When we counts outs, we are making the assumption that the outs are not already dealt. This can be correct in the heads up, but I guess that in the full game this will not be correct (due to the biggest probability of one or more outs are already dealt). Does exists a corrected formula for counting outs in the full game and short game?

  • 1
    I was thinking of exactly the same thing today ! Small world :) ... Feb 14, 2013 at 21:43
  • 1
    All cards that you do not know are unknowns. Feb 21, 2013 at 21:44

5 Answers 5


You can't adjust the "outs" because you don't have enough information (that's what poker is, a game of incomplete information) but that's perfectly ok because the outs themselves do mean absolutely nothing.

If I tell you, you have 16 outs to win... are you ahead or are you most likely dead? You don't know, 16 outs means you're ahead in the flop but you're not in the river and you're definitevely drawing dead in the national lotery... with the same outs. We're so familiar with counting outs that we forget about the implicit information: the game and the street.

What's really important though it's the odds of winning and that DOES take into account the already dealt cards.

The theory, summarazing, is that the probability of an event doesn't get affected by unknown factors (let me clarify this, it doesn't get affected as long as you include those unknown factors into the total number of possible events). That way, if you have a flush draw in the flop you have 9 outs (to simplify we'll assume it's a nut flash draw as if you're in a 5 way flop and you're counting you're 2-5 as a winner you need to revise your strategy), there're 9 cards that will help you but to calculate the odds of you winning you multiply that by four and you get the magic number of 36%. But how did you came up with that number??

What you really did was (by means of a simple rule that gives approximate results) was the probability of getting one of those cards by taking 2 random cards out of a deck of 47 cards, 47!

That 47 magic number is the key, in a 6-max table the actual number of cards left in the deck is 37, as ten other cards have been dealt to the other players but to calculate the odds you're including that as if those cards were still in the deck and the effect is exactly the same. You could calculate it as:

  • The odds of taking two random cards over a 47 deck with 9 remaining hearts and one being a heart or
  • The odds of taking two random cards over a 37 deck with x remaining hearts and one being a heart minus the odds of dealing x hearts when picking 10 random cards when taking x cards from a deck of 47.

The second formula is equivalent to the first (produces the same results) but it's harder to calculate so we use the first.

The main point is those dealt hand are taken into account by the fact that when you calculate the odds you count the deck as having all the cards except the ones in the flop and in your hand, that's equivalent as including the dealt cards to the other players, those cards are unknown and so it's as if they were still on the deck.

Of course one thing that DO change the odds is the way the hand has been played.


I think you should look at it slightly different. instead of counting the number of cards you need to win a hand, you should look at the chances you'll hit one of them.

Let's say that after the flop and turn there are 10 cards that can help you.

  1. In heads-up, your chances to hit one of the 10 are around 20%. (I assume your opponent might have one of your cards, two of your cards or none of them).
  2. In a game with 9 players (including you), there are 30 cards in the deck ( 52 - (2*9) - 4). now, the options are your opponents might hold all your cards, 9 of your cards, 8 of them and so on until non of them is help by the other players. your chances to draw one of your cards in the worst case is 0% (all your cards are dealt to the other players). your chances to draw one of your cards in the best case is 33.33% (all your cards are still in the deck).

if you'll make an average on your chances to get one of your cards with 9 players, you will see that it is the same as getting one of your cards playing heads-up.



Unless you know something about your opponents range, it doesn't matter. Consider this simplified version of the problem:

Shuffle a deck of cards. What are the chances of an Ace being on top? Because there are 4 aces and 52 cards in the deck, this is clearly 4/52. So lets say we remove two cards from the top of the deck and set them aside. What is the chance of the next card being an Ace? Note we are now choosing one of 50 cards, because two have been set aside. However, the odds are still 4/52. One way to visualize this is to imagine the two cards on top being removed, and then looking at the third card. As long as you know nothing about the first two cards, the chance of the third one being an Ace remains 4/52.

This same situation can be extended to a poker game, where 18 cards are removed from the deck and placed in 9 player's hands.


