many other cards and adding 4 times that number.
I'm ahead of you on this one.
I get how to count the outs. I know about checking straights on the board.
I'd like to see the hard math on WHY 21 outs was chosen, because you CANNOT discount that the dealer gets 2 cards and not just one.
I derived Mike's 21 outs result towards the bottom of the following post. I take "runner-runner" into consideration in my derivation.Quote: ZimdogI'd like to see the hard math on WHY 21 outs was chosen, because you CANNOT discount that the dealer gets 2 cards and not just one.
Here it is ...
Clearly if the player folds, then his EV is -2.
Let N be the number of outs under consideration for the dealer to beat the player. Then the probability that the dealer’s first card is an out is p = N/45. For his second card, the dealer who whiffed on his first card most likely has 3 additional “pair outs” to pair his first card and beat the player. He may also generate new straight or flush outs (call these 1 additional “out,” so-called “runner-runner”). So, the probability of the dealer beating the player by hitting an out on his second card is approximately (N + 4)/44.
Overall, the probability that the dealer beats the player is then,
p = N/45 + [(45 – N)/45]*[(N + 4)/44].
Simplifying, we get:
p = (-N^2 + 85 N + 180)/(45*44)
Note that if the dealer doesn’t hit an out, then he won’t qualify. It follows that the EV for the player who raises 1x on the Turn/River bet is:
EV = p*(-3) + (1-p)*(1) = 1 – 4p.
We make the raise whenever EV > -2. That is, 1 – 4p > -2. Solving for p gives
p < 3/4.
That is, the player raises 1x when his chance of beating the dealer is 25% or higher.
Combining the two expressions for p, we see that EV > -2 whenever
(-N^2 + 85 N + 180)/(45*44) < 3/4.
Simplifying gives the quadratic equation,
N^2 – 85N + 1305 > 0
Solving this quadratic equation gives roots:
(1/2)*(85 + sqrt(2005)) = 64.9
(1/2)*(85 – sqrt(2005)) = 20.1
For the quadratic equation to be positive, N must be either larger than both roots or smaller than both roots. That is, either N ≥ 65 or N ≤ 20. The first case is the “impossible solution,” leading to the conclusion that there can be at most 20 dealer outs that can beat the player.
Read the comments in the link I posted. In the special case where the dealer already qualifies because the board is paired you should raise 1x unless the dealer has 23 or more outs that can beat you. You can find many other special cases that modify the 21 outs conclusion, but 21 outs is optimal as an overall single number.Quote: ZimdogHowever, if the board already has a pair, the dealer still qualifies. Your explanation includes the statement that the dealer doesn't qualify, yet he would. Is that also considered or are we looking at only 5 unique cards on the board?
Quote: teliotI derived Mike's 21 outs result towards the bottom of the following post. I take "runner-runner" into consideration in my derivation.
Here it is ...
Clearly if the player folds, then his EV is -2.
Let N be the number of outs under consideration for the dealer to beat the player. Then the probability that the dealer’s first card is an out is p = N/45. For his second card, the dealer who whiffed on his first card most likely has 3 additional “pair outs” to pair his first card and beat the player. He may also generate new straight or flush outs (call these 1 additional “out,” so-called “runner-runner”). So, the probability of the dealer beating the player by hitting an out on his second card is approximately (N + 4)/44.
Overall, the probability that the dealer beats the player is then,
p = N/45 + [(45 – N)/45]*[(N + 4)/44].
Simplifying, we get:
p = (-N^2 + 85 N + 180)/(45*44)
Note that if the dealer doesn’t hit an out, then he won’t qualify. It follows that the EV for the player who raises 1x on the Turn/River bet is:
EV = p*(-3) + (1-p)*(1) = 1 – 4p.
We make the raise whenever EV > -2. That is, 1 – 4p > -2. Solving for p gives
p < 3/4.
That is, the player raises 1x when his chance of beating the dealer is 25% or higher.
Combining the two expressions for p, we see that EV > -2 whenever
(-N^2 + 85 N + 180)/(45*44) < 3/4.
Simplifying gives the quadratic equation,
N^2 – 85N + 1305 > 0
Solving this quadratic equation gives roots:
(1/2)*(85 + sqrt(2005)) = 64.9
(1/2)*(85 – sqrt(2005)) = 20.1
For the quadratic equation to be positive, N must be either larger than both roots or smaller than both roots. That is, either N ≥ 65 or N ≤ 20. The first case is the “impossible solution,” leading to the conclusion that there can be at most 20 dealer outs that can beat the player.
This is great stuff, Eliot. Thanks for taking the time!
Found at a casino in Wisconsin that the Trips only pays if you bet (not a regardless of folding). So if the board trips, you need to play to win the Trips bet even if you have dogs in the hole.
How does this change the payout for Trips bet....The pay out is table 3.
1) it's awful. 2) I think it's illegal. You'd have to check with the creators of the Trips side bet, but it's supposed to be (like all other side bets) a completely separate bet from the hand. The casino is more than likely INCORRECTLY enforcing this rule.Quote: ZimdogOk, one more....
Found at a casino in Wisconsin that the Trips only pays if you bet (not a regardless of folding). So if the board trips, you need to play to win the Trips bet even if you have dogs in the hole.
How does this change the payout for Trips bet....The pay out is table 3.
Quote: Romes1) it's awful. 2) I think it's illegal. You'd have to check with the creators of the Trips side bet, but it's supposed to be (like all other side bets) a completely separate bet from the hand. The casino is more than likely INCORRECTLY enforcing this rule.
If the conclusion comes from a single dealer, could be just the dealer didn't know
If from the pit boss, more of a problem, but ...
I don't believe a casino can change the rules willy-nilly btw.
Quote: odiousgambitFlush pays 3 to 2, Straight 1 to 1, for the Blind bet.
What does that have to do with his question of 2x'ing the Play bet?
Quote: ZimdogOk, one more....
Found at a casino in Wisconsin that the Trips only pays if you bet (not a regardless of folding). So if the board trips, you need to play to win the Trips bet even if you have dogs in the hole.
How does this change the payout for Trips bet....The pay out is table 3.
UPDATE: It isn't printed on the board, and not every dealer tells the player their option, but they do allow you to fold the hand, but still play the trips. It just isn't advertised.
Procedure is:
Dealer pulls the board out of the shuffler, distributes all player hands, then pulls the dealer hand.
In this particular hand, all the cards are out, but when the dealer pulls his own hand, he exposes both cards accidentally. He calls floor, and floor declares misdeal and takes all the cards in.
I didn't have anything, and his hand had me crushed, so I didn't mind, but if I'd had something strong, I would've been pretty upset.
I saw this happen at 3CP where the end box was only playing Pair+ and got AKQ clubs - now in this case the dealer's hand is irrelevant. In all of these types of games I usually only look at my hand when the dealer has finished dealing- that way I have no idea what I had if there's a misdeal - much simpler and it can slow the game down a bit.Quote: offTopic...accidentally...[misdeal called]...if I'd had something strong, I would've been pretty upset.
Quote: charliepatrickI saw this happen at 3CP where the end box was only playing Pair+ and got AKQ clubs - now in this case the dealer's hand is irrelevant. In all of these types of games I usually only look at my hand when the dealer has finished dealing- that way I have no idea what I had if there's a misdeal - much simpler and it can slow the game down a bit.Quote: offTopic...accidentally...[misdeal called]...if I'd had something strong, I would've been pretty upset.
I've seen many casinos where this waiting is required on TCP, UTH and PGP, for this exact reason.
For a paired board (dealer qualifies) you only have to win 1 in 5 hands to be even
Assuming a $10 bet, 5 folds = 5X$20 = $100 loss,
5 bets with one win = 4X$30 = $120 loss + 1X$20 (Ante & Play) + $0 (push for blind)+ = $20 win
$120 loss + $20 win = $100 loss
The same math for an unpaired board (dealer doesn't qualify) you have to win 1 in 4 hands to be even
For folding: $80 loss (4x$20)
For betting: $90 loss (3X$30) + $10 win (only play bet pays) = $80 loss
Would love some numbers crunched as to how this affects game play, starting hands played as well expected values. Feel like this give the house a SIGNIFICANT advantage in comparison.
Main Blind Pay Table I'm used to seeing:
Royal - 500 to 1
Straight Flush - 50 to 1
Full House - 3 to 1
Flush - 3 to 2
Straight - 1 to 1
Thus, if you MUST use 2 cards from your hand, you'll get paid on these less of the time... which would take away from the overall payback to the player and therefor up the house edge.
What is your blind pay table?
This would also be my next question. As long as the dealer must also use both their cards, then the win% of the hands should remain exactly the same, as the dealer will also make less flushes/etc/etc.Quote: HunterhillDoes the dealer also have to play both hole cards?
