How many outs do we REALLY have?

Does anyone know of a website, a poker research company or software that can answer the following question:

When on a draw (minimum 6 outs), how often are we missing at least one of our outs due to someone holding it, folding it or the dealer burning it? What about two or more outs?

Thanks!

Oh wow we're getting into Quantum Physics math here. Probability and Odds equations... I've never seen a formula for that.

I would assume its irrelevant given that you have no idea what people held and what has been burned.

Outs are based on what you know, not what you don't (and can not) know. Imagine that you have 4d 4c and flop comes Ad 2s 3d.

Your outs would be either of 2 remaining fours (for a set), and 4 of the 5s (for a str8) and 10 remaining diamnonds (for a flush)... i think.

Just because you don't know whether the out cards are available (burned or in another hand), you still have "possible" 16 outs.

I may be really showing my ignorance here, if so I'm sure someone will straighten me out.

Something just occured to me, someone maybe could figure out way to tweak the outs a little. In the example i assumed that all the diamonds were still in the deck (best case), but there are probablities that could be calculated for and against a certain number of the diamonds still being in the deck... hmmmm.... maybe not such a crazy idea Fixit... dagnabbit, u got me thinking :0

You're starting to catch on. Of course I know the conventional method is to assume you have all your outs when you're on a draw, but I'm publishing two poker books next year and I'm looking into the risks of playing a draw. One such risk is that you don't actually have a lot of your outs, which reduces the value of playing a draw. I really just wanted a good research website.

Hmm sounds really complex I don't think having information like this would be worth having. This is where the chaos theory would come into play which you cannot ever account for.

You may have trouble finding anyone doing this type of analysis. I bet (no pun intended) that a high school or college statistics/probabilty course teacher would be willing to use the question as a challenge for their students.

I don't think it will take strange atractors and fractal analysis to figure this out. Probability should be able to.

All you have to adjust to your "best case" outs for the actual probability that the cards you need are still available i.e. the 10 diamonds that can fill your flush are certainly NOT all in the deck... well not certainly, but THAT CHANCE can be calculated.

I'm no mathmetician but it sure seems like a simple calculation. It may turn out to be completely useless, if the analysis figures out that your number of "best case" outs have to be adjusted downward by .2% it's not much use, but if your outs adjusted by 30% it certainly may help in making decisions (maybe not really ten outs, but only 7).

You can't adjust in the manner which you describe. We'll use a nut draw to a flush for an example. Only the river to come, you have 4 diamonds and if one more comes you'll win.

As most people know, there are 7 known cards, 45 unknown. Of the 45 unknown 9 are diamonds, 36 are not. Your odds are 36:9 or 4:1 to hit the flush. So you'll make it 20% of the time.

Now you want to try and adjust. You know it's likely that someone had at least one diamond. So really you have 44 unknown because there's one less diamond and therefore 36:8 or 4.5:1 to hit. But if you begin down this path, then it's even more likely that cards that are not diamonds have been ditched. After all, there are three times as many other suits as diamonds.

It's going to get way too hard way too fast. Mathematics in poker isn't about formulas and getting answers to several decimal places. You have incomplete information most of the time anyway, and all that is needed is a rough approximation.

Math in poker is definitely a case of KISS

I agree with much of this. To accurately assess the situation, there are just too many variables. You must take into account the hands being dealt. Then which hands each player will play. Then what % of the time their playing of a hand contains one of your outs. Further, player's ranges will change based on players which have entered the pot ahead of them, stack sizes, metagame factors, and I'm sure lots of things I'm not going to think of right now.

I don't have any desire to work out the math, but I will tell you that the given current equation is much like EV. You're not going to have 4:1 every time. Sometimes all the diamonds are gone and you're drawing to 0 outs. Other times every single one is in there and you have really good odds of hitting it. But on average, over many trials, regardless of what cards have been folded or not, you will hit that flush 20% of the time.

i have no idea, but maybe you can calculate the amount of "dead cards" either the burn cards or cards other players have and work out how many times that a diamond is going to be in their hand, on average

so there is 4 players including yourself so if on the flop you have 9 "outs" and there is 7 cards that are dead, including the burn card, perhaps work out how many times a diamond is likely to be in one of those 7 cards

but again, i really dont have a clue

Quick and simple:
Having a decreased chance of hitting your outs because they are burned or being held by someone else is exactly offset by your *increased* chance of hitting those outs when the burn cards or your opponents do not have them.

In other words, it equals out.

Since we do not know the burn or opponents cards, we calculate out outs on the known factors.

Don't worry about things you cannot know and calculate your outs vs the rest of the deck.

Im not sure I totally agree with this. We figure outs for the maximum number of outs. and yes sometimes others are holding some or some are burned by the dealer. So those hands have less actual outs. Its never more than the maximum outs. So how would it even out.


So actual outs on a 10 outer on average is probably only an 8 outer.

I would be interested in seeing a formula and some tests with this theory

0, so hurry up and fold already.

In terms of odds, which to me would be the simplified way to approach it would be to calculate the following (odds of winning hand on a particular street)*(1-the odds your opponents/dealer have the cards you need).

Statistically speaking, I think you always have to assume the cards are available to be drawn. Yes, that will always be a "best case" scenario- but random theory doesn't care whether or not the cards have been burned, dealt or at the bottom the deck.

^^this. Some of you are missing the point, boys and girls. It doesn`t matter where the card you are looking for actually is on any particular deal. All that matters is that it is potentially available because it is not among the cards visible to you.