Well, I guess you could find a way if you have a flush draw. It's the easiest.

Let's say you have A:diamonds: J:diamonds: and the board is 7:diamonds: 5:spades: Q:diamonds: 2:hearts: . You need to hit one of the 9 remaining diamonds to make a flush.

If you're playing 6max, then it means there were 10 cards dealt to the other 5 players + 2 burn cards. This is a total of 12 cards.

Now, on average there is always an even distribution of suits, no matter how many players there are. See here to learn why.

What this means is that, on average, out of those 12 cards, there will be 12 / 4 suits = 3 diamonds.

This implies that your 9 outs just turned into 9 - 3 = 6 outs. This is a significant difference (- 33 %) and, if you take it into consideration, then it can drastically affect the size of the hypothetical bet you're willing to call with your hand.

I sometimes take this into consideration when I deal with flush draws. You can do it too from time to time, but do it with caution. It's relatively safe, not unlike the scenario below:

Let's move away from flush draws and go to other draws.

Let's say you have 9:diamonds: T:diamonds: . This time, the board is A:spades: 7:hearts: J:clubs: Q:spades: . Now, this time you have an open-ended straight draw but you feel that your super tight and suddenly very aggresive opponent is way ahead (maybe Aces up or KT for a broadway or Pocket Jacks etc.). What you would need to win this is:

  • either bluff it by putting him all-in and throwing him off his hand or
  • hit a King or an Eight to make a straight.

Since we're talking about outs, we're going to focus on the second option.

Now, typically you have 8 outs: 4 Kings and 4 Eights. But if you want to make this correction, you have to take into consideration the chances that:

  • your opponent has a King or an Eight OR
  • one of the other opponents folded a King or an Eight.

Now let's apply the same logic as with the flush draw: Assume you're playing 6 handed, just like before. This means that there were 10 cards dealt to the other players + 2 burn cards = 12 cards.

If we consider the same even distribution, then 12 cards / (52 cards / 8 outs) = 12 / 6.5 = 1.84 outs. What this means is: in those 12 cards, on average, there will be 1.84 of your outs. This implies that your outs went from 8 to 8 - 1.84 = 6.16. This can again be significant difference.

I strongly recommend that you never ever rely on the second correction (with the straight draw). It's just waaay too unreliable. It's extremely hard, next to impossible that this is even remotely close to the real value. Why ? Because that 6.16 is an average value with HUGE fluctuations from hand to hand.You'll only be guessing and most of the time you would be wrong.

If you want to apply such outs corrections to your game, do it only to the flush draws, their values are usually very close to average. Like a Bell curve, if you're familiar with the concept.

  • In your example with the outs dropping from 9 to 6: When counting with the result 9, you say you need 1 of 9 cards out of 47. In your method where you get the result 6, you need 1 of 6 cards out of 35 (52 - 12 pocket cards - 3 community cards - 2 burn cards). So together with the number of outs, you have also reduced the size of the card pool where the cards are drawn, resulting in a similar probability. The only reason to do such a detailed computation is when you think there is a non-average probability of some of your outs being in some pocket cards (see the answer of Cory Kendall).
    – azimut
    Apr 5, 2013 at 22:38
  • 1
    I rarely downvote but I had to here because I believe the conclusion of your answer is wrong- and misleading. You wrote: "if you take it into consideration, then it can drastically affect the size of the hypothetical bet you're willing to call with your hand." That is not true. If you calculate the remaining outs the way you described, the probability of hitting the flush or straight does not change at all. One thing I do agree with in your answer is maybe the possible assumption that your opponent has a King because of the way he is playing his hand.
    – TTT
    Apr 9, 2013 at 14:15
  • @TTT I won't be mad, I promise :) . Apr 9, 2013 at 14:17

There is now way of calculating that as far as I am aware.

Unless you openly know your opponents cards.

The idea of counting the outs is just to give you a rough idea of what percentage of the time you are likely to get the cards you need. The more outs you have the better your chances of hitting.

I am not sure its really possible to calculate your chances of hitting your outs if you try and work out your opponents chances of having one of your cards.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.