After yesterday's post, a reader asked for some examples and clarification of what "pot odds" really are. Rather than respond directly to him in the comments or in an email, I thought my response might be better served by posting some examples directly to the blog. For all examples, assume the limits are $0.50/$1.00.
Example 1
You hold 8s7s on the button. The board is K-6-5-A, no flush possible. A total of $5 went into the pot before and after the flop. On the turn, there are three of you left in the hand. The action goes: bet $1, call $1 and you are the last to act. If you catch your straight, you will have the nuts. Should you call?
In order to answer this question, we need to look at your expected odds, your pot odds, and to a lesser extent your implied odds. Pot odds are the easiest to calculate, so lets start there. At the time of your decision, there is $5+1+1=$7 in the pot. You have to call $1. Thus, you can say the pot is laying you odds of 7:1 on your call. Your pot odds are 7:1. Simple enough, right?
Next, we need to consider the expected odds of making your straight. After the turn card comes [K-6-5]-Q, there are 46 cards in the deck that are "unknown" to you (you obviously know the two in your hand, and the four on the board). 8 of the 46 unknown cards will make your hand ("outs") -- 4 nines, 4 fours -- while 38 don't help you, or at least probably won't help you enough to win the pot. Thus, the expected odds of making your draw on the next card are 38:8, or 4.75:1. For sake of convenience, let's just say your expected odds are approximately 5:1.
Back to the initial question. Should you call? Yes! In this instance, it is correct for you to call, because you're getting a good overlay on your draw (7:1 pot odds v. 5:1 expected odds). If you were to call six times, you would expect to miss on the river five times, thus costing you $1x5=$5. However, you expect to catch your card once, giving you the winning hand and a $7 pot. Thus, over the long-run, if you call every time in this situation, you can expect to net $2 every six hands.
Now that we understand that, it's time to make things more complex.
Example 2
Similar situation. You hold 8s7s. This time, the flop comes Kh-6h-5d and the turn is the Ac. There is $5 in the pot after the flop. You are last to act and the turn betting goes bet $1, call $1. Your action. Should you call?
Start with pot odds again. 7:1. Nothing has changed.
Next, expected odds. Your expected odds to make a straight are also the same at approximately 5:1. This time, however, you won't always know you have the best hand if you make your straight. Do you see why? Suppose the river is [Kh-6h-5d-Ac]-9h. You have made your straight, but it's quite possible that someone else has a flush! Sklanksy would call this "negative implied odds" or outs that are not "clean"; cards that will make your hand, but also make a bigger hand for someone else (and thereby cost you several big bets).
How should you account for this situation? One way is to assume that the 9h and 4h are not "outs" for you when calculating your expected odds. Thus, you have only 6 cards (not 8) that make your draw, versus 40 (not 38) of the total 46 "unknown" cards that do not. 40:6 is 6.7:1, or approximately 7:1 expected odds. This is a *much* closer decision than Example 1.
Consider this though -- the hand doesn't end when the river card comes. There's still a round of betting. Any bets you expect to collect after you have made your hand are part of your implied odds. For example, assume that if the river is the 9d, one player will bet, the player on the heart flush draw will fold, you now raise and the first player will pay you off. On the river, then, you expect to collect an extra $2. Now your $1 call on the turn nets you a pot of $9 when you hit. Thus, on the turn when deciding to call, your pot odds are 7:1 and your implied odds are 9:1. Your implied odds have just turned a borderline call/fold into a definite call (9:1 implied odds v. 7:1 expected odds). Remember, though, that implied odds are just that: "implied". They're not guaranteed.
Example 3
Now let's turn it around. Three players limp in to see a flop of Td-6c-2d, and you are second to act with AcTs. Assume that you know the player to act after you has Ad 9d. Do you see why you should raise the flop if the first player bets? It is not *only* because you probably have the best hand and are raising for value. It is also to charge the draws to stick around or, ideally, get them to fold.
There is $1.50 in the pot. If the first player bets and you simply call, you will be giving the third player 5:1 pot odds ($0.50 call to win $2.50) to catch his flush, which has expected odds of 4.2:1. On the other hand, if you raise, he will have to call $1 into a $3 pot for pot odds of 3:1. It is now incorrect for him to call, because the pot is not laying him the right price (3:1 pot odds v. 4.2:1 expected odds). If he does call and the first player folds, there will be $4 in the pot. If you bet the turn when it comes a blank, he will be making another mistake to call, especially since -- being the savvy player you are -- you will not pay him off if the flush hits on the river.
Addenda
There are several upshots to all of this. Number one: the larger the pot, the more correct it is for the draws to stay in. This is what Sklansky calls the "schooling" phenomenon. Imagine a situation where 8 players limp into a pot for $0.50 each. There is $4 in the pot, and on the button you flop a draw to the nut flush. The betting goes: bet, fold, fold, raise, fold, fold, reraise. You're faced with a $1.50 bet, but there is $4+0.50+1+1.50=$7 in the pot. You can see that you are getting about 4.5:1 pot odds on a draw that is also about 4.2:1. Factor in implied odds, and you're set. Even if you miss on the turn, if both other players call the flop raises, there will be $10 in the pot on the turn. If the action is check, bet, raise, you are still getting 13:2 or 6.5:1 on the same 4.1:1 draw. It would be a mistake to fold, unless you are sure a check-raise or reraise is coming from one of the EP players.
Again, this assumes that all of your outs are "clean". In the situation as given, it is probable that one of the players is on a set, which would reduce your outs by two (the two spades that pair the non-spade cards on board). If that's the case, your expected odds are really 5.5:1. It's still a call -- especially since the set will pay you off on the river -- but you should understand that it's not as big of an overlay as it appears at first blush.
Upshot Number Two: any time you can manipulate the size of the pot, whether you are drawing or are the one in the lead, you should do so. For example, it is sometimes correct to keep the pot small on the flop, like when you have a big hand and your raise will probably not force many players out. By raising on the flop in this situation, all you are doing is making it more correct for players to call on the turn when they call your flop raise. However, if you wait until the turn to raise, you stand a better chance of inducing them into making an incorrect decision or, better yet, folding. It may reduce your profit for the hand, but it will save you a bet when they hit the turn (remember, they were going to call regardless) and will increase the certainty of winning the pot when they miss. Some might call this tight-weak, but at low limits I think it's sometimes the way to play. Ramming and jamming is not always the best policy.
Conclusion
Well, that's all for now. I could go on and on, making the examples progressively more complex by looking at calls/raises on the flop, but this is just supposed to be an introductory primer to pot odds. Feel free to ask questions in the comments.
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