No, your outs are *always* against the pool of unknown cards.
You're outs are always against the 50 unknown cards pre-flop.
You're outs are always against the 47 unknown cards on the flop.
You're outs are always against the 46 unknown cards on the turn.

Times that other people are holding your outs you have a decreased likelihood of hitting, and times when they do not hold your outs you have in increased likelihood of hitting. These even out because it's always the same pool of 50 cards that are randomly distributed.

The odds of other people holding cards you need does not change. It is a static number. Your opponent *always* has 2 random cards of 50 unknown cards. VS eight opponents, they *always* have 16 random cards of the unknown 50 cards.

Over the long run people will be holding your outs as often as they should, and they won't be holding your outs as often as they should. You cannot see their cards so you cannot readjust those numbers. You can only calculate known information, and that information is the pool of cards you cannot see which are randomly distributed.

Well said. Here's what I'm doing. I'm running my own test. I'm dealing out 9 hands open, burning a card open and dealing a flop. I'm on hand #140 and the results are staggering. Small sample size so far but well worth the time.

"Times that other people are holding your outs you have a decreased likelihood of hitting, and times when they do not hold your outs you have in increased likelihood of hitting".

Isn't this saying the same thing that he disagreed with in the first place? With 9 flush card outs and one card to come you have about a 20% chance to hit. However, built into that formula is the assumption that you ALWAYS have 9 outs...in other words, only the times they do not hold your outs, so there is no evening out. If tests show that on average you miss 2 outs on a flush draw, then the conventional theory of hitting cards and pot odds is in question, or just needs to be tweaked.

Fixit, I'm not sure what you're hoping to accomplish, but I applaud your dilligence. What are you hoping to accomplish? I think what you're observing is not so much a relevation in statistical relevance, but more about how odds are altered when you know that circumstances have changed, ala the monte hall gambit. See the example below.

http://www.stat.sc.edu/~west/javahtm...MakeaDeal.html

if figure the likely hood someone has your outs then you will know the likely hood they are in the deck. say someone did have your outs 15% of the time that means 85% of the time they don't. so whatever % is taken away that 15% ot the time will be added 85% of the time. so it all equals out in the end. it's just like when the run it twice it's the exact same odds whether run one or twice.

It's 20% because there are 9 outs in all unknown cards in the deck (45 after the turn). In reality if all 9 outs are still in the deck after dealing the turn card, then in a 9 handed game there are 24 cards gone. 18 dealt to players, 3 on the flop, 1 on the turn, and 2 cards were burned on the flop and turn. This leaves 28 cards in the deck, 9 good 19 bad. Therefore the likelyhood of hitting your draw is 19:9 or 2.1:1. Now you're hitting your flush 32.25% of the time.

I want to better understand the risk associated with playing drawing hands, and I'm publishing two poker books next year in which I want to address this. If we find out in a test that in those 'unseen' cards, let's say 50% of the time there is an average of 1 or more outs missing in a draw then we would need to consider altering the conventional odds of hitting. At the very least, players should consider the risks of not having all their outs available when playing drawing hands.

This is just incorrect.

+1 Nailed It

Now it sounds like you're saying that the 20% to hit a 9-out draw with one card to come already takes into account the fact that outs might be missing. Is that what you're saying? Is the 20% merely an average?

Wait, shouldn't you have done 28:9 for remaining cards in the deck to 9 good cards, which is 3.1 to 1 or less than 25%?

It's irrelevant whether or not the outs have been dealt or not. It's only relevant that you don't have them and you don't know who has them- be it with players, the burn pile or the active deck.

No, there would be 28 cards remaining. 9 of them are good, 19 of them are not. In straight odds it is 19:9 against.

And yes, that's correct. 20% can be described as the result of taking missing outs and such into account.

i used the 15% and 85% as an eg. but no it's not incorrect. those times you take outs away will be compensated for those times you take non-outs away.

You're looking at it wrong.

What matters is "outs in relationship to the cards that are left." If you only had a two-outer, it's no big deal if there are only three cards left in the deck.

If one of the other eight people at the table has a heart and you're going for a heart flush, that is good! Yes, you are down one out, but that means that there are fifteen fewer bad cards in the pile to draw from.
That is, on a 2 heart flop instead of having 9 outs you might have only 8 outs, but 9 outs in a pool of 47 remaining cards (15%) is worse than 8 outs in a pool of 31 (26%).

I'd say you have about a 100% chance of your two books being horrible based on this thread, but who knows maybe you still have outs.

Unkind, but sadly true. It`s usually considered wise to develop a grasp of the basic principles of one`s topic before writing about it.

oh my god this thread is hilarious. publish your book asap sir

No, your missing the context of those outs.
Tthe calculation is based on the relationship between the outs you need and the number of cards you have to draw them from.

A two outer is bad when there are 47 cards to draw from.
A two outer is excellent if there are only three cards to draw from.

In fact, with that formula you are already *assuming* that your opponents have some of your outs.

If you flop a flush draw, you have 9 outs. Against 8 other people, you're assuming that their 16 cards hole cards have as many hearts and they would on average. If they have more, your chances of winning drop. If they have fewer, you chances of winning increase. But you do not know, so you just calculate based on known factors.

Sometimes you will have a worse chance than you calculate, sometimes you will have move, but in the long run you will be pretty close.

adsgjhsdfkgjsdfk wv stole the 'insert joke about published books not working out' post before i could get here