Quick question gentlemen.

9 handed game, You hold AhQh with a 8h5hQd flop. From what i understand, the odds of you getting you flush on the next card are slightly better than 2to 1 (9 hearts in the deck that make your flush and 22 cards that dont.) Every book i have ever read seems to "assume" that all of the remaining hearts are still in the deck.

Obviously, this is not possible, but how do you adjust your calculations so that you are closer to right?

If your 9 handed do you assume on average that there will be x amount left?

5 handed?

HU?

Maybe I am looking at this all wrong, but it seems to me that you would take an average of the possible number of cards and not just "assume" that all the cards you need still lie in the deck.

Thanks in advance.

MillConsole542

First, this gets asked kind of often...

What? 9 hearts make your hand, yes, but that leaves 38 cards that don't make your hand. There are 52 cards in the deck. You have 2 and the flop has 3 so there are 47 remaining. You'll make the flush 9/47 times = 19%

Not really. Let's start with a thought experiement. The turn and river cards are going to come from the top of the deck right? It doesn't matter whether a card is in your opponent's hand, or at the bottom of the deck, if it's not at the top of the deck, it ain't getting dealt. So, we can treat all t he unknown cards as being in the deck, with your opponent's cards being at the bottom. Either way it makes no difference, they can't be dealt. So then the question just becomes, out of this meta-deck, how often is there a heart at the top?

I'm not typing it again for the whatevereth time ..... The following is a post of mine from this thread: http://forumserver.twoplustwo.com/15...d-outs-428674/




Why does this come up every couple weeks?


Yes some flush outs are probably gone ... and non-flush out cards are also gone.



You hold a flush draw on the flop in a 6 player game so there's 10 cards the other players hold that you don't know what they are:


First let's do it without considering other players' cards:

You know your cards and the flop .... so that leaves 47 cards and 9 that make you a flush. So 9/47.




Now let's think about the other player's cards:

Now in the 47 remaining cards there are 9 of your suit. That means about 19% of the remaining cards are of your suit. So now let's look at the 10 cards the other players are holding. 19% of 10 is about 1.9 so the other players hold on average 1.9 of the cards that are the suit you need for the flush and 8.1 of them will be of other suits ..... so now we have 7.1/37





9/47 = 7.1/37. It's the same.

Thank you gentlemen for your insights and have a good day.


MillConsole542

I think you are having trouble with realizing that since poker is a game of incomplete info, you have to assume that all of your outs are in the deck - there's no way to know that some of your outs are in your opp's hand. For that reason alone you don't exclude your opp's cards out of calculation. You count all cards that are known to you: your hand and the board (plus any accidentally/otherwise exposed cards).

Others have answered but I'll give my brief version too. Nothing at all is assumed, and the calculation is actually *exactly* right.

If you know they are all still available, then your probability to hit on the next card is better than 9/47 (19.1%). With 8 opponents whom you know don't hold your cards, your chances go way up to 9/31 (29%). If you knew that 3 of your outs were held by other players, your chance goes down to 6/41 (14.6%). These are all conditional probabilities. But the weighted sum of all the possibilities is 9/47, and it is the unconditional probability.

The actual frequency that you will actually hit the flush over time, is exactly 9/47. There is no assumption about available cards at all.
Absolutely not. See above.

Yes, completely wrong wording. I mentally just put all the cards in one deck/pile, therefore disregarding all of the opponents. But after actually reading lego's mathematical explanation, it makes complete sense.

The responses above are correct - but I do want to add a little something that isn't really about the original question, but is related. It doesn't always come into play in holdem, but in Omaha it is much more common. And that is that if you have 9 outs to hit your flush, not all of them make you a winner. Assuming you have the nut flush draw (Ah9h) on a Kh7h5c2c board, and someone is betting aggressively - it is very possible that the 2h and 5h are not really outs for you, since if someone has a set, those cards will make your flush and you will still be crushed. Similarly, on a paired board, it is always very dangerous to assume that your flush outs are really winners.

WOOOOOW I JUST.....WOW GREAT

bump to prevent archiving

I came here from the FAQ. Of cause OP's idea was just plain wrong, but it also not true that you make no assumptions at all. You make the assumption that your opponents cards are evenly distributed. That is actually a very strong assumption in most cases. Let's say you put your opponent on a worse flush draw. Now you do indeed have less outs, because there is a higher likelihood these outs are in your opponents hand.