Odds question...
 

Hi guys, I am trying to explain to my friend pot-odds, and I can't give him a good explanation of why when you calculate pot odds, you don't include the chances that someone else at the table either has one of your outs or it was already folded. Can someone give me a quick and easy explanation for why you do not include cards that were dealt to other people to adjust your outs, and please dont say because there is an equally likely chance that they do have the card as if they don't. Thanks.

-Obi-

Re: Odds question...
 

OK...let's do a flush draw. You have T [img]/images/graemlins/diamond.gif[/img]9 [img]/images/graemlins/diamond.gif[/img], board is A [img]/images/graemlins/diamond.gif[/img]Q [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/diamond.gif[/img]2 [img]/images/graemlins/heart.gif[/img].

There are 9 diamonds left in the deck and 46 unknown cards. So you have a 9/46 chance of hitting.

Your friend is confused about counting outs because your opponent might have one of your diamonds?

OK...he MIGHT have one of your diamonds. He MIGHT have two of your diamonds. He MIGHT have none of your diamonds. You have no idea what he has, so you just have to count the cards your opponent has with what is left in the deck.

Re: Odds question...
 

He understands pot odds Kyro. Your answer didn't really explain why you don't take into account the probability that other people at the table could have been dealt one of your outs.... He really wants to know why there is not adjustment for the possibility that since there are so many cards that are out of the deck in the muck pile or in someones hand, why you just ignore that...

Re: Odds question...
 

Yes I did.

[ QUOTE ]
OK...he MIGHT have one of your diamonds. He MIGHT have two of your diamonds. He MIGHT have none of your diamonds. You have no idea what he has, so you just have to count the cards your opponent has with what is left in the deck.


[/ QUOTE ]

Let me try it another way.

Let's say you have 9 outs with 46 cards to come. It's a 10-handed game. So on average, the other 9 people at your table will have 3.5 diamonds in their hand. So there are on average 5.5 diamonds left, but now you only have 28 cards left to come. 5.5/28 = 9/46. Is that easier?

Re: Odds question...
 

it doesn't really make a difference whether the cards are in someone else's hand or not. what is the difference between some of the outs being in someone else's hand and some of the outs being at the bottom of the deck? either way, you are not going to hit those outs, and you should think of them in the same way. you'd only take into account what other people are holding if you actually KNOW what they have for a fact.

in other words, saying "maybe somebody else is holding some of my diamonds" is no different than saying "maybe some of my diamonds are at the bottom of the deck." so you just count them both.

Re: Odds question...
 

28x2=56, not 46
haha

Re: Odds question...
 

you are good at math. so obviously you'll be able to agree with me when i say that 28+18=46.

Re: Odds question...
 

He explained it as best as it could be explained I think. Ask your friend this -- how should the probability be adjusted when *nothing is known* about the cards that are in other player's hands or in the muck? Sure, they are dead and are not in the deck, but you know nothing of what those cards are. If you don't know what they are, you have no idea how the remaining content of the deck has changed. Therefore, you can't make any adjustments to the probability because you have no new information.

Re: Odds question...
 

I would also agree with you that 1+1=2, what is your point?

Re: Odds question...
 

Well this would explain why you can't help your friend out.

Go read my second post again. Especially the last part of it. If you can't figure out how that applies to the situation, then I give up and I hope that someone else will have better luck with you.

Re: Odds question...
 

My friend replied to your post mickianthony:

you cant say that other people having the cards is the same as them being at the bottom because the probabilities of those are very different...

Help?

Re: Odds question...
 

[ QUOTE ]
My friend replied to your post mickianthony:

you cant say that other people having the cards is the same as them being at the bottom because the probabilities of those are very different...

Help?

[/ QUOTE ]

I would love to hear the logic behind this one.

Re: Odds question...
 

Friend: but the closer they are to the top, the more likely they get played

Re: Odds question...
 

