Ultimate Texas Hold Em
The strategy proposed by the Wizard is very easy to master, expect for the "less than 21 outs" rule. See the game and rules here: https://wizardofodds.com/games/ultimate-texas-hold-em/
Here is an example. I have pocket 3s4d and the board is 10s, 10h, Qd, Jd, 9d.
I counted 19 outs: (Q+J + Nine) *3 + Tens * 2 + (A + K) * 4.
So I chose to call, but the game tells me the that's wrong play and I should fold.
What am I counting wrong? Sorry if my lingo is incorrect.
However my understanding is the outs rule only counts the effect of single cards being dealt, not combinations, so the potential pocket pair or 2 to a flush would not factor in. I may be wrong.
Quote: AceThanks, four eights bring the total to 23 and into fold territory. My rookie mistake.
However my understanding is the outs rule only counts the effect of single cards being dealt, not combinations, so the potential pocket pair or 2 to a flush would not factor in. I may be wrong.
Actually it does. When counting single, as in a hole card games, you call with 10 (edit: or 11. Been so long I've forgotten) bad outs or less that either give you a win or push.
For the 21 outs, you count potential pairs, and not straights and flush possibilities, which is not to say they don't matter. To avoid the wrong play, you need to use a better strategy than the Wizard's ... he would agree with that I think.
Usually counting the outs, it is an end decision about whether your kicker is good enough. Using the Wizard strategy, you have to get a gut feeling about potential dealer straights and flushes and modify your decisions.
Fold
You can be beaten by Ace, King, Queen, Jack, Ten, 9, 8
If Dealer is 7-high or less (no two diamonds) PUSH
Chances (23/45 + 22/44) Dealer wins, (22/45 + 16/44) Push
Looks like a negative EV
98
How can you get beaten by 8?Quote: 98ClubsHolding 3-4 off-suit?
Fold
You can be beaten by Ace, King, Queen, Jack, Ten, 9, 8
If Dealer is 7-high or less (no two diamonds) PUSH
Chances (23/45 + 22/44) Dealer wins, (22/45 + 16/44) Push
Looks like a negative EV
98
if you mean 3 of spades and 4 of diamonds is what you held, I have no idea why you would think there was some prayer counting outs. There must be something I am not getting.
Maybe I’m not getting it. I’m new to this.Quote: odiousgambiti thought you meant you had pocket 3s, i.e. a pair of 3s ... and kind of ignored the rest that didnt make sense
if you mean 3 of spades and 4 of diamonds is what you held, I have no idea why you would think there was some prayer counting outs. There must be something I am not getting.
My understanding is that I’m playing the board and my pocket 3-4 is inconsequential. Lowest card on the board is 9 so excluding the potential straight there are 19 outs.
Quote: Ace2Maybe I’m not getting it. I’m new to this.
My understanding is that I’m playing the board and my pocket 3-4 is inconsequential. Lowest card on the board is 9 so excluding the potential straight there are 19 outs.
yep, you're right
it's just a case where the strategy is too simple in the face of the possible straight I think.
sorry, I added more confusion than anything
A one card hit ?Quote: beachbumbabsYou really do have to count the 8s and fold. A one-card hit should be among your considerations. 23 against you is too many.
Quote: BenjiexThere is no way it could be a one card hit
In the OP, if the dealer has 8,2 offsuit, you lose to a one card hit.
Dealer 8,2 would be a push.Quote: IbeatyouracesIn the OP, if the dealer has 8,2 offsuit, you lose to a one card hit.
Quote: AceI am new to all forms of poker but I just started playing this (practicing online first).
The strategy proposed by the Wizard is very easy to master, expect for the "less than 21 outs" rule. See the game and rules here: https://wizardofodds.com/games/ultimate-texas-hold-em/
Here is an example. I have pocket 3s4d and the board is 10s, 10h, Qd, Jd, 9d.
I counted 19 outs: (Q+J + Nine) *3 + Tens * 2 + (A + K) * 4.
So I chose to call, but the game tells me the that's wrong play and I should fold.
What am I counting wrong? Sorry if my lingo is incorrect.
8,2 would give the dealer a straight. And you a loss.
You should not count dealer having to use both their cards to make a hand in counting outs. There might be a AK975 rainbow on the board. You have a Jx. There are 19 outs the dealer has with a single card. Bet the jack. When the dealer shows up with 6-8o, s$!# happens. You were still correct in betting the J.
Quote: CharmedQuarkI think with this example, you play for the push. You are dead with your hole cards, so the board is all you got. Yes, your have 23 outs, but this is a paired board where the cutoff is 22 (actually 22.44) outs or less according to teliot. So you are sitting there with two bets (ante-blind) to protect. At 23 outs, to me it's almost a 50-50 shot - you get lucky or not. I always play for the push on these 'edgy' situations.
Who is the teliot you mention?
Quote: zrlcsxWho is the teliot you mention?
This is him:
https://www.google.com/amp/s/www.888casino.com/blog/apheat/all-apheat-posts%3famp
Quote: IbeatyouracesWhat many players don't see, is you have to look at the ante/blind bet as if they are the pot in a live Holdem game. Those bets are no longer your money and are in the pot you're trying to win.
Another way to look at it is: What loss percentage is required to break even?
With a paired board and you can beat the board (dealer qualifies you win ante and play)
- only need to win 1 in 5 to break even or 80%
5 folds loses 5*2 or -10 units
5 bets with 4 losses (-4*3) and one win (1*2) also loses 10 units (-12+2 = 10)
With an unpaired board (dealer doesn't qualify and you only win ante)
- only need to win 1 in 4 to break even or 75%
4 folds loses 4*2 or -8 unitsins
4 bet with 3 loses (-3*3) and one win (1*1) also loses 8 units (-9 +1)
Playing the board (you can only push)
- only need to win 1 in 3 to break even or 66.7%
3 folds loses 3*2 or -6 units
3 bets with 2 losses (-2*3) and one push (1*0) also loses 6 units (-6+0)
Quote: PapaChubbyThere is something which is not being addressed in this thread, and it has me curious. The Wizard's strategy says to make the 1x bet if there are less than 21 dealer outs that beat you. I infer (maybe incorrectly) that the alternative to being beaten in this case is (usually) to win. However, in the OP's example, the player has no chance to win and is strictly playing for a push. Since the reward for a favorable outcome is diminished, it seems to me that a risk-reward analysis would recommend a fold even if the dealer has considerably fewer outs. I think. Does this make sense?
I'm pretty sure the calculation that resulted in the 21out rule took pushes into consideration, as well as dealer not qualifying (in the OP example, dealer qualifies with the board 10s, but in general). If the OP example did not include 4 to an inside or outside straight, it would be correct to play for the push. As it is, the 4 8s outstanding make the count 23, and recommend a fold.
As said above, there is value (computed into the 21 out rule) in playing 1 to save 2, because of the Blind. In other circumstances (no board pair), you may still rescue the Ante, which has some value, though it could be argued it's a wash if you lose the Play anyway.
Fold, you always lose Ante +Blind. The 21out rule saves under only slightly advantageous rules, and at their lowest are well below a 50% chance of success. But pay just often enough to be +ev when compared to folding, even though you incur additional risk with the Play.
