# Example of a hand where pot odds aren't compared to breakeven odds

I'm reading Ed Miller's "Getting Started in Hold 'Em". The book goes over a concept - comparing pot odds with breakeven odds - via this example:

I'm 3 off the button in a limit hold 'em hand. One player limps, I look down and see: J♠ T♠. This is suited and connected, so I call. The next player raises, everyone folds to the big blind, who calls. The limper calls.

So far the pot size is 8.5 small bets. The flop is A♥ Q♦ 4♣.

Everyone checks to the preflop raiser, who bets. The big blind and limper call again. Pot size = 11.5 small bets.

Of course I'm going for a straight draw, for which there are 4 outs - any of the Kings. Approx. breakeven odds for 4 outs is 10.5 to 1, while the pot odds are 11.5 to 1. So I should call.

So far, so good. But now I reach another example, as follows:

This time I'm under the gun. I get K♠ Q♠. This is again suited and connected, plus high cards, so I raise. Two players call, as do both blinds. Pot size = 10 small bets.

The flop is K♥ 8♦ 6♦. Quoting from the book: "You've flopped top pair, and your kicker is very strong. Even though two diamonds are on the board, and the eight and six can combine with another card on the turn or river to make a straight possible, you have a solid holding. You should assume that you have the best hand until the action suggests otherwise."

More action: small blind bets, big blind folds.

The book suggests that I raise at this point. Suppose I do, and one player behind me calls both bets. The flop bettor calls too. Pot size = 8 large bets. No raises by anyone else, so the action doesn't seem to suggest anybody else has a stronger hand.

Here's the confusing part: "Any bet on the turn will offer pot odds of at least 9-to-1, more than the breakeven odds for a five-out draw (e.g. a hand with a pair made from one of the small board cards like 9♥ 8♠ or A♣ 6♣)"

Where did these 5 outs come from? The best hand I can hope for immediately after the flop would be King trips - which makes for 2 outs. The next best thing is a two pair, which I can get through either of a Queen, eight or six - but that definitely makes for more than 3 outs.

What's causing the confusion is the perspective flip between the two examples. (Which isn't helped by the author using similar starting hands in both. Even though they look similar, they'll play very differently with their respective flops.)

In the first example, you are the drawing hand trying to work out how to play against the made hand. You need to make the straight to win against any hand that would raise (maybe some Ace hand that has made top pair, maybe two pair or a big pocket pair or trips...). How many outs do you have to make that straight? 4. So your opponent raises and you decide whether or not to call based on your odds.

However. In the second example you are the made hand, with top pair and a good kicker. (There are better potential hands but not many: the winning hand in Texas Hold-em is just a single pair a lot of the time, and the only single-pair hand that beats you is AK, which probably would have reraised preflop.)

So you are the made hand, trying to work out how your raise will affect potential drawing hands. Your opponents with those hands will count how many outs they have and react appropriately. (This is the opposite side of the scenario to the first example.)

Ed Miller suggests a raise size that would give pot odds of 9 to 1. That's close to the odds of a hand with five outs, so hands with more than five outs will call and less than five outs will fold. What sort of hand has five outs against you? One example is a hand with a lower pair, which is losing to you at the moment and can only beat you by improving to trips (2 outs: the other two cards of their pair) or two pair (the three cards that would matched their unpaired hole card).

So that's what the author is talking about: not whether you have 5 outs, but whether your opponent has 5 outs to beat your made hand.

• That is a really good explanation! Thanks a lot. So here's my takeaway - essentially for no. of outs (for opponent) calculation, we're assuming that they have have a hand which gives them a reason to continue, but not so good a hand that they'd want to raise. If they'd made anything better than a pair on the flop, they'd have raised. So we assume that they have a pair and accordingly do the outs calculation as you showed. Is that correct? Commented Nov 2, 2022 at 7:33
• As a follow up to my previous comment, the author gives this scenario: "Say you have Q♥ T♥. The board is Q♦ J♦ 7♣ 4♠, and someone bets. Your hand is decent, but the bettor could easily have you beaten. If the pot is large (e.g. 10 bets or more), and there are other players in the hand, you should typically raise." If the bettor could easily have me beaten, why on earth would I want to raise? Commented Nov 2, 2022 at 8:12
• An aside: Miller seems to be talking about a variant where the raise size is fixed, not chosen by the raiser. (I guess as a lead in to gradually introduce concepts to you, if this is an introductory book to poker.) So we're not really guessing the opponent's hand and doing the odds calculation based on that: the odds are determined by the fixed raise size, and we classify opponent's hands based on the odds that we get. Which is sort of backwards. Personally, I'd prefer a larger raise to give the 8-9 out hands of the flush or straight draws something to think about. Commented Nov 2, 2022 at 14:00
• As for your follow-up, I think that wants to be its own question. Commented Nov 2, 2022 at 14:00
• You're also increasing the amount they need to pay to stay in the pot, which lowers the incentive to call. Commented Nov 2, 2022 at 16:18

KQ vs 98 on K86

Suppose the turn is a 2, you still have the best hand, and you bet. How many outs does Villain have to improve? Three 9s will give him two-pair, and two 8s will give him three of a kind. In total, that's the five outs.