Board: K♥ 9♥ 7♥ 4♥ 3♣
The dealer outs are the 9 hearts and a 4, 7, 9, or K in any of the other 3 suits = 9 + (4*3) = 9 + 12 = 21
I should have clarified my original question. I was looking at hands where all i had was a kicker (no pair).
The whole point of the "21 outs" strategy is to only consider one-card outs.
"If the dealer has X and Y" is not a one-card out.
Quote: JBYou: Q♠ 3♦
Board: K♥ 9♥ 7♥ 4♥ 3♣
The dealer outs are the 9 hearts and a 4, 7, 9, or K in any of the other 3 suits = 9 + (4*3) = 9 + 12 = 21
You: Q♠ 3♦
Board: K♥ 9♥ 7♥ 4♥ 3♣
Raise EV = -1.9909
However, if you change the 4♥ to 5♥
You: Q♠ 3♦
Board: K♥ 9♥ 7♥ 5♥ 3♣
Raise EV = -2.0273
I always find these kind of close-call hands interesting to know.
Quote: gordonm888Quote: JBYou: Q♠ 3♦
Board: K♥ 9♥ 7♥ 4♥ 3♣
The dealer outs are the 9 hearts and a 4, 7, 9, or K in any of the other 3 suits = 9 + (4*3) = 9 + 12 = 21
You: Q♠ 3♦
Board: K♥ 9♥ 7♥ 4♥ 3♣
Raise EV = -1.9909
However, if you change the 4♥ to 5♥
You: Q♠ 3♦
Board: K♥ 9♥ 7♥ 5♥ 3♣
Raise EV = -2.0273
I always find these kind of close-call hands interesting to know.
More possible dealer straights by changing the 4 to a 5
Quote: JBYou: Q♠ 3♦
Board: K♥ 9♥ 7♥ 4♥ 3♣
The dealer outs are the 9 hearts and a 4, 7, 9, or K in any of the other 3 suits = 9 + (4*3) = 9 + 12 = 21
This hand brings up a unique situation. Requardless of my kicker to the three there are always 21 outs (I think). Yet only Q or Ace kicker have an expected value greater than the minus 2 for folding.
Values from calculator on wizzards site. ( are you the JB that developed that application? )
A♠ 3♦ = -1.976
Q♠ 3♦ = -1.991
J♠ 3♦ = -2.006
10♠ 3♦ = -2.021
8♠ 3♦ = -2.036
The wizards basic strategy says to bet 1x with a hidden pair ( I understand it is not optimal). A simple tweak would be to fold low pair with 4 flush on board. Not sure it is practical to devise a rule for when to bet low pair with four flush on board.
Quote: zrlcsx[This hand brings up a unique situation. Requardless of my kicker to the three there are always 21 outs (I think). Yet only Q or Ace kicker have an expected value greater than the minus 2 for folding.
Values from calculator on wizzards site. ( are you the JB that developed that application? )
A♠ 3♦ = -1.976
Q♠ 3♦ = -1.991
J♠ 3♦ = -2.006
10♠ 3♦ = -2.021
8♠ 3♦ = -2.036
The wizards basic strategy says to bet 1x with a hidden pair ( I understand it is not optimal). A simple tweak would be to fold low pair with 4 flush on board. Not sure it is practical to devise a rule for when to bet low pair with four flush on board.
Yes.
For this question I just threw together some code which searched for 21 outs by shuffling, dealing the player's 2 cards + the 5 board cards, iterating through every possible remaining card as if it were the dealer's only card, and determining who won (your best 5-of-7 hand versus the dealer's best 5-of-6 hand).
All of the 21-out situations appear to meet the following criteria:
- You have a pair (pocket, or you paired a board card)
- There is a 4-flush on board
- 4 of the 5 ranks on the board, if paired, would beat your pair
Back to your original question, I don't think there are (I didn't find) any situations where you have Ace-high or lower and the dealer has exactly 21 outs.
Another question. Where does the 21 outs come from? I am assuming it predicts the dealer's chances of beating you. I tried to calculate it and came up with 75.76%. The dealer will beat you 75.76% of the time, not accounting for filling a three card straight or flush. The actual chance would be slightly higher when there is a three card straight or flush on the board.
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 when
(-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.