Zimdog
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January 29th, 2016 at 1:24:31 PM permalink
I'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.

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!!!
odiousgambit
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January 29th, 2016 at 2:05:08 PM permalink
Quote: Zimdog

I'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.

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.

https://wizardofodds.com/games/poker/

Quote:

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?

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 page
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.

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].
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
beachbumbabs
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January 30th, 2016 at 12:25:38 AM permalink
I agree with OG's assessment on all of this. I would say, though, that either I don't understand your question on the 21 outs, or you're using it wrong.

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!
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Zimdog
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January 30th, 2016 at 6:15:20 AM permalink
I understand the idea of outs...but even if you don't consider straights or flushes, if the dealer has 21 outs as indicated, he only needs one of his two cards to be one of those 21 cards. So, we need to calculate the probability that neither of his cards is one of those 21 cards. Since there are 45 unrevealed cards, P (0) would be (24/45)×(23/44) = .278 chance of winning. Even with only 15 outs is P (0) = .439

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.
odiousgambit
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January 30th, 2016 at 7:03:43 AM permalink
Quote: Zimdog

we 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.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
Zimdog
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January 30th, 2016 at 10:25:08 AM permalink
And 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.
98Clubs
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February 4th, 2016 at 11:07:57 PM permalink
While the 21-out rule is a good basic strategy, there are 3 exceptions to using it... all are obvious (?) >21-outs, but a noob needs to know that there are more outs to count.

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
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ThatDonGuy
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February 5th, 2016 at 8:42:02 AM permalink
Quote: Zimdog

And 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.
Zimdog
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February 5th, 2016 at 8:56:11 AM permalink
See....my point is that I think 21 outs is way too high.

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.
teliot
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February 5th, 2016 at 9:04:20 AM permalink
Quote: Zimdog

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.

http://apheat.net/2015/06/01/ultimate-texas-holdem-basic-strategy-and-house-edge/

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.
Last edited by: teliot on Feb 5, 2016
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Zimdog
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February 5th, 2016 at 9:30:51 AM permalink
Gotcha.....I was overlooking that were dealing with more beyond W/L. The non_qualifying dealer on some hands leads to pushs. However, 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?
teliot
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February 5th, 2016 at 11:02:17 AM permalink
Quote: Zimdog

However, 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?

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.
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beachbumbabs
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February 5th, 2016 at 9:48:05 PM permalink
Quote: teliot

I derived Mike's 21 outs result towards the bottom of the following post. I take "runner-runner" into consideration in my derivation.

http://apheat.net/2015/06/01/ultimate-texas-holdem-basic-strategy-and-house-edge/

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!
If the House lost every hand, they wouldn't deal the game.
Zimdog
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February 7th, 2016 at 10:00:12 PM permalink
Ok, 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.
Romes
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February 8th, 2016 at 9:14:27 AM permalink
Quote: Zimdog

Ok, 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.

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.
Playing it correctly means you've already won.
odiousgambit
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February 8th, 2016 at 9:23:41 AM permalink
Quote: Romes

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.



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.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
Ibeatyouraces
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February 8th, 2016 at 10:44:00 AM permalink
I've posted before that MotorCity has this same policy.
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mcallister3200
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February 8th, 2016 at 10:44:51 AM permalink
Wisconsin casinos are all native American casinos. Nuff said regarding legality
BTLWI
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February 8th, 2016 at 2:02:35 PM permalink
Quote: odiousgambit

Flush 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?
Zimdog
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February 9th, 2016 at 10:22:33 PM permalink
Quote: Zimdog

Ok, 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.
odiousgambit
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February 10th, 2016 at 3:08:34 AM permalink
so if you are a ploppy you get screwed. Nice attitude at that casino.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!”   She is, after all, stone deaf. ... Arnold Snyder
offTopic
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March 7th, 2016 at 8:31:34 AM permalink
Saw this happen last week - standard?

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.
Ibeatyouraces
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March 7th, 2016 at 8:40:38 AM permalink
Had that happen a few times. A misdeal is normal procedure.
DUHHIIIIIIIII HEARD THAT!
Zimdog
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March 7th, 2016 at 8:54:05 AM permalink
Misdeal will happen.
charliepatrick
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March 7th, 2016 at 9:00:52 AM permalink
Quote: offTopic

...accidentally...[misdeal called]...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.
Deucekies
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March 7th, 2016 at 1:51:59 PM permalink
Quote: charliepatrick

Quote: offTopic

...accidentally...[misdeal called]...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.



I've seen many casinos where this waiting is required on TCP, UTH and PGP, for this exact reason.
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zrlcsx
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April 19th, 2016 at 12:50:15 PM permalink
I've simplified the math in my own head:

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
Chuck
Zimdog
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April 20th, 2016 at 2:25:55 PM permalink
I'm in New Zealand and was looking forward to possibly playing some UTH'em at the casino, but they have one VERY distinct difference. You HAVE to play your two hole cards. This significantly limits the number of flushes (have to be dealt 2) as well other bigger hands.

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.
Hunterhill
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April 20th, 2016 at 2:42:57 PM permalink
Does the dealer also have to play both hole cards?
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Romes
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April 20th, 2016 at 2:44:53 PM permalink
It gives the house an advantage over your blind bet (assuming the blind pay table is the same), but not your hands themselves.

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?

Quote: Hunterhill

Does the dealer also have to play both hole cards?

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.
Playing it correctly means you've already won.
Zimdog
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April 20th, 2016 at 2:46:10 PM permalink
I believe it was that
Hunterhill
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April 20th, 2016 at 2:47:54 PM permalink
Duplicate post
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Zimdog
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April 20th, 2016 at 2:51:55 PM permalink
Win % same but EV would seem to drop.
Commish
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July 20th, 2016 at 9:52:16 AM permalink
The blind bet pay table cannot change. It is copyrighted. Only the trips pay table can change.
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