[ QUOTE ]
Friend: but the closer they are to the top, the more likely they get played

[/ QUOTE ]

Your friend doesn't have what it takes to win at poker. I'm honestly not saying that to be a dick, but it's true. That statement is so completely irrelevant to the problem.

Re: Odds question...
 

Your not saying that to be a dick? All I'm trying to do is explain something to somebody that is confused about a question and you throw that out. That is so completely pompous and rude that you shoudn't be allowed to post on this conference. Your obviously too insecure and STUPID to be answer this question in a reasonable way, so don't attempt to make his question sound stupid when it is you in actuality. By the way, if you ever want to lose money, take me on heads up, any stakes any time tough guy. I'll take your whole bankroll bitch.

Re: Odds question...
 

I thought so...

Re: Odds question...
 

[ QUOTE ]
Your obviously too insecure and STUPID to be answer this question in a reasonable way, so don't attempt to make his question sound stupid when it is you in actuality.

[/ QUOTE ]

I answered your question twice in the thread. What more do you want from me?

[ QUOTE ]

By the way, if you ever want to lose money, take me on heads up, any stakes any time tough guy. I'll take your whole bankroll bitch.

[/ QUOTE ]

If you were as good as you thought you were, you wouldn't be coming on here to ask for help explaining a sickeningly easy concept.

Re: Odds question...
 

Be nice, he's just trying to understand the problem.

But yeah, it is pretty irrelevant... lol.
I'm actually trying to understand why he's thinking this and why it applies to the problem. I actually don't think nickianthony's way of visualising this is the BEST way to look at it, which may be the source of his thinking here. There is obviously a difference if the "outs" are in someone else's hand versus in the deck -- they can be dealt out if they're in the deck. The "depth" those cards are at is unknown though so you can't think of it in those terms

It sounds like your friend is thinking of this as a conditional probability problem -- he's trying to change the probability because certain cards are "dead". The problem is that you know nothing about the condition, so the probability doesn't change.

Please look at my other post, I'm kind of curious what he sees that is unclear about that logic, or why it's wrong. Maybe I can think of a better explanation or analogy.

Re: Odds question...
 

Nice way to avoid my challenge. You don't belong here if you can't accept a challenge from somebody that you don't think is as good as you... ESPECIALLY if you say something as irrational and pointless as calling somebody learning the game hopeless... DICK.

Re: Odds question...
 

[ QUOTE ]
Nice way to avoid my challenge. You don't belong here if you can't accept a challenge from somebody that you don't think is as good as you... ESPECIALLY if you say something as irrational and pointless as calling somebody learning the game hopeless... DICK.

[/ QUOTE ] Kyro answered your question in a polite way. He also explained it as well as it can be explained. You were rude to him in about 5 different posts. And I agree with Kyro--if you and your friend can't understand this simple concept, then you shouldn't play poker.

Re: Odds question...
 

How about this...

You calculate pot odds based on the information available to you at the time you're making the decision. Since you usually only know the two cards in your hand and the three on the flop, that is all you can go on. The other 47 (be they still in the deck, or in your opponents' hands are all unknown and have to be treated the same. As unknown.

If you believe your opponent(s) has/have some of your outs then take them out of your calculation. If you have 8 outs for an open-ended straight draw, but there are two clubs on the board and you believe your opponent is holding two clubs, then remove those two cards that makes your straight and give him the flush. Therefore you have about a 12% chance of hitting your straight on each of the turn and the river.

Tell him to remember that "outs" are cards that improve your hand. They are not necessarily cards that give you the winner.

About his outs being folded...if he didn't see it being folded, he can't possibly know it's gone. Therefore it's just like the cards in the deck and cards being held by your opponents...unknown.

Re: Odds question...
 

The odds scale.

When you say, "Well I have 2 opponents, 1 of them has a diamond because they have 4 cards" that means you must also say, "Well I know 3 cards AREN'T diamonds" and your flush draw percent doesn't change one bit*

*However since you're good at math you also know that when you have 2 diamonds and there's two on the flop then a diamond won't be distributed in other players hands exactly 1 out of every 4 cards.