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Wizard
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Wizard
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May 27th, 2015 at 4:10:53 PM permalink
Quote: odiousgambit

So the question is, do you want to figure the outs that give straights and flushes, or does he mean to only count what he shows in the example?



No. My strategy is complicated enough as it is. Just forget about the two-card draws. I admit it is not a perfect strategy.
It's not whether you win or lose; it's whether or not you had a good bet.
odiousgambit
odiousgambit
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May 29th, 2015 at 4:08:19 PM permalink
I have some more input; been practicing like a son-of-a-B on wizard's trainer/game. Ironically, probably would have to be lucky to find the game up and running as it's going to be mid-week and maybe slow for them.


Anyway, seems clear to me this is less daunting than it seems. I like the 21 out thing. My input: If playing much you should be able to memorize the outs of "situations", here is my list so far:

what I mean by "the board" is the community cards, by 'high card' I mean cards higher than your kicker


*no pairs and no high cards on board: the board counts 15 outs. Your kicker can only take 1 dealer high card's outs = 19 for a player King kicker minimum [likely you went 4x]


*no pairs and high cards on board: the board still counts 15 but qualifying kickers open up because you are not going to consider the high cards twice. This now becomes practical counting the available high cards to dealer [speed matters to me]. 4 outs per high card, clearly 2 cards is 'it'


BOARD PAIRS


*a pair and no board high card reduces board to 11 outs. The 10 more allowable outs make Jack or better good

*a pair and high cards on board, you still start with the 11 and counting outs you just now keep in mind the 10 more allowable outs, the kicker possibilities get better again.


BOARD TWO PAIR or TRIPS

*board is down to 7 notice there is a pattern here



BOARD 4 OAK


*board is down to 3! 18 [16] more outs to go [4 higher ranks] ... if the 4 oak is composed of a high rank, it matters. A 9 kicker may qualify for a shot. 4 oak with ace means K, Q, J, 10 beat you if you don't have them. Other possible situations don't apply, if you like high variance it may be time to go for the 8 kicker. I am convinced you can't win in the long run without the Blind bet paying off nicely once in a while, so if you are willing to risk higher variance by sacrificing HE ... [unwizardly thing to do though]

your thoughts?
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
CharmedQuark
CharmedQuark
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June 1st, 2015 at 4:26:25 PM permalink
Odious - the rule states 'less' than 21 outs. To me that means 20 or less. You're counting 21 outs. Maybe I'm missing something. . . but I'm a newbie and still learning.
odiousgambit
odiousgambit
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June 1st, 2015 at 5:21:37 PM permalink
Quote: CharmedQuark

Odious - the rule states 'less' than 21 outs. To me that means 20 or less. You're counting 21 outs. Maybe I'm missing something. . . but I'm a newbie and still learning.



well, when you come down to the remaining outs, the ones not represented on the community board, couting them, it's four outs at a time and you are either starting with 15, 11, 7, ... you only get to 19 before you have to stop, seems to me. Clear in my mind anyway LOL

here is my current evaluation. I'm sure I'm only re-inventing the wheel ... others are way ahead of me ... BUT, I like this:

*the community board has nothing, that means level one: King kicker is needed
-Ace or King in board, now Queen is good.
-etc.

*community board has a pair, second level: Queen is needed
-Ace in board, now Jack is good
-Ace and King, or Ace and Queen, etc, now 10 is good

etc, etc, etc

I just count them down after deciding if it is level one, level two, level 3 ... hope you can follow that, it becomes simple. The wizard trainer seems to support the thinking, I'm not getting warned off that the evaluation is wrong while playing

PS, yeah, re-reading the prior post, you were right, examples are a card too far, sorry. The current eval. seems to be working
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
teliot
teliot
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Thanks for this post from:
WTflush
June 1st, 2015 at 6:44:59 PM permalink
Here is my rough-justice proof of the "21 outs" statement in Mike's strategy.

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 inequality,

N^2 – 85N + 1305 > 0

Solving the 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.
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pokerface
pokerface
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June 1st, 2015 at 7:51:53 PM permalink
Quote: teliot

Here is my rough-justice proof of the "21 outs" statement in Mike's strategy.


Very nice and professional.
I wish all the other strategies are based on math like this.
winning streaks come and go, losing streak never ends.
pokerface
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June 1st, 2015 at 8:04:54 PM permalink
Quote: Ibeatyouraces

This is why is so much easier when you know one of the dealers cards :-)


At least twice I had the opportunity of knowing dealer's card, but didn't win any of the hands, talking about luck.
winning streaks come and go, losing streak never ends.
bigfoot66
bigfoot66
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Thanks for this post from:
WTflush
June 2nd, 2015 at 12:35:20 AM permalink
When there is 4oak on the board you should raise very liberally. Let's say board is QQQQ2 and you have 7 high. Here there are 24 outs to beat you but if you win you get paid 10:1 on the blind. In effect you are wagering 1 unit to win 14. Because of the huge payout it's gotta be right to call here.
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odiousgambit
odiousgambit
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June 2nd, 2015 at 2:05:46 AM permalink
Quote: bigfoot66

When there is 4oak on the board you should raise very liberally. Let's say board is QQQQ2 and you have 7 high. Here there are 24 outs to beat you but if you win you get paid 10:1 on the blind. In effect you are wagering 1 unit to win 14. Because of the huge payout it's gotta be right to call here.



I'm thinking this too. By the end of a session, whether or not you've had some nice payouts on the blind bet is huge!

your example has a 2 for the board kicker - a player should be alert for the other kind of 4 oaks with a high kicker on the board - excellent chances to be tied - you don't want to fold - you also don't want the dealer to use his lower-than-board kicker and claim you lost
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
odiousgambit
odiousgambit
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June 2nd, 2015 at 4:03:16 AM permalink
Quote: teliot

Here is my rough-justice proof of the "21 outs" statement in Mike's strategy.



I don't think Mr. Jacobson takes too much for granted. Awesome.
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

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