Hi all,

Something I have been curious about for some time and interested in opinions.

Should we adjust our outs to include mucked cards? What I mean by this is say we have a flush draw, according to accepted theory we assume we have 9 outs. But in reality, how likely is it that all of those 9 outs are still in the deck? Between all the hands mucked pre flop, our opponents holding and at least one burn card, it seems hugely unlikely that at least one of our outs won't now be dead.

I know that if we are basing things purely off math then we must assume that all 9 outs are live, but it seems like wishful thinking.

If we were to remove one of our outs from our calculations (bringing things in line with a more realistic outlook), do you think we would be making more accurate and profitable decisions in the long run? This would mean we would have less chance of making our flush, so in turn be folding more (when basing decisions purely off pot odds), and in turn making more profitable calls.

Let me know what you think. Thanks

Unless you know which cards are more likely to have been mucked this is flawed logic

Here is a handy thread for your reference.

Are your non outs not getting mucked?

Here's a little experiment for you to try. Take a deck of cards and deal out poker hands to some number of places (doesn't matter how many). Deal yourself 2 spades, and put out 2 spades and 1 diamond and 1 heart on the board. You now have one card to come and you need a spade for a flush. Take one card off the top of the deck and place it face down. Pick up all the cards you dealt, put them with the rest of the deck, then put all of those cards in your pocket.

What is the probability that the one card sitting there is a spade?

You know that you have put a bunch of spades in your pocket. And also a bunch of hearts, diamonds and clubs. So what possible method could you use to estimate the probability that this is a spade?

You know 6 of the cards - the 2 you have and the 4 on the board. Of the remaining 46 how many are spades? You know there is only one card remaining - it can't be 9 spades. But it also can't be a bunch of diamonds or hearts of clubs. So essentially that one card is, mathematically, a fraction spades, a fraction clubs, a fraction diamonds and a fraction hearts. Specifically, it is 9/46 spades, 12/46 diamonds, 12/46 hearts and 13/46 clubs. If you do this exercise hundreds of times, about 9 out of every 46 that one remaining card will be a spade.

This is the actual situation that you have in a poker hand. All cards in the deck that are NOT going to be the river have basically been thrown out. The cards that have been mucked, have been thrown out. The cards in other people's hands have been thrown out (although depending on the action, it is possible that one or more might also be on a flush draw and you need to consider that). That one river card, or the single turn card and single river card, will be the only card(s) that actually matter. What you want to know is the probability that the card helps you, and you can only go by the information available to you at the time.

Yes this really is a nice way to think about it. Basically we can consider the entire remaining deck to be also 'mucked' cards and therefore nothing has changed probability wise. Thanks for this and the other replies here too.

Here is a simple example that expands on Anero’s response.

Assume 2 players and the deck is Aces, Kings and Queens.

The flop is one card and a 3-flush wins: You are dealt As Ks. You need the Qs to win.
There are 12-4 = 8 cards remaining in the deck

Ignoring opponent holding:

Pr(win) = Pr(flop is Qs) =1/8

Considering opponent holding:

Pr(win) = Pr(Vill has Qs)*Pr(win |V has Qs) + Pr(V doesn’t have Qs)*Pr(win|V no Qs)

=(1/8)* 0 + (7/8) * (1/7) = 1/8

If cards are mucked randomly from the deck, it does not affect your outs. If 1/4 of cards in the deck are your outs, then 3/4 of the time the deck mucks a non-out, and 1/4 of the time it mucks an out, and everything evens out at the end.

What if cards are not mucked randomly though? What if people tend to fold low cards more often than high cards, skewing the distribution?

Shameless plug: https://www.holdemresources.net/blog...ching-effects/

Lets work it out.

The traditional view is that you you have 9 outs in 47 cards (3 flop 2 hole cards) or a 9/47 chance of hitting the flush on the turn.

First, lets see if the burn card makes a difference.
There is a 9/47 chance the burn card is a flush card you need which leaves you with 8/46 chance of flush, or 9/47 * 8/46 chance.
There is a 38/47 chance it isn't a flush card which means you would have a 9/46 chance of a flush or 38/47 * 9/46.

If we sum up the probability we get:
9/47*8/46 + 38/47 * 9/46 = 72/2162 + 342/2162 = 414/2162 = 9/47

So we know the burn card makes no difference as the probabilities work out the same.

We can repeat the same calculation for the first card dealt to your opponent, then the seconds card, then the first card dealt to first opponent that folded etc. In each case we end up with the same calculation.

There are things which can make a difference but it is to small to worry about. For example if you are counting Aces as an out (for an over pair for example) then you can be pretty certain that nobody folded AA preflop which increases the chance of an ACE by a small fraction of a percent.

Thanks for all the responses guys, it's amazing how maths just demolishes what you think is common sense or "realistic"
Cheers

Suit removal is a thing in multiway pots.

Care to elaborate?

lets pretend you always think someone has an ace when they raise, so you always fold your hand if an ace hits the flop. Sure, you might lose less but you'll also win less too.

It affects, one reason collusion exists and is bad for the game. Should it matter in a normal honest game? Depends on the spot and how narrow the ranges are, not sure if can be accurately calculated esp in a 30 sec in game decision spot.

I'm not the one you asked, but I can elaborate.

Example: there is a double suited flop, and you have the ace high flush draw. You count 9 outs to make a flush, and a few less to end up with the nuts.

But now someone (you or someone else) makes a bet, and 6 people call. This is pretty unusual, and would indicate it is very likely that someone else in the hand also has a flush draw. Even though you will still win if the flush comes, you may want to consider that it is likely you have only 7 outs to make your hand instead of 9. Of course you can't know for sure, but the more players who see the turn, the more likely it is that some of your outs are dead. It also becomes more likely that someone has a set or two pair, in which case some of your flush outs which pair the board will result in you losing the hand, and possibly a lot of money, which is also a reason to be more cautious.

True but the value of hitting the nut flush goes up enormously and decreased equity is made up for in implied odds if you overflush someone or make it against someone who can't fold a set.

Sometimes It does; I play mostly LHE myself so that doesn't really apply as much.

Hmm very interesting. Thanks again

Here and there the ace is pretty significant, or when you don't have